Integers are great for counting whole numbers, but sometimes we need to store *very* large numbers, or numbers with a fractional component. A **floating point** type variable is a variable that can hold a real number, such as 4320.0, -3.33, or 0.01226. The *floating* part of the name *floating point* refers to the fact that the decimal point can “float”; that is, it can support a variable number of digits before and after the decimal point.

There are three different floating point data types: **float**, **double**, and **long double**. As with integers, C++ does not define the size of these types. On modern architectures, floating point representation almost always follows IEEE 754 binary format. In this format, a float is 4 bytes, a double is 8, and a long double can be equivalent to a double (8 bytes), 80-bits (often padded to 12 bytes), or 16 bytes.

Floating point data types are always signed (can hold positive and negative values).

Category | Type | Minimum Size | Typical Size |
---|---|---|---|

floating point | float | 4 bytes | 4 bytes |

double | 8 bytes | 8 bytes | |

long double | 8 bytes | 8, 12, or 16 bytes |

Here are some definitions of floating point numbers:

1 2 3 |
float fValue; double dValue; long double dValue2; |

When we assign literal numbers to floating point numbers, it is convention to use at least one decimal place. This helps distinguish floating point values from integer values.

1 2 3 |
int n(5); // 5 means integer double d(5.0); // 5.0 means floating point (double by default) float f(5.0f); // 5.0 means floating point, f suffix means float |

Note that by default, floating point literals default to type double. An f suffix is used to denote a literal of type float.

**Scientific notation**

How floating point variables store information is beyond the scope of this tutorial, but it is very similar to how numbers are written in scientific notation. **Scientific notation** is a useful shorthand for writing lengthy numbers in a concise manner. And although scientific notation may seem foreign at first, understanding scientific notation will help you understand how floating point numbers work, and more importantly, what their limitations are.

Numbers in scientific notation take the following form: *significand* x 10^{exponent}. For example, in the scientific notation `1.2 x 10`

, ^{4}`1.2`

is the significand and `4`

is the exponent. This number evaluates to 12,000.

By convention, numbers in scientific notation are written with one digit before the decimal, and the rest of the digits afterward.

Consider the mass of the Earth. In decimal notation, we’d write this as `5973600000000000000000000 kg`

. That’s a really large number (too big to fit even in an 8 byte integer). It’s also hard to read (is that 19 or 20 zeros?). In scientific notation, this would be written as `5.9736 x 10`

, which is much easier to read. Scientific notation has the added benefit of making it easier to compare the magnitude of two really large or really small numbers simply by comparing the exponent.^{24} kg

Because it can be hard to type or display exponents in C++, we use the letter ‘e’ or ‘E’ to represent the “times 10 to the power of” part of the equation. For example, `1.2 x 10`

would be written as ^{4}`1.2e4`

, and `5.9736 x 10`

would be written as ^{24}`5.9736e24`

.

For numbers smaller than 1, the exponent can be negative. The number `5e-2`

is equivalent to `5 * 10`

, which is ^{-2}`5 / 10`

, or ^{2}`0.05`

. The mass of an electron is `9.1093822e-31 kg`

.

In fact, we can use scientific notation to assign values to floating point variables.

1 2 3 4 5 |
double d1(5000.0); double d2(5e3); // another way to assign 5000 double d3(0.05); double d4(5e-2); // another way to assign 0.05 |

**How to convert numbers to scientific notation**

Use the following procedure:

- Your exponent starts at zero.
- Slide the decimal so there is only one non-zero digit to the left of the decimal.
- Each place you slide the decimal to the left increases the exponent by 1.
- Each place you slide the decimal to the right decreases the exponent by 1.
- Trim off any leading zeros (on the left end)
- Trim off any trailing zeros (on the right end) if the original number had no decimal point. We’re assuming they’re not significant unless otherwise specified.

Here’s some examples:

Start with: 42030 Slide decimal left 4 spaces: 4.2030e4 No leading zeros to trim: 4.2030e4 Trim trailing zeros: 4.203e4 (4 significant digits)

Start with: 0.0078900 Slide decimal right 3 spaces: 0007.8900e-3 Trim leading zeros: 7.8900e-3 Don't trim trailing zeros: 7.8900e-3 (5 significant digits)

Start with: 600.410 Slide decimal left 2 spaces: 6.00410e2 No leading zeros to trim: 6.00410e2 Don't trim trailing zeros: 6.00410e2 (6 significant digits)

Here’s the most important thing to understand: The digits in the significand (the part before the E) are called the **significant digits**. The number of significant digits defines a number’s **precision**. The more digits in the significand, the more precise a number is.

**Precision and range**

Consider the fraction 1/3. The decimal representation of this number is 0.33333333333333… with 3’s going out to infinity. An infinite length number would require infinite memory to store, and we typically only have 4 or 8 bytes. Floating point numbers can only store a certain number of significant digits, and the rest are lost. The **precision** of a floating point number defines how many *significant digits* it can represent without information loss.

When outputting floating point numbers, cout has a default precision of 6 -- that is, it assumes all floating point variables are only significant to 6 digits, and hence it will truncate anything after that.

The following program shows cout truncating to 6 digits:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
#include <iostream> int main() { float f; f = 9.87654321f; std::cout << f << std::endl; f = 987.654321f; std::cout << f << std::endl; f = 987654.321f; std::cout << f << std::endl; f = 9876543.21f; std::cout << f << std::endl; f = 0.0000987654321f; std::cout << f << std::endl; return 0; } |

This program outputs:

9.87654 987.654 987654 9.87654e+006 9.87654e-005

Note that each of these is only 6 significant digits.

Also note that cout will switch to outputting numbers in scientific notation in some cases. Depending on the compiler, the exponent will typically be padded to a minimum number of digits. Fear not, 9.87654e+006 is the same as 9.87654e6, just with some padding 0’s. The minimum number of exponent digits displayed is compiler-specific (Visual Studio uses 3, some others use 2 as per the C99 standard).

However, we can override the default precision that cout shows by using the std::setprecision() function that is defined in a header file called iomanip.

1 2 3 4 5 6 7 8 9 10 11 |
#include <iostream> #include <iomanip> // for std::setprecision() int main() { std::cout << std::setprecision(16); // show 16 digits float f = 3.33333333333333333333333333333333333333f; std::cout << f << std::endl; double d = 3.3333333333333333333333333333333333333; std::cout << d << std::endl; return 0; } |

Outputs:

3.333333253860474 3.333333333333334

Because we set the precision to 16 digits, each of the above numbers is printed with 16 digits. But, as you can see, the numbers certainly aren’t precise to 16 digits!

The number of digits of precision a floating point variable has depends on both the size (floats have less precision than doubles) and the particular value being stored (some values have more precision than others). Float values have between 6 and 9 digits of precision, with most float values having at least 7 significant digits (which is why everything after that many digits in our answer above is junk). Double values have between 15 and 18 digits of precision, with most double values having at least 16 significant digits. Long double has a minimum precision of 15, 18, or 33 significant digits depending on how many bytes it occupies.

Precision issues don’t just impact fractional numbers, they impact any number with too many significant digits. Let’s consider a big number:

1 2 3 4 5 6 7 8 9 10 |
#include <iostream> #include <iomanip> // for std::setprecision() int main() { float f(123456789.0f); // f has 10 significant digits std::cout << std::setprecision(9); // to show 9 digits in f std::cout << f << std::endl; return 0; } |

Output:

123456792

123456792 is greater than 123456789. The value 123456789.0 has 10 significant digits, but float values typically have 7 digits of precision. We lost some precision!

Consequently, one has to be careful when using floating point numbers that require more precision than the variables can hold.

Assuming IEEE 754 representation:

Size | Range | Precision |
---|---|---|

4 bytes | ±1.18 x 10^{-38} to ±3.4 x 10^{38} |
6-9 significant digits, typically 7 |

8 bytes | ±2.23 x 10^{-308} to ±1.80 x 10^{308} |
15-18 significant digits, typically 16 |

80-bits (12 bytes) | ±3.36 x 10^{-4932} to ±1.18 x 10^{4932} |
18-21 significant digits |

16 bytes | ±3.36 x 10^{-4932} to ±1.18 x 10^{4932} |
33-36 significant digits |

It may seem a little odd that the 12-byte floating point number has the same range as the 16-byte floating point number. This is because they have the same number of bits dedicated to the exponent -- however, the 16-byte number offers a much higher precision.

*Rule: Favor double over float unless space is at a premium, as the lack of precision in a float will often lead to inaccuracies.*

**Rounding errors**

One of the reasons floating point numbers can be tricky is due to non-obvious differences between binary (how data is stored) and decimal (how we think) numbers. Consider the fraction 1/10. In decimal, this is easily represented as 0.1, and we are used to thinking of 0.1 as an easily representable number. However, in binary, 0.1 is represented by the infinite sequence: 0.00011001100110011… Because of this, when we assign 0.1 to a floating point number, we’ll run into precision problems.

You can see the effects of this in the following program:

1 2 3 4 5 6 7 8 9 10 11 |
#include <iostream> #include <iomanip> // for std::setprecision() int main() { double d(0.1); std::cout << d << std::endl; // use default cout precision of 6 std::cout << std::setprecision(17); std::cout << d << std::endl; return 0; } |

This outputs:

0.1 0.10000000000000001

On the top line, cout prints 0.1, as we expect.

On the bottom line, where we have cout show us 17 digits of precision, we see that d is actually *not quite* 0.1! This is because the double had to truncate the approximation due to its limited memory, which resulted in a number that is not exactly 0.1. This is called a **rounding error**.

Rounding errors can have unexpected consequences:

1 2 3 4 5 6 7 8 9 10 11 12 13 |
#include <iostream> #include <iomanip> // for std::setprecision() int main() { std::cout << std::setprecision(17); double d1(1.0); std::cout << d1 << std::endl; double d2(0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1); // should equal 1.0 std::cout << d2 << std::endl; } |

1 0.99999999999999989

Although we might expect that d1 and d2 should be equal, we see that they are not. If we were to compare d1 and d2 in a program, the program would probably not perform as expected. We discuss this more in section 3.5 -- Relational operators (comparisons).

One last note on rounding errors: mathematical operations (such as addition and multiplication) tend to make rounding errors grow. So even though 0.1 has a rounding error in the 17th significant digit, when we add 0.1 ten times, the rounding error has crept into the 16th significant digit.

**NaN and Inf**

There are two special categories of floating point numbers. The first is **Inf**, which represents infinity. Inf can be positive or negative. The second is **NaN**, which stands for “Not a Number”. There are several different kinds of NaN (which we won’t discuss here).

Here’s a program showing all three:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
#include <iostream> int main() { double zero = 0.0; double posinf = 5.0 / zero; // positive infinity std::cout << posinf << "\n"; double neginf = -5.0 / zero; // negative infinity std::cout << neginf << "\n"; double nan = zero / zero; // not a number (mathematically invalid) std::cout << nan << "\n"; return 0; } |

And the results using Visual Studio 2008 on Windows:

1.#INF -1.#INF 1.#IND

INF stands for infinity, and IND stands for indeterminate. Note that the results of printing Inf and NaN are platform specific, so your results may vary.

**Conclusion**

To summarize, the two things you should remember about floating point numbers:

1) Floating point numbers are great for storing very large or very small numbers, including those with fractional components, so long as they have a limited number of significant digits (precision).

2) Floating point numbers often have small rounding errors, even when the number has fewer significant digits than the precision. Many times these go unnoticed because they are so small, and because the numbers are truncated for output. Consequently, comparisons of floating point numbers may not give the expected results. Performing mathematical operations on these values will cause the rounding errors to grow larger.

**Quiz**

1) Convert the following numbers to C++ style scientific notation (using an e to represent the exponent) and determine how many significant digits each has:

a) 34.50

b) 0.004000

c) 123.005

d) 146000

e) 146000.001

f) 0.0000000008

g) 34500.0

**Quiz Answers**

2.6 -- Boolean values and an introduction to if statements |

Index |

2.4a -- Fixed-width integers and the unsigned controversy |

It is far less confusing after reading this. Thank you. In class we covered this in about ten minutes and moved on to the next thing.

I’ve been to a few sites already trying to get a grasp on these (floats/doubles) and this summation really did the trick.

Thanks!

What do the f’s after some of the float and double values mean?

By default, if you type a floating point value into C++ it’s typed as a double. Consequently, if you do something like this:

You’re assigning a double to a float, which loses precision, and the compiler will probably complain.

Putting an “f” after the value means that you intend that value to be a float, not a double. Then when you do this:

You’re assigning a float value to a float variable, which makes more sense.

I missed something. How is it that 4.53 is a double and not a float?

The 4.53 is a literal constant of type

`double`

by default. When you add the f suffix to it like 4.53f it then becomes a literal constant of type`float`

it might be good if you add this explanation in the lesson instead of just a comment in the code. I was wondering myself about the usage of the f suffix due overlooking the comment.

Good idea. Done.

Alex,

It seems kind of dumb to inatialize a variable as a float, and then specify that you want a float to be output, by adding the f suffix. {example 9.87654321f }

Except for the fact that all floats are defaulted as double in C++. I’ve ran your example with all the numbers with the f suffix and without and get the same output. My PC has i7 processor. Why use float at all, since both use four bytes?

Your lesson are great, Thanks

I’m not quite sure I understand what you’re getting at. Doubles are generally 8 bytes, and floats are normally 4.

Is there something wrong with my code?

The compiler brings up a problem with setprecision()..

Thanks

You need to include iomanip.h to use setprecision() that way.

See the lesson on ostream for more info about output manipulators and stuff.

Coming form a Java background, I wonder if anyone can advise me a C++ library with a similar function as Java’s BigDecimal.

Preferably one that works on Linux with gcc(so not the decimal type from Visual C++)

So with all the rounding errors and precision problems, how do programmers deal with operations that need to display something that would end up with a precision or rounding error? Or am I just over-thinking things?

Most of the time it’s simply not necessary to display a number to the number of significant digits where precision/rounding errors creep in. Generally with floating point numbers, programs will truncate the display to 2-5 decimals.

For applications which need highly precise and extreme ranges, developers rely on libraries developed specifically to handle big and high precision numbers. Phrases describing them include: arbitrary-precision arithmetic, bignum arithmetic, multiple precision arithmetic, or sometimes infinite-precision arithmetic. That’s how they get past issues of rounding and precision errors because they are written to maintain accuracy using a variety of techniques. Beyond this tutorial.

this is very help full site

how can i make the value a user inputs into a float?

then when i run the program from main() and i put in 2 values like eg. x = 10 y = 3 then the answer is 3 instead of 3.333333

You are already storing the user input values as a float. The problem is that your function is returning an integer, so it’s truncating the result of x/y. Change your function to return a float and you will be good.

Hi Alex, really great guide. Enjoying the challenges its throwing up so far. With regards to functions returning integer’s, how do you change a function to actually return a float? My little "add" program is working but it cuts of the decimal? so if I have x = 4.5 + y = 5.5 it will give me the answer of 9! im probably missing something ridiculously obvious but if you could help that would be great!

Just change the return type of the function from int to float (or even better, double).

However, if x and y are ints, and you try to assign fractional values to them (e.g. 4.5 or 5.5) the fractions will get lost. So you may need to make those floats/doubles as well.

I set the precision level to 4, and added cout for the 2 values, fValue1 + fValue2.

I got fValue1 IS actually rounded off to 1.345 and fValue2 IS actually 1.123, expecting now the get the result of 2.468, but still reports ‘fTotal is not 2.468’

Why is that?

Chris

Rounding error. The numbers printed on your screen by cout are rounded in this case, so you’re not seeing the full representation. However, when you do the comparison, it does so with the actual numbers, not the rounded ones, which can lead to rounding issues.

How do I convert a Float say

x = 1234.567890123456789

to

y = 1234.5678901234 (small float ..10 decimal places only)

Something similar to setPrecision, to use NOT for display/Printing, but to use as a value for calculations / pass it on to a Database etc ?

I’m not sure what the best way to do this is. For small numbers, you can multiply by 10^x, cast to an integer to drop the remaining decimals, then divide by 10^x. However, if your number is too large you’ll overflow the int when you do the casting so I won’t say this is foolproof.

Didn’t understand how 0.1 is represented in binary by 0.00011001100110011…

In decimal, .1 is tenths, .01 is hundredths, .001 is thousandths and so on. Likewise, in binary, .1 is halves, .01 is quarters, .001 is eights, and so on.

0.000110011… would be equal to 1/16 + 1/32 + 1/256 + 1/512 + …

Ok, so in binary, we have to approximate the 1 of decimal 0.1 with an infinite sum. What I still don’t understand is why we don’t use all the “weights”, that is, all the powers of 2, but only 1/16, 1/32, 1/256, 1/512 and so on, that is, the 4th position (2^4 = 16), the 5th, the 8th, the 9th, and so on. In other words, why don’t we have 0.011111111…….. which is equal to 1/2 + 1/4 + 1/8 + 1/16 + …? It also approaches 1! (I am referring of course to the decimal part of 0.1, that is, the 1.)

That’s exactly why you can’t use every power of 1/2. The infinite sum would add up to 1, which is ten times the number we require. In order for the sum to add up to 0.1, you would need to add Sum[(1/2)^4n + (1/2)^(4n+1)], taking n from 1 to infinity. You can try it yourself if you want.

Yes, you are right, this sum indeed converges to 0.1, whereas the sum I used converges to 1.0. Thank you for the clear and concise explanation!

There is also a good explanation in Wikipedia (yes, sometimes - not often though - Wikipedia has good articles):

“Fractions in binary

Fractions in binary only terminate if the denominator has 2 as the only prime factor. As a result, 1/10 does not have a finite binary representation, and this causes 10 × 0.1 not to be precisely equal to 1 in floating point arithmetic. As an example, to interpret the binary expression for 1/3 = .010101…, this means: 1/3 = 0 × 2^(-1) + 1 × 2^(-2) + 0 × 2^(-3) + 1 × 2^(-4) + … = 0.3125 + … An exact value cannot be found with a sum of a finite number of inverse powers of two, and zeros and ones alternate forever.”

Follows a table of the conversion (fractional approximations) for fractions from decimal to binary. For the ones who are interested:

http://en.wikipedia.org/wiki/Binary_numeral_system

When I run the following code, the values seem really wrong when output. What is going wrong here?

your C ++ compiler has a tendency to roundoff 8th precision onwards.

For any value lesser then 8;

It will display 1 lesser than called for.

Hi there! Congratulation, very good explanation!!! Just what I was looking for.

Thank you.

When I run this code, I get z = 0.333333 and q = 0

float x = 1;

float y = 3;

float z = (x/y);

float q = (1/3);

Can someone explain why? I realize that if I write

float q = (1.0/3.0);

that this problem doesn’t occur, but I’m just wondering why I can’t use (1/3) since q is defined as a float. This page says it’s just a convention to have the decimal point.

Think it through as follows:

float x = 1 reads “put INT 1 into FLOAT x.” This changes its type from int to float. The same is true for float y = 3.

Thus float z = x/y divides two floats and returns a float.

However for float q = (1/3), this is a two part statement.

The first part (1/3) reads “divide INT 1 by INT 3”. Since this is division of two integers, this means it must return an integer (the floor), which in this case is 0.

The second part is then q = 0, which reads “put INT 0 into float q.”

An important thing to keep in mind is that division on a float is different than division on an integer. The literal 1 is read as an integer, however, the literal 1.0 is read as a float/double. This is why q = (1.0/3.0) is different than q = (1/3).

Hope this helped.

thank you!

Yup, 1 / 3 performs integer division (which gives an answer of 0, as the fractional component is dropped) , whereas 1.0 / 3.0 performs floating point division (which gives an answer of 0.333333…)

This is a very good article on the floating-point computation issue: “Microsoft Visual C++ Floating-Point Optimization”, by Eric Fleegal, MSDN, 2004

http://msdn.microsoft.com/en-us/library/aa289157(v=vs.71).aspx

Hi people,

I am new to C++ so please don’t flame me 🙂

I wrote a simple prog. to test this course but something isn’t really working well and I can’t figure out why…

#include<iostream>

#include<string>

#include <iomanip> // for setprecision()

using namespace std;

main()

{

cout<<setprecision(7); //7 decimals

float v = 1;

float j = 3;

float cc;

cc = v/j;

//TEST with FLOAT NUMMERS

float ff = 0.3333333; // 7 decimals as set in "setprecusion(7);"

if(cc<ff) {

cout<<"cc is smaller then ff"<<endl;

}

else if(cc>ff) {

cout<<"cc is bigger then ff"<<endl;

}

else{

cout<<"cc equals to ff"<<endl;

}

cout<<cc<<" = cc"<<endl;

cout<<ff<<" = ff"<<endl;

`return 0;`

}

The output gives me that cc is bigger than ff…

I don’t understand why as I set precision to 7 and my var ff has also 7 decimals.

They should both be equal.

Any suggestions where I made an error?

Thanks!!

I Have Tasted Your Code.

No, You didn’t make any wrong.

I think setprecision() function is only for setting the precision at time of showing your variable when you use cout.

I mean setprecision() cant change your variable.Like in your code cc = v/j so cc is stored as 0.33333333333333333333333333………. & setprecision() cant change this.You stored ff as 0.3333333.

0.3333333333333333…….. is greater than 0.3333333 isnt it?So your code show cc is bigger than ff.

Read the “Comparison of floating point numbers” part of this tutorial.It dosent say that you can use setprecision() for Comparison of floating point numbers.

Thank You

How to control the numbers after decimal point in C++?

Like in C language if i take a floating variable f = 123.4567 and i want to show only 2 numbers after decimal point than i will use printf(“%0.2f”).Then it will show 123.45.

In C++ i have to use setprecision().But It determines total numbers not just numbers after decimal point.So, it makes problem.Like if i dont know that what numbers my floating variable will contain after calculation it can contain 123.123 or 1234.123 so if i set precision to 5 for first case it will show 123.12 and for second case it will show 1234.1!But i always want to show 2 numbers after decimal points for every case.How can i do that in c++?

i have same question like you

Use the

`std::fixed`

stream manipulator and the member function`std::precision`

.For example if you want to display with 2 decimal places:

double pi=3.14159;

std::cout.precision(2);

std::cout << "Today's price for a slice of pi is $" << std::fixed << pi << std::endl;

and it should print:

`Today's price for a slice of pi is $3.14`

I'm perplexed! I'm working through Bjarne S's book and have done one of the drills but I don't understand what's going on with varying results.

Here's some output;

9.99

10

smaller is a 9.99 the larger is b 10

result 0.01

they are almost equal

99.99

100

smaller is a 99.99 the larger is b 100

result 0.01

199.99

200

smaller is a 199.99 the larger is b 200

result 0.01

they are almost equal

Why doesn't entering 99.99 and 100 give the message "they are almost equal"?

This is an interesting example of rounding errors.

10 - 9.99 = 0.01, but due to rounding error, C++ is representing this as 0.0099999999999997868

100 - 99.99 = 0.01, but due to rounding error, C++ is representing this as 0.0100000000000005116

One of this is larger than 0.01, and one is smaller.

Argggh! I almost went insane yesterday evening try to see where I'd gone wrong.

How does one avoid rounding errors? I can think of any number of applications where rounding errors even at the level of precision in my example might be disastrous.

There are a couple of ways to “avoid” rounding errors:

1) Avoid use of floating point numbers altogether (sometimes this is possible, sometimes it isn’t).

2) Don’t do raw comparisons like you’re doing. In section 3.5 -- Relational operators (comparisons), we discuss how to tell if floating point numbers are equal. These can be extended to handle less then/greater than cases. This would help avoid the case you see above.

3) Ensure that when you use floating point numbers, you only treat them as accurate to a certain level of precision.

I don’t know why all outputs are same.

Please someone help me.(VS2015,64 bit Windows)

Thank you!!

Your float doesn’t have enough precision to store the entire number, so it’s being truncated. If you change f to a double (and remove the f suffix on the literals) then you’ll see that the numbers print differently.

Sorry, I still feel confused.

So even i use setprecision, i still cannot change precision.

Is my understanding correct?

Thanks for your answering!

setprecision can’t show more significant digits than the underlying number has. A float only has 6 to 9 digits of precision, so you’re generally only going to get between 6 and 9 digits of precision even if you ask for more.

The number you picked (9876543.21f) has 9 significant digits, but because you tried to put it in a float, it got truncated to 7 significant digits (9876543) internally. So regardless of whether you ask for 7, 8, or 9 significant digits of precision, you’re only going to get 7 because that’s all the underlying number that got stored can offer.

Got it.

Thank you Alex!

whats the difference between:

cout<<precision(17)

&

cout.setprecision(17)

With Visual Studio, cout << setprecision() calls cout.precision() internally. So they essentially do the same thing. Calling cout.precision() is probably slightly more efficient, but unless you're calling it hundreds of times, it won't really matter.

SIR PLZ TEL ME HOW TO PRINT ON OUTPUT SCREEN 14 ND 0.5 IF THE GIVEN VALUE IS 14.5 BASICALLY I WANT TO PRINT IT SEPARATELY ON OT PUT SCREEN BY FLOAT ND INT DATA TYPE

Casting your float to an int via static_cast will drop the remainder, allowing you to print the integer. You can then subtract the integer from the floating point value to get just the remainder.

Hello, I’m not sure if you’ll read this but I’m confused about some of the outputs:

9.87654e+006

9.87654e-005

Why is 006 and 005 after the e?

How does that make sense?

I don’t get what 9.87654e+006 means… I thought you could only have numbers like 6 or -3 and stuff after the e not numbers like 006.

I’m confused now… Shouldn’t it just be 9.87654321e6 ? Why is it 006 ?

PS: I noticed your site is built on WordPress and I think it would be really helpful if you got a plugin that would email commenters when someone replies, would save me a lot of time, lol.

Anyway, thanks a lot, I’m finding this site really helpful 🙂

9.87654e+006 is the same as 9.87654e6. Visual Studio always prints exponents with at least 3 digits. Other compilers print exponents with at least 2 (as required by C99). In other words, it’s implementation specific.

Oh right, Ok I understand thanks a lot!

So… Just to clarify, the varying results seen in the above examples are caused by varying error(s) in rounding?

Eg, each time you round to the hundreds/hundredths place, it will be different from rounding to the tens/tenths place. And with that, the higher you the place you round to, the more accurate the answer will be (As with standard math). Is this correct?

Also, I’m getting slightly different results than what’s presented above. For instance, what’s 3.333333333333334 to you is 3.333333333333333 on my machine, despite rounding to the same place. Just thinking about that makes me uncomfortable!

Edit: it’s also worth noting that despite the fact that the comment box says we can use certain HTML tags, it automatically encodes the brackets.

I’m not sure which examples you’re referring to. But generally speaking, the fact that floating point numbers have limited precision leads to rounding errors, especially in cases where a number is represented as an infinite sequence in binary (like 0.1).

re: The HTML tags issue, I used to allow HTML on the site, but I’ve installed a plugin to treat all text literally so that people who post code snippets won’t have lines such as “#include” treated as invalid HTML. I’ll see if I can remove/disable the misleading text.

In your 0.1 example you forgot to add "#include <iostream>" at the top

Fixed.

I understood everything perfectly except one thing: why are we leaving trailing zeroes untrimmed when the original number has a decimal point?

Good question. If we write a large number with no decimal point, e.g. 21,000,000,000, it’s not clear whether we mean “exactly 21 billion”, or “somewhere around 21 billion”. More often than not, we mean “Somewhere around 21 billion” (because in most cases, the difference between 21,000,000,000 and 21,000,000,001 is insignificant).

So when we convert 21,000,000,000, we assume that the trailing zeros are not significant.

However, when the number has a decimal point, e.g. 21,000,000,000.01, or even 21,000,000,000.0, it’s clear that this number is exact, otherwise we wouldn’t have provided the decimal point. So in the case where a decimal is provided, we assume all of the trailing zeros are significant.

Make sense?

Your explanation makes sense, but does the compiler care? What does it do differently? Or is this just to let the code document what we know about the number?

I’m guessing that the compiler would consider 21,000,000,000.0 to be equivalent to 21,000,000,000. So although we would probably say one has a high precision and one doesn’t, the compiler likely treats them both as low-precision numbers.

I have to point this out, for the sake of correct math:

First off, infinity is not a number, it’s a concept.

Second, x/0 is not and never will be infinity or -infinity.

Numberphile did a good video that explains the division by 0 problem: https://www.youtube.com/watch?v=BRRolKTlF6Q

Ah mathemagicians and their concepts. Almost as bad as physicist.

Outputs:

0.10000000000000001

0.99999999999999989

In my machine. So, when I use double without setprecission(17), It shows the correct results: 0.1 and 1

So my question is, without using precission(17), do they store the same value? Or double without precission stores 0.10000000000000001 but somehow does a rounding to 0.1?

Greaat tutorials by the way! You are helping a lott!!

Cout rounds the numbers for display based on the precision. Internally, they’re stored as 0.10000000000000001 and 0.99999999999999989.

This is driving me crazy! Do floats have 6 or 7 digits of precision?

"When outputting floating point numbers, cout has a default precision of 6 -- that is, it assumes all variables are only significant to 6 digits, and hence it will truncate anything after that."

"The value 123456789.0 has 9 significant digits, but float only supports about 7 digits of precision"

"4 bytes ±1.18 x 10-38 to ±3.4 x 1038 7 significant digits"

The answer to this question is complicated due to the way floating point numbers are stored. Most floats have (at least) 7 digits of precision, however there is a subset of the floating point numbers that only have 6 digits of precision. I’ll update the article to indicate 6 instead of 7.

f = 9876549.21f in this case would scientific notation be 9.87655e006 because of after 6th specific digit there is 9? Mathematically 0.49 can be equal to 5.So could it be same in c++ ?

The precise scientific notation would be 9.87654921e006.

However, because std::cout rounds to 6 significant digits, 9.87654921e006 would be rounded to 9.87655e006.

This is something that is easy to verify yourself with a simple program:

Compiler says:

"error: ‘setprecision’ was not declared in this scope."

but when I delete "cout" from "std::cout << setprecision(x) " (e.g. std::setprecision(x)), it works as expected. Pretty confusing for beginners like me.

I updated the lesson to make it more clear that std::setprecision() lives in the std:: namespace.

Dear Alex, Is there a way to compare integer value with float type. Like If I want to write a code which takes input number from user and finds out if its a prime/even/odd number? Consider if for prime, division by any number other than 1 and the entered integer, would result in a fraction Now if i had taken input from user in Int and it just got converted to a fraction (which should be stored in a Float variable) how would I knoe if any thing like that has happened? OfCourse one can use as manu If else statements to compare the division or even the inserted number but I want to write smalest code for frist 100 or 200 prime numbers.

Hi waqas,

> Now if i had taken input from user in Int and it just got converted to a fraction

I’m having a hard time understanding what you’re asking.

Instead of doing a floating point division on an integer to see if the result is a fraction or not, you’d be better off using the modulus operator (%) to see if doing an integer division yields a modulus. You can use a for loop to loop through all of the possible divisors to see if any of them have a modulus of 0. Modulus is covered in chapter 3, and loops are covered in chapter 5.

I wanted to ask about the division by zero. While I am sure you understand the mathematics behind why it is undefined, here in your code it indicates that it will return infinity. Is it the double having a decimal that causes this to happen rather than just giving an error instead? I ask because I want to understand what the code is doing under the hood so that I never mistakenly fall into this trap. Thanks and great set of lessons!

The short answer is that it works that way because that’s how IEEE 754 (the standard to which floating point number implementations adhere) defines it that way. See http://grouper.ieee.org/groups/754/faq.html#exceptions.

Thanks! I will bear this in mind when I wish to reprogram my coffee pot! 😉

It does make sense though from that perspective, hardware can be stupid and this has to be adaptable to that.

After mixing my stupid experiments with your great lessons on floating point numbers, I ended up with this conclusion. Please correct me if I m wrong somewhere:

Floating numbers are different in a machine from what we expect. For example: 1.11 is not stored as 1.11 in a machine. Floating point numbers can’t take more than 4,8,12 or 16 bytes in memory. That is why numbers going out of Infinity has to be rounded up or down to an approximate value. Setting precision shows how numbers are stored in the machine but can’t result the exact number because it has to truncate the number according to the given limit or due to limited memory. After truncating, it returns an approximate value (approximate according to machine). However, we can see exact 1.11 to be printed out to the console because cout displays expected results when precision is set under the precision range of a type given to the object (e.g. double x; // x is okay with precision 15,16 or 17 when cout prints it’s value. If precision is set to 18 for x, cout will display unexpected digits after 15,16 or 17 significant digits) type and fails only when floating numbers are compared to each other using relational operator, or when evaluated as a result of an expression, using arithmetic operators. 50 is the largest parameter setprecision can hold (setprecision (50)). I m a bit confused here. Why 50. Does a number with 51 digits goes out of 16 byte (maximum a floating point number can reserve). Or setprecision () has a rule that this object can’t display more than 50 significant digits.

You’re pretty much correct. Floating point numbers are not stored in decimal format internally (1.11 is not stored as “1.11”), they are stored as some magic combination of bits that gets reconstituted into 1.11). Just like decimal numbers, some floating point numbers can be represented precisely, and some can not. For example 1/3rd can’t be represented precisely in decimal format (0.333333… you have to truncate somewhere), but 1/10 can (0.1).

It’s unclear to me whether setprecision() having a max value 50 is a limitation of your compiler or something else.

cout prints maximum 50 significant digits for all setprecision values greater than 50 in my machine. I m using code::blocks.

float f is printed out as 8e-010. I am using visual studio. Is there a mistake in initializing the variable? ignore the random long double.

No. 8e-10 and 8e-010 are the same number.

But rather than giving out 0.0000000008 it gives out 8e-10.

Yup. std::cout will print some numbers in scientific notation, particularly if they are large or small.

A good example which made me to understand the precision error is "Patriot Missile Failure" during GULF war. The internal clock was multiplied by 1/10 to get real time.

This calculation was doing a fixed point 24 bit multiplication. And over 100 hours this multiplication was yielded a drift of 0.34 sec which was enough for the missile to go undetected by the Radar.

Details at :

Hello, I can’t get one thing - is there any difference with rounding errors depending on which math operations we use.

double d = 0.1;

cout << d << endl; // shows 0.10000000000000001

double d2 = (d + d + d + d + d + d + d + d + d + d);

cout << d2 << endl; // shows 0.99999999999999989

double d3 = 10 * d;

cout << d3 << endl; // shows 1 ?? why ??

return 0;

Why after multiplying the result is without rounding error ?

Thanks upfront for answer and great tutorial!

Yes, the amount of rounding error can depend both on which mathematical operation you and how many times you use them. The more times you use them, the more errors tend to grow. The plus case has more error because we used plus 10 times.

Hi Alex,

if we want to store a 60 digits number in a variable or print it in output without scientific notation, what can we do ?

for example in python we can calculate 9999 to the exponent 9999 and it will print the result.

is there any way to do it ?

Thanks.

Since C++ doesn’t have support for arbitrarily large integers, probably the best solution here would be to install a library that implements large integers (e.g. https://mattmccutchen.net/bigint/).

Hi, I’m trying to write a program that evaluates a fraction from the user, does a math operation on it and spits out another fraction (in fractional format, not decimals). Is it possible to do this in C++? I ran a simple program with inputting a fraction into a float but the program doesn’t work, I suspect it’s not possible with floats but how else would I do it?

Yes, it is definitely possible, but in order to do that you’ll need to manage the numerator and denominator yourself (e.g. store them as integers) and write the operations to add, subtract, multiple, and divide them yourself. Floats won’t be of use for this.

Why?

45.5+45.5=90

or

123.456/654.321=0.188073

It should be 123.456/654.321= 0.1886780342 (right)

Thank Alex.

main.cpp

add.cpp

add.h

You should heed the warnings of your compiler. Your compiler should be warning you that enterNumber() is returning a float, but the return type is an integer. That’s why you’re getting incorrect answers: all the fractional numbers you’re entering are getting truncated to integers.

As a side note, your summation function is misnamed, as it doesn’t do any summation. You’re doing the summation in the function argument. As written, the function should be named “printResult” or something.

You can edit my post again. to give me a better idea.

Change:

to

oh ^^ thanks Alex.you are a great support.

I just have to say: Amazing tutorial! THANK YOU!!!!!!!!!!!

Hi Alex

look at the following snipped:

result: 0.12345

but the following snipped:

result: 0.123457

How can I fix it and why it happens?

std::cout has a default precision of 6 digits, but your expected value of 0.123456789 is 9 digits of precision. You need to tell std::cout to print more digits of precision:

Last two lessons…. very tough for me. Just like a lot of information to take in. And it’s very hard to remember without connecting to something meaningful.

I understand that there are rounding errors due to conversions between binary and decimal. Why don’t these errors propagate consistently?

For example, in your example of 0.1, the compiler prints 0.10000000000000001 (which is slightly greater than 0.10000…) when precision is set to 17 sig figs. So why do you get a number slightly less than 1 when you add 0.1 ten times? Logically, it would seem like the compiler might give a number like 1.00000000000000010 instead of 0.99999999999999989.

The errors do propagate as you’d expect. However, when a number can’t be stored exactly in floating point representation, it will generally be rounded to the closest floating-point representation with the given precision. This number could be larger or smaller than the actual number. It also means two numbers that “round down” may sum to one that “rounds up”, or vice versa.

Consider the following program:

This prints:

Three numbers that are larger than expected add up to one that is smaller than expected.

Hi, there.

Its me again.

Based on the following codes

Header file:

C++ source file:

Why does the ans return "inf" instead of a proper one, that is 1.43?

Thanks in advance

It works fine for me. I ran your example and got 1.43. It looks like you might be running an older compiler (since you’re using conio.h). Maybe try upgrading your compiler to a more modern one?

I’m having some trouble with this in Visual Studio (Community 2015 edition) concerning user-defined literal operators. I have no idea what they are, though. I tried to do this:

And when I mouse-over it, I get a note saying, "Error: user-defined literal operator not found". I don’t know how to fix this at all. Please help.

Try this instead:

123456789 is an integer literal, so I guess the compiler is refusing to do a conversion in this case.

It compiled when I took away the "f" suffix. I don’t know why, though, as it compiled fine with that suffix included on any other code I wrote.

Because then it treated the literal as an integer, and converted the integer literal to a float when the initialization took place.

I’m a newby but I disagree with a) and b) solutions. We physicists would assume that the zeroes on the right are not significant digits.

Really? Are you sure?? Is that your final answer???

I think you might be eligible for a refund on your degree.

Hi,

I’m a bit confused by the bit regarding the f appended to some numbers. I did some googling but am still confused.

"Note that by default, floating point literals default to type double. An f suffix is used to denote a literal of type float."

However, when you initialize or define a variable, you would have to specify its type as "float" or "double" wouldn’t you? Or do the two for some reason initialize the same type, and the f is the only way to initialize a non-double float?

Thanks for the great tutorial and for still keeping the comment section updated after all these years! 🙂

Even though you must assign your variable a type, C++ won’t infer what kind of literal you’re intending based on the type of variable you’re initializing. Therefore, you must use a suffix to specify what type the literal is.

So I can call it a float and use it as a double? And if I assign it a "double" type, but add the f suffix, it will be treated as a single float?

If you do something like this:

The compiler will implicitly convert the float literal to a double before initialization.

I decided to do a small test but… I still don’t seem to understand the trailing "f"…

I’m not sure how the format turns out, so here’s a pastebin:

http://pastebin.com/TU7R91qS

The output is:

fValue1: 4 bytes

fValue2: 4 bytes

dValue1: 8 bytes

dValue2: 8 bytes

ldValue1 16 bytes

ldValue2 16 bytes

It looks like the size of the float remains 4 bytes regardless of using the trailing f or not. Does it have to do with the compiler? Is using the f the standard, and hence "safe" way to do it, or am I just not getting something?

Sorry for all the questions, and thanks for your patience. 🙂

The f suffix exists to tell C++ that the literal number (in this case 0.1) should be treated as a floating point number rather than a double. It doesn’t have any impact on the size of the variable you assign that value to, but you may see differences in precision:

Hello,

Alex.

Could you please clarify below this question.

double d1 =3123456712345678912345678912345678912345.6789;

std::cout<<d1<<‘\n’;// it prints output- works okay.

double d2 =3123456712345678912345678912345678912345.6789f

std::cout<<d1<<‘\n’;// but this is not. it shows infinity.

As you have stated before (above comment) :

double d = 0.333333f; // converts a float literal to a double

My question- if the suffix converts a float literal to a double then in my case why my d2 variable doesn’t print which is converted from a literal to a double? As i understood, it is converted to double not a float.i.e the size of double could have cout the d2 value, but this acts as if it has float size (4 byte). This same situation between long double and double.

Thanks in Advance.

I think you have it backwards -- the f suffix causes the compiler to treat the literal as a float, not a double. So d1 is being initialized with a double literal, whereas d2 is being initialized with a float literal. Your very large number appears to be in range for a double, but not for a float, so the compiler says, 3123456712345678912345678912345678912345.6789f is out of range for a float, we’ll treat this as infinity. That gets assigned to d2.

34500.0 is equal to 3.45000e4 (6 significant digits)

But why 123456789.0 has 9 significant digits not 10 significant digits in above?

It should have been 10. I’ve updated the example.

can someone please explain to me what a "significand" or "exponent" is

The significand is the part of a number in scientific notation that contains the significant digits.

The exponent is the same exponent used in standard math. It determines how many times the base is multiplied. E.g. 2^4 means 2 is multiplied 4 times to get 16.

So in the number 1.2 x 10^4, 1.2 is the significand and 4 is the exponent. This evaluates to 1.2 * 10000 = 12000.

I don’t understand why you need to use the f suffix

isn’t a double basically the same as a float, but with more breathing space (memory)? and if so, why don’t you need to use suffixes for the different types of integers?

Generally speaking, literals don’t know what type they are -- numbers are just numbers devoid of context. The suffixes are used to specify what type the literal is intended to be. If you don’t specify a suffix, the compiler will make an assumption about what your intent is.

With integers, the compiler will automatically find the smallest integer that fits your integer literal and assume your integer literal is of that type. This works because integers do not suffer any precision issues, so the compiler only needs to worry about the range of the type. If you then assign the integer to a larger type, no problem.

However, with floating point numbers, there are precision issues, and that complicates things. A floating point literal gets represented differently as a float or a double, with a possible loss of precision, so the compiler can’t just put your floating point literal inside the smallest floating point type that it fits (e.g. a float) and call it good. But it has to do something if you don’t specify a prefix, so it assumes you meant double in that case.

Consider what happens if you do something like this:

4.3 is a decimal number with no suffix, so the compiler assumes you meant a double. 4.3 gets converted to a double, with some loss of precision. That double is then assigned to a float, which incurs another loss of precision. Double precision loss!

If 4.3 had an f suffix, then you only get one loss of precision, when 4.3f is converted to a float. The assignment doesn’t incur an additional loss of precision because the types are already identical by that point.

Make sense?

Hi, the word significand is correctly written with a t at the end -> significant

I put this page on the witelist of my AdBlocker Plus but stil get a message that ABP is active.

Any idea?

Greetings from Switzerland, Ronald Hofmann

--

No, significand is actually the correct term. It’s a mathematical term dealing with numbers in scientific notation (along with the mantissa and exponent).

Unsure about Adblock Plus, I’ve never used it so I can’t advise on how to fix issues with it. 🙁

Yes, you are right. I realized that 5 seconds after my comment, sorry for that.

Well, it was the first time I put an address on the white list.

Still trying

Greetings from Switzerland, Ronald Hofmann

--

I have a question.

Lets say I am setting up a clock for the hours and minutes and seconds and even miliseconds.

how would I write my code so as to show a maximum number for my miliseconds as well as a minimum?

lets say I want it to show all numbers up to the 100ths so it would say .50 or .55, how would I write?

If you want to display a fixed number of decimal places for floating point numbers, you need to tell std::cout to print numbers in fixed-point format. To do that:

The other alternative is to store your hours, minutes, seconds, and microseconds as integers and do your own formatting.

cool, thank you.

Another question, is there a way to capture a specific part of the data?

lets say with 123.4567. I only want to keep the .4567 and discard the rest.

The way I am thinking is to create an integer, making it equal to the float so it captures the whole numbers, then subtracting it from the float. Would that work?

The easiest way is as you suggest, cast the floating point number to an integer and subtract that from the floating point number. e.g.

Alex, me again.

As a correction note, when explaining how to convert numbers to scientific notation, on the first example, on step two (trimming leading zeros) I believe you trimmed the trailing zero by mistake. That could lead to a misunderstanding of the "protocol".

Where one reads:

"Start with: 42030

Slide decimal left 4 spaces: 4.2030e4

No leading zeros to trim: 4.203e4

Trim trailing zeros: 4.203e4 (4 significant digits)"

It could be:

"Start with: 42030

Slide decimal left 4 spaces: 4.2030e4

No leading zeros to trim: 4.2030e4 //no change on trailing zeroes yet

Trim trailing zeros: 4.203e4 (4 significant digits)"

Best regards.

Mauricio

Thanks! Fixed.

In C langauge I was taught to force truncate any unnecessary significant digits by hand before comparing a float for i.e. a while loop condition. Now I understand that if I didn’t, I could’ve created an infinite loop.

I don’t quite remember how to do that in C, but that shouldn’t matter for C++ now, problem is I don’t know it for C++ either yet. Will this be covered in a future lesson? If so I’ll wait patiently and read along, else I’d like to have a quick heads up, because I tend to try alot of crazy things and wouldn’t want to melt my CPU doing them.

Anyway, thanks for the tutorial so far. It’s been really helpful already!

I’m not sure what you mean by “truncate any unnecessary significant digits”. If they’re significant, then they’re not unnecessary, are they?

I do talk in the next chapter about how to properly compare floating point numbers, so if that’s the topic of interest here, keep reading.

Oh my bad there, I tend to call any digit that isn’t a zero before the first significant digit significant even though they’re unsignificant or even unwanted.

In example, if my float stores 0.1000000000000000001 instead of 0.1 flat, I’ve been taught to tell the compiler he has to cut off any digits that go beyond the 1st decimal (in C laungauge). Is this process explained / still used in C++ ?

Also I started wondering about the origin of the error, does it have to do with how we convert decimals to binary? In example: 0.5f = 2^-1 or 0.125f = 2^-3 etc. Not gonna break my head over it, but just wondered if that has something to do with it.

Just out of curiosity, how are you telling the compiler to 0.10000000000001 to 0.1? What if the number get represented as 0.999999999999 instead?

The origin of precision issues do generally have to do with how we convert decimals to binary, as explained in the section “Rounding errors” above.

Your table at the beginning of this lesson has different minimum sizes for floating point numbers than your table in lesson 2.3. In this lesson you have float 4, double 4, and long double 4.

In lesson 2.3 you have float 4, double 8, long double 8.

Thanks for pointing that out. I’ve updated the article accordingly.

thank you very much Mr. Alex, I really appreciate this tutorial.

please teach me how to solve this question. i have tried but not sure of the answers am getting.

I will be very grateful if anyone help me out.

A,b,c are variables of type int while d, e, f, and g are of type float.

•If a= 4, b = 6 and c -3, Find

•d = 3a + -ac

•f = d/a

•e = f+2c

Hello

Alex, for asking the question again, again.

You have stated above- Long double has a minimum precision of 15, 18, or 33 significant digits depending on how many bytes it occupies.And the table above is stated as-

------------------------------------------------

80-bits (12 bytes) ±3.36 x 10-4932 to ±1.18 x 104932 18-21 significant digits

-----------------------------------------------

16 bytes ±3.36 x 10-4932 to ±1.18 x 104932 33-36 significant digits

------------------------------------------------

#include<iostream>

int main()

{

std::cout << std::setprecision(22);

long double d1(1.44+1.69);

std::cout << d1 << std::endl;//output 3.1299999999999998933419

long double d2(0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1 + 0.1); // should equal 1.0

std::cout << d2 << std::endl;//output:0.9999999999999998889777

std::cout<<sizeof(d2);// output 16

}

In my case the long double has 15 precision as you see in program. the size of long data type is 16 bytes. I didn’t understand what you said that minimum precision of 15, 18, or 33 significant digits depend on how many bytes it occupies. Mine is 16 bytes as it cout in program.What does prevent getting minimum 33 significant bytes? is there anything that i missed,Thanks in advance.

Your request to display 22 bits of precision isn’t being honored. You need to std::cout << std::fixed before setting the precision.

Also don’t forget to #include iomanip.

can someone explain this code for me

a,b,c are variables of type int while d, e f and g are of type float.

•If a= 4, b = 6 and c -3, Find

•d=3a + -ac

•f = d/a

•e = f+2c

•g = f/e

thank you…

Hi,

Why does the following code compile at all?

I get a compile error when dividing 0.0/0.0 directly, while the outcome of the above is -nan(ind).

Is this because of the variable initialization, the usage of which C++ does not to limit and cause unexpected compile errors?

Thanks,

Garnik

If you try to divide by a 0.0 literal, the compiler can easily determine that you’re trying to divide by 0.0, and disallow it.

If you try to divide by a variable, the compiler can’t easily determine that the variable may hold the value 0.0. Therefore, it allows it, and you get a runtime issue.

Hi Alex,

Thanks for your reply, and like everyone doing your tutorials, I am very much grateful for the work you have done, as well as for being so responsive and supportive.

Best Regards,

Garnik

Want to know more about floating point numbers work? See here: https://en.wikipedia.org/wiki/Floating_point.

Learn more about NaN here: https://en.wikipedia.org/wiki/NaN.

On Visual Studio 2015, Windows 10, Intel Core i3-2130 @3.40 GHz, 8 GB RAM, x64 processor, when I run:

I get the output:

inf

-inf

-nan(ind)