In programming, a constant is a fixed value that may not be changed. C++ has two kinds of constants: literal constants, and symbolic constants. We’ll cover literal constants in this lesson, and symbolic constants in the next lesson.
Literal constants (usually just called literals) are values inserted directly into the code. For example:
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return 5; // 5 is an integer literal bool myNameIsAlex { true }; // true is a boolean literal std::cout << 3.4; // 3.4 is a double literal |
They are constants because their values can not be changed dynamically (you have to change them, and then recompile for the change to take effect).
Just like objects have a type, all literals have a type. The type of a literal is assumed from the value and format of the literal itself.
By default:
| Literal value | Examples | Default type |
|---|---|---|
| integral value | 5, 0, -3 | int |
| boolean value | true, false | bool |
| floating point value | 3.4, -2.2 | double (not float)! |
| char value | ‘a’ | char |
| C-style string | “Hello, world!” | const char[14] |
Literal suffixes
If the default type of a literal is not as desired, you can change the type of a literal by adding a suffix:
| Data Type | Suffix | Meaning |
|---|---|---|
| int | u or U | unsigned int |
| int | l or L | long |
| int | ul, uL, Ul, UL, lu, lU, Lu, or LU | unsigned long |
| int | ll or LL | long long |
| int | ull, uLL, Ull, ULL, llu, llU, LLu, or LLU | unsigned long long |
| double | f or F | float |
| double | l or L | long double |
You generally won’t need to use suffixes for integer types, but here are examples:
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unsigned int value1 { 5u }; // 5 has type unsigned int long value2 { 6L }; // 6 has type long |
By default, floating point literal constants have a type of double. To make them float literals instead, the f (or F) suffix should be used:
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float f { 5.0f }; // 5.0 has type float |
New programmers are often confused about why the following doesn’t work as expected:
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float f { 4.1 }; // warning: 4.1 is a double suffix, not a float suffix |
Because 4.1 has no suffix, it’s treated as a double literal, not a float literal. When C++ defines the type of a literal, it does not care what you’re doing with the literal (e.g. in this case, using it to initialize a float variable). Therefore, the 4.1 must be converted from a double to a float before it can be assigned to variable f, and this could result in a loss of precision.
Literals are fine to use in C++ code so long as their meanings are clear. This is most often the case when used to initialize or assign a value to a variable, do math, or print some text to the screen.
String literals
In lesson 4.11 -- Chars, we defined a string as a collection of sequential characters. C++ supports string literals:
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std::cout << "Hello, world!"; // "Hello, world!" is a C-style string literal std::cout << "Hello," " world!"; // C++ will concatenate sequential string literals |
String literals are handled very strangely in C++ for historical reasons. For now, it’s fine to use string literals to print text with std::cout, but don’t try and assign them to variables or pass them to functions -- it either won’t work, or won’t work like you’d expect. We’ll talk more about C-style strings (and how to work around all of those odd issues) in future lessons.
Scientific notation for floating point literals
There are two different ways to declare floating-point literals:
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double pi { 3.14159 }; // 3.14159 is a double literal in standard notation double avogadro { 6.02e23 }; // 6.02 x 10^23 is a double literal in scientific notation |
In the second form, the number after the exponent can be negative:
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double electron { 1.6e-19 }; // charge on an electron is 1.6 x 10^-19 |
Octal and hexadecimal literals
In everyday life, we count using decimal numbers, where each numerical digit can be 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9. Decimal is also called “base 10”, because there are 10 possible digits (0 through 9). In this system, we count like this: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, … By default, numbers in C++ programs are assumed to be decimal.
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int x { 12 }; // 12 is assumed to be a decimal number |
In binary, there are only 2 digits: 0 and 1, so it is called “base 2”. In binary, we count like this: 0, 1, 10, 11, 100, 101, 110, 111, …
There are two other “bases” that are sometimes used in computing: octal, and hexadecimal.
Octal is base 8 -- that is, the only digits available are: 0, 1, 2, 3, 4, 5, 6, and 7. In Octal, we count like this: 0, 1, 2, 3, 4, 5, 6, 7, 10, 11, 12, … (note: no 8 and 9, so we skip from 7 to 10).
| Decimal | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| Octal | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 10 | 11 | 12 | 13 |
To use an octal literal, prefix your literal with a 0:
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#include <iostream> int main() { int x{ 012 }; // 0 before the number means this is octal std::cout << x; return 0; } |
This program prints:
10
Why 10 instead of 12? Because numbers are printed in decimal, and 12 octal = 10 decimal.
Octal is hardly ever used, and we recommend you avoid it.
Hexadecimal is base 16. In hexadecimal, we count like this: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F, 10, 11, 12, …
| Decimal | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 |
| Hexadecimal | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | A | B | C | D | E | F | 10 | 11 |
To use a hexadecimal literal, prefix your literal with 0x.
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#include <iostream> int main() { int x{ 0xF }; // 0x before the number means this is hexadecimal std::cout << x; return 0; } |
This program prints:
15
Because there are 16 different values for a hexadecimal digit, we can say that a single hexadecimal digit encompasses 4 bits. Consequently, a pair of hexadecimal digits can be used to exactly represent a full byte.
Consider a 32-bit integer with value 0011 1010 0111 1111 1001 1000 0010 0110. Because of the length and repetition of digits, that’s not easy to read. In hexadecimal, this same value would be: 3A7F 9826. This makes hexadecimal values useful as a concise way to represent a value in memory. For this reason, hexadecimal values are often used to represent memory addresses or raw values in memory.
Prior to C++14, there is no way to assign a binary literal. However, hexadecimal pairs provide us with a useful workaround:
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#include <iostream> int main() { int bin{}; bin = 0x01; // assign binary 0000 0001 to the variable bin = 0x02; // assign binary 0000 0010 to the variable bin = 0x04; // assign binary 0000 0100 to the variable bin = 0x08; // assign binary 0000 1000 to the variable bin = 0x10; // assign binary 0001 0000 to the variable bin = 0x20; // assign binary 0010 0000 to the variable bin = 0x40; // assign binary 0100 0000 to the variable bin = 0x80; // assign binary 1000 0000 to the variable bin = 0xFF; // assign binary 1111 1111 to the variable bin = 0xB3; // assign binary 1011 0011 to the variable bin = 0xF770; // assign binary 1111 0111 0111 0000 to the variable return 0; } |
C++14 binary literals and digit separators
In C++14, we can assign binary literals by using the 0b prefix:
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#include <iostream> int main() { int bin{}; bin = 0b1; // assign binary 0000 0001 to the variable bin = 0b11; // assign binary 0000 0011 to the variable bin = 0b1010; // assign binary 0000 1010 to the variable bin = 0b11110000; // assign binary 1111 0000 to the variable return 0; } |
Because long literals can be hard to read, C++14 also adds the ability to use a quotation mark (‘) as a digit separator.
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#include <iostream> int main() { int bin{ 0b1011'0010 }; // assign binary 1011 0010 to the variable long value{ 2'132'673'462 }; // much easier to read than 2132673462 return 0; } |
If your compiler isn’t C++14 compatible, your compiler will complain if you try to use either of these.
Printing decimal, octal, hexadecimal, and binary numbers
By default, C++ prints values in decimal. However, you can tell it to print in other formats. Printing in decimal, octal, or hex is easy via use of std::dec, std::oct, and std::hex:
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#include <iostream> int main() { int x { 12 }; std::cout << x << '\n'; // decimal (by default) std::cout << std::hex << x << '\n'; // hexadecimal std::cout << x << '\n'; // now hexadecimal std::cout << std::oct << x << '\n'; // octal std::cout << std::dec << x << '\n'; // return to decimal std::cout << x << '\n'; // decimal return 0; } |
This prints:
12 c c 14 12 12
Printing in binary is a little harder, as std::cout doesn’t come with this capability built-in. Fortunately, the C++ standard library includes a type called std::bitset that will do this for us (in the <bitset> header). To use std::bitset, we can define a std::bitset variable and tell std::bitset how many bits we want to store. The number of bits must be a compile time constant. std::bitset can be initialized with an unsigned integral value (in any format, including decimal, octal, hex, or binary).
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#include <iostream> #include <bitset> // for std::bitset int main() { // std::bitset<8> means we want to store 8 bits std::bitset<8> bin1{ 0b1100'0101 }; // binary literal for binary 1100 0101 std::bitset<8> bin2{ 0xC5 }; // hexadecimal literal for binary 1100 0101 std::cout << bin1 << ' ' << bin2 << '\n'; std::cout << std::bitset<4>{ 0b1010 } << '\n'; // we can also print from std::bitset directly return 0; } |
This prints:
11000101 11000101 1010
We can also create a temporary (anonymous) std::bitset to print a single value. In the above code, this line:
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std::cout << std::bitset<4>{ 0b1010 } << '\n'; // we can also print from std::bitset directly |
creates a temporary std::bitset object with 4 bits, initializes it with 0b1010, prints the value in binary, and then discards the temporary std::bitset.
Magic numbers, and why they are bad
Consider the following snippet:
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int maxStudents{ numClassrooms * 30 }; |
A number such as the 30 in the snippet above is called a magic number. A magic number is a literal (usually a number) in the middle of the code that does not have any context. What does 30 mean? Although you can probably guess that in this case it’s the maximum number of students per class, it’s not absolutely clear. In more complex programs, it can be very difficult to infer what a hard-coded number represents, unless there’s a comment to explain it.
Using magic numbers is generally considered bad practice because, in addition to not providing context as to what they are being used for, they pose problems if the value needs to change. Let’s assume that the school buys new desks that allow them to raise the class size from 30 to 35, and our program needs to reflect that. Consider the following program:
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int maxStudents{ numClassrooms * 30 }; setMax(30); |
To update our program to use the new classroom size, we’d have to update the constant 30 to 35. But what about the call to setMax()? Does that 30 have the same meaning as the other 30? If so, it should be updated. If not, it should be left alone, or we might break our program somewhere else. If you do a global search-and-replace, you might inadvertently update the argument of setMax() when it wasn’t supposed to change. So you have to look through all the code for every instance of the literal 30, and then determine whether it needs to change or not. That can be seriously time consuming (and error prone).
Fortunately, better options (symbolic constants) exist. We’ll talk about those in the next lesson.
Best practice
Don’t use magic numbers in your code.
 4.13 -- Const, constexpr, and symbolic constants |
 Index |
 4.11 -- Chars |
Hi there! I have one question about magic numbers. The basic solution would be to put them into variables right? But won't we start taking up memory once we put every single magic number into a variable? That could cause optimisation problems.
Generally if you put them into const variables, the compiler will be able to optimize the variables away. That gives you all of the maintainability benefits for no performance cost.
your saying long and int are literals yet they can be changed in the program
You can have int literals and long literals. They can be changed, but you'll have to recompile afterward. You can't change them at runtime.
Is 0x10 a hexadecimal literal or a binary literal? How do we differentiate between the two in the versions prior to c++14?
Hex. Binary literals start with 0b.
HOW CAN WE PRINT OR RETURN VALUES IN ANY OTHER NO. SYSTEM THAN DECIMAL.
See this article.
Literal constants (usually just called “literals”) are values inserted directly into the code. They are constants because you can’t change their values. For example:
int x = 5;
But isnt this a variable that i can change anytime? im sry, im confused in this one, literal constants are variables too? why they are called constant then?
Please help me :(
5 is the literal, not x (x is a variable). 5 has the value 5. You can't change that.
I see, thanks
Was also confused. I would suggest something like that:
bool myNameIsAlex = true; // boolean variable 'myNameIsAlex' is assigned a boolean literal constant 'true'
int x = 5; // integer variable x is assigned an integer literal constant 5
std::cout << 2 * 3; // 2 and 3 are integer literals
Updated per your suggestion.
Looks like someone has had a bad time with magic numbers :)
In the code fragment on hexidecimal below
on line 15
you have the incorrect binary value for B. it should be 1011 not 1010
Thanks, fixed.
Typo towards the end of section "Literal constants":
"By default, floating point literal constant have a type of double."
"Constant" should be plural.
Fixed, thanks.
Hi Alex,
There are many vital pieces of information about "Binary" in section 2.1 & 2.3 (... all data on a computer is just a sequence of bits.....When you assign a value to a data type, the compiler and CPU take care of the details of encoding your value into the appropriate sequence of bits for that data type. When you ask for your value back, your number is “reconstituted” from the sequence of bits in memory....)
In this section, Octal and Hexadecimal are introduced...but I don't know what they are for?
Thanks, Have a nice day.
As I mention in the lesson, octal is hardly ever used, so you can basically forget about it.
Hexadecimal values are used a lot though, mainly because two hexadecimal values cover 8 bits, which is a byte. Therefore, when we talk about the contents of a memory address (which are a byte), instead of representing those contents as 8 binary digits, it's much easier to represent them as 2 hexadecimal digits.
I believe you meant to use "ways" in the following line:
While boolean and integer literals are pretty straightforward, there are two different way to declare floating-point literals:
thanks - appreciate all your hard work!
Thanks for the typo notification. Fixed!
does gcc support C++14?
Yes, some newer versions do. But you have to pass it a flag to enable that functionality (-std=c++14 or -std=c++1y, depending on gcc version)
I'm CLICKING ON EVERY AD! You're welcome.
Well that's one way to live your life.
another is to read through the comments trying to find ones you can write pithy replies to :))
Hello, I tried to do a converter decimal/hexadecimal or decimal/octal with his code :
But it seems that using 0x01 and 0number doesn't work if you write it in the consol. Is their any way to create a converter like that ?
Yes, use this instead:
I don't really understand the definition of litteral, because in math I found this definition :
"Literal numbers are the letters which are used to represent a number."
But here it seems to have a totally different signification, but which ?
Literals in programming are values typed directly into the code. For example:
4
6.0f
"hello"
Your syntax highlighter certainly makes a mess of the infix apostrophes in binary literals...I guess it's not C++14 compliant :-)
Yeah, it definitely butchers those. Hopefully it'll get fixed in a future version.
alex, thx for all these best tuts. i want to ask u, are u now going to write tuts for some other platforms also or not??....it will be best for me if u will write javaEE.
Nope. I barely have enough time to write these tutorials and answer questions, let alone try and do this for two languages. :)
I'm confused about the float type default of double. Why isn't the default of float set to float? It seems weird to have to use a suffix to specify that the type should be float for a type that's already been defined as float. Where does the double come in?
Are you asking why for the following:
4.0 isn't assumed to be a float since it's assigned to a float variable? I presume because:
* C doesn't have type inference (C++11 does, but this was inherited from C).
* Type inference works from right to left, not left to right.
* This gives the programmer has explicit control over what 4.0 means (type double) regardless of what's on the left-hand side of the literal.
Why 10 instead of 12? Because numbers are stored in decimal, and 12 octal = 10 decimal.
I think it should be:
Because numbers are being printed in... (or something)
They actually stored in binary.
Good call. Updated the wording.
Thanks for the tutorial
HI there Alex,
I am a bit confused as to why one needs to type in the suffixes of the data types? They are not necessary right?
The suffixes of the data types tell C++ how to interpret a literal.
For example, if you type in 5.2, C++ knows this is a floating point literal, but it doesn't know whether you meant a float or a double. So it assumes double.
If you wanted/needed 5.2 to be a float, you're better off specifying the literal as 5.2f, so C++ knows you meant a float, not a double.
Otherwise, if you do this:
C++ will convert 5.2 from a double to a float before assigning to variable f, and you may lose some precision.
Suffixes are only needed if the default type for a literal isn't sufficient for your needs. Generally, when using literals, it's a good idea to ensure your literal has the same type as the variable it's being assigned to, to minimize the chance of something unexpectedly going wrong somewhere.
A question: why FF in hexadecimal is 255 in decimal?
That's simple hexa coding, you definitely should re-read about it.
"F" in hexa stands for a "15" in decimal (1,2,3,4,5,6,7,8,9,A,B,C,D,E,F), so when you have 0xFF to convert it to decimal:
0xFF = F * 16^1 + F * 16^0 = 15 * 16 + 15 = 240 + 15 = 255.
Alejandro
Hi Alex
Can you add ";" to your examples?
Done. Thanks for pointing that out.
A good thing to have in mind counting in oct and hex:
oct is base 8, so starting the count is 0, 1, 2, 3, 4, 5, 6, 7
all 8 digits were used, so now we add 1 to the left: 10, 11, 12, 13, 14, 15, 16, 17
20, 21, 22, 23, 24, 25, 26, 27 etc.
same for hex, base 16:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F
add 1 left:
10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B, 1C, 1D, 1E, 1F
this principle is for decimal as well, just an easy way to count...
I think you made a typo here
above code should be
-Kanchana
I sure did. Thanks for noticing.
The last example seems to have wrong comments:
should be:
Thanks for the tutorial !
Yup, you're correct. Thanks for noticing!
Above you have: Octal is base 8 — that is, the only digits available are: 0, 1, 2, 3, 4, 5, 6, 7, 8.
That would be base 9, though, right?
Yup! Good catch.
Octal is base 8 - because the digits available are : 0,1,2,3,4,5,6,7.
Hi! I'm not totally new to C++, but new to this course. (Yes, I'm calling it a "course" 'cause it's better than all other C++ teaching I've come across in the past - THANK YOU SO MUCH!)
Anyhow... I've been digging through the tutorials very seriously. All of the stuff is really well explained and if a piece of information happens to be missing, it is usually covered by the comments. However, here (in 2.8), I'm a bit lost... Can you explain to me, why we'd want to add an "L" to
long nValue2 = 5L;
or an f to
float fValue = 5.0f;
...?
They have already been declared as being "long" and "float"... So what's the point here...???
Thx!
There aren't a whole lot of reasons you'd need to specify the L prefix.
But lets say you had two functions:
void doSomething(int);
void doSomething(long);
if you called doSomething(5), you'd get the int version instead of the long version. Using doSomething(5L) would get you version that takes the long parameter.
There are probably other obscure examples.
The f prefix is used more commonly, because floating point numbers have weird truncation/rounding issues.
Consider the following code:
This prints "not equal"! Why? When 0.67 gets assigned to f, it gets truncated to 0.670000017. When you compare that truncated value to the double value 0.67, it doesn't match!
This one works as expected:
Hi Alex,
I'm new to c++. My question may be dump to you but what's the difference between "int = 5" or "int == 5".
there is no such thing like "int = 5" or “int == 5?
"int" is a keyword used to define variables type
thanks a lot dear.
this is very nice tutorial..
i m not new to c++ also not proficient but getting good concepts from this tutorial
Alex:
Ref.: it is a good idea to try to avoid using literal constants that aren't 0 or 1.
So I should use only literal constants that are 0 or 1? As for above.
Generally I only use the literal constant 0 -- anything else is generally defined as a symbolic constant.
Literal constants are literal numbers inserted into the code. They are constants because you can't change their values.
int x = 5; // 5 is a literal constant
I don't understand this, you can change 5 to 6, how is it unchangable?
If you use the number 6 instead of 5, you are using a different literal, not changing the value of a literal. In other words, literals are constants because the symbol 5 always has the value 5. You can't change the symbol 5 to the value 6, or any other number.
When declaring an integer variable that isn't const, the value of the variable x can be reassigned to, later on.
int x = 5; // declares x as an integer variable and assigns 5 to x
x = 2; // 2 is now assigned to x, instead of 5
x = 4; // 4 is now assigned to x, instead of 2
In the above example, x is declared and the number 5 is assigned to x. We can then assign another number to x later on, which will then change the value of x.
We can check that the values have changed by printing them to the console:
using namespace std;
int x = 5;
cout << x << endl; // the value of x displayed here is 5
x = 2;
cout << x << endl; // the value changes to 2
x = 4;
cout << x << endl; // now it is 4
This outputs:
5
2
4
However when we declare a variable as const, and assign a value to it, we cannot assign another value later on:
const int x = 5; // declares x as an constant integer variable and assigns 5 to x
x = 2; // compiler error, as we cannot assign another value to x
Although the compiler comes up with an error "you cannot assign to a variable that is const", this is misleading, as you can assign (initialize) a literal constant to a variable only once. Consequently its value remains constant throughout the entire program.
P.S. I'm also a newbie learning C++, but I'm just getting this logic from what Alex has written in the past, along with some practise on Visual C++ 2010 Express!
P.P.S. I'm loving these tutorials Alex ;)