In order to understand the bit manipulation operators, it is first necessary to understand how integers are represented in binary. We talked a little bit about this in section 2.4 -- Integers, and will expand upon it here.

Consider a normal decimal number, such as 5623. We intuitively understand that these digits mean (5 * 1000) + (6 * 100) + (2 * 10) + (3 * 1). Because there are 10 decimal numbers, the value of each digit increases by a factor of 10.

Binary numbers work the same way, except because there are only 2 binary numbers (0 and 1), the value of each digit increases by a factor of 2. Just like commas are often used to make a large decimal number easy to read (e.g. 1,427,435), we often write binary numbers in groups of 4 bits to make them easier to read (e.g. 1101 0101).

As a reminder, in binary, we count from 0 to 15 like this:

Decimal Value | Binary Value |
---|---|

0 | 0 |

1 | 1 |

2 | 10 |

3 | 11 |

4 | 100 |

5 | 101 |

6 | 110 |

7 | 111 |

8 | 1000 |

9 | 1001 |

10 | 1010 |

11 | 1011 |

12 | 1100 |

13 | 1101 |

14 | 1110 |

15 | 1111 |

**Converting binary to decimal**

In the following examples, we assume that we’re dealing with unsigned integers.

Consider the 8 bit (1 byte) binary number 0101 1110. 0101 1110 means (0 * 128) + (1 * 64) + (0 * 32) + (1 * 16) + (1 * 8) + (1 * 4) + (1 * 2) + (0 * 1). If we sum up all of these parts, we get the decimal number 64 + 16 + 8 + 4 + 2 = 94.

Here is the same process in table format. We multiply each binary digit by its digit value (determined by its position). Summing up all these values gives us the total.

Converting 0101 1110 to decimal:

Binary digit | 0 | 1 | 0 | 1 | 1 | 1 | 1 | 0 |

* Digit value | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |

= Total (94) | 0 | 64 | 0 | 16 | 8 | 4 | 2 | 0 |

Let’s convert 1001 0111 to decimal:

Binary digit | 1 | 0 | 0 | 1 | 0 | 1 | 1 | 1 |

* Digit value | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |

= Total (151) | 128 | 0 | 0 | 16 | 0 | 4 | 2 | 1 |

1001 0111 binary = 151 in decimal.

This can easily be extended to 16 or 32 bit binary numbers simply by adding more columns. Note that it’s easiest to start on the right end, and work your way left, multiplying the digit value by 2 as you go.

**Method 1 for converting decimal to binary**

Converting from decimal to binary is a little more tricky, but still pretty straightforward. There are two good methods to do this.

The first method involves continually dividing by 2, and writing down the remainders. The binary number is constructed at the end from the remainders, from the bottom up.

Converting 148 from decimal to binary (using r to denote a remainder):

148 / 2 = 74 r0

74 / 2 = 37 r0

37 / 2 = 18 r1

18 / 2 = 9 r0

9 / 2 = 4 r1

4 / 2 = 2 r0

2 / 2 = 1 r0

1 / 2 = 0 r1

Writing all of the remainders from the bottom up: 1001 0100

148 decimal = 1001 0100 binary.

You can verify this answer by converting the binary back to decimal:

(1 * 128) + (0 * 64) + (0 * 32) + (1 * 16) + (0 * 8) + (1 * 4) + (0 * 2) + (0 * 1) = 148

**Method 2 for converting decimal to binary**

The second method involves working backwards to figure out what each of the bits must be. This method can be easier with small binary numbers.

Consider the decimal number 148 again. What’s the largest power of 2 that’s smaller than 148? 128, so we’ll start there.

Is 148 >= 128? Yes, so the 128 bit must be 1. 148 - 128 = 20, which means we need to find bits worth 20 more.

Is 20 >= 64? No, so the 64 bit must be 0.

Is 20 >= 32? No, so the 32 bit must be 0.

Is 20 >= 16? Yes, so the 16 bit must be 1. 20 - 16 = 4, which means we need to find bits worth 4 more.

Is 4 >= 8? No, so the 8 bit must be 0.

Is 4 >= 4? Yes, so the 4 bit must be 1. 4 - 4 = 0, which means all the rest of the bits must be 0.

148 = (1 * 128) + (0 * 64) + (0 * 32) + (1 * 16) + (0 * 8) + (1 * 4) + (0 * 2) + (0 * 1) = 1001 0100

In table format:

Binary number | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 |

* Digit value | 128 | 64 | 32 | 16 | 8 | 4 | 2 | 1 |

= Total (148) | 128 | 0 | 0 | 16 | 0 | 4 | 0 | 0 |

**Another example**

Let’s convert 117 to binary using method 1:

117 / 2 = 58 r1

58 / 2 = 29 r0

29 / 2 = 14 r1

14 / 2 = 7 r0

7 / 2 = 3 r1

3 / 2 = 1 r1

1 / 2 = 0 r1

Constructing the number from the remainders from the bottom up, 117 = 111 0101 binary

And using method 2:

The largest power of 2 less than 117 is 64.

Is 117 >= 64? Yes, so the 64 bit must be 1. 117 - 64 = 53.

Is 53 >= 32? Yes, so the 32 bit must be 1. 53 - 32 = 21.

Is 21 >= 16? Yes, so the 16 bit must be 1. 21 - 16 = 5.

Is 5 >= 8? No, so the 8 bit must be 0.

Is 5 >= 4? Yes, so the 4 bit must be 1. 5 - 4 = 1.

Is 1 >= 2? No, so the 2 bit must be 0.

Is 1 >= 1? Yes, so the 1 bit must be 1.

117 decimal = 111 0101 binary.

**Adding in binary**

In some cases (we’ll see one in just a moment), it’s useful to be able to add two binary numbers. Adding binary numbers is surprisingly easy (maybe even easier than adding decimal numbers), although it may seem odd at first because you’re not used to it.

Consider two small binary numbers:

0110 (6 in decimal) +

0111 (7 in decimal)

Let’s add these. First, line them up, as we have above. Then, starting from the right and working left, we add each column of digits, just like we do in a decimal number. However, because a binary digit can only be a 0 or a 1, there are only 4 possibilities:

- 0 + 0 = 0
- 0 + 1 = 1
- 1 + 0 = 1
- 1 + 1 = 0, carry a 1 over to the next column

Let’s do the first column:

0110 (6 in decimal) + 0111 (7 in decimal) ---- 1

0 + 1 = 1. Easy.

Second column:

1 0110 (6 in decimal) + 0111 (7 in decimal) ---- 01

1 + 1 = 0, with a carried one into the next column

Third column:

11 0110 (6 in decimal) + 0111 (7 in decimal) ---- 101

This one is a little trickier. Normally, 1 + 1 = 0, with a carried one into the next column. However, we already have a 1 carried from the previous column, so we need to add 1. Thus, we end up with a 1 in this column, with a 1 carried over to the next column

Last column:

11 0110 (6 in decimal) + 0111 (7 in decimal) ---- 1101

0 + 0 = 0, but there’s a carried 1, so we add 1. 1101 = 13 in decimal.

Now, how do we add 1 to any given binary number (such as 1011 0011)? The same as above, only the bottom number is binary 1.

1 (carry column) 1011 0011 (original binary number) 0000 0001 (1 in binary) --------- 1011 0100

**Signed numbers and two’s complement**

In the above examples, we’ve dealt solely with unsigned integers. In this section, we’ll take a look at how signed numbers (which can be negative) are dealt with.

Signed integers are typically stored using a method known as **two’s complement**. In two’s complement, the leftmost (most significant) bit is used as the sign bit. A 0 sign bit means the number is positive, and a 1 sign bit means the number is negative.

Positive signed numbers are stored just like positive unsigned numbers (with the sign bit set to 0).

Negative signed numbers are stored as the inverse of the positive number, plus 1.

**Converting integers to binary two’s complement**

For example, here’s how we convert -5 to binary two’s complement:

First we figure out the binary representation for 5: 0000 0101

Then we invert all of the bits: 1111 1010

Then we add 1: 1111 1011

Converting -76 to binary:

Positive 76 in binary: 0100 1100

Invert all the bits: 1011 0011

Add 1: 1011 0100

Why do we add 1? Consider the number 0. If a negative value was simply represented as the inverse of the positive number, 0 would have two representations: 0000 0000 (positive zero) and 1111 1111 (negative zero). By adding 1, 1111 1111 intentionally overflows and becomes 0000 0000. This prevents 0 from having two representations, and simplifies some of the internal logic needed to do arithmetic with negative numbers.

**Converting binary two’s complement to integers**

To convert a two’s complement binary number back into decimal, first look at the sign bit.

If the sign bit is 0, just convert the number as shown for unsigned numbers above.

If the sign bit is 1, then we invert the bits, add 1, then convert to decimal, then make that decimal number negative (because the sign bit was originally negative).

For example, to convert 1001 1110 from two’s complement into a decimal number:

Given: 1001 1110

Invert the bits: 0110 0001

Add 1: 0110 0010

Convert to decimal: (0 * 128) + (1 * 64) + (1 * 32) + (0 * 16) + (0 * 8) + (0 * 4) + (1 * 2) + (0 * 1) = 64 + 32 + 2 = 98

Since the original sign bit was negative, the final value is -98.

If adding in binary is difficult for you, you can convert to decimal first, and then add 1.

**Why types matter**

Consider the binary value 1011 0100. What value does this represent? You’d probably say 180, and if this were standard unsigned binary number, you’d be right.

However, if this value was stored using two’s complement, it would be -76.

And if the value were encoded some other way, it could be something else entirely.

So how does C++ know whether to print a variable containing binary 1011 0100 as 180 or -76?

Way back in section 2.1 -- Basic addressing and variable declaration, we said, “When you assign a value to a data type, the compiler and CPU takes 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.”

So the answer is: it uses the type of the variable to convert the underlying binary representation back into the expected form. So if the variable type was an unsigned integer, it would know that 1011 0100 was standard binary, and should be printed as 180. If the variable was a signed integer, it would know that 1011 0100 was encoded using two’s complement (assuming that’s what it was using), and should be printed as -76.

**What about converting floating point numbers from/to binary?**

How floating point numbers get converted from/to binary is quite a bit more complicated, and not something you’re likely to ever need to know. However, if you’re curious, see this site, which does a good job of explaining the topic in detail.

**Quiz**

1) Convert 0100 1101 to decimal.

2) Convert 93 to an 8-bit unsigned binary number.

3) Convert -93 to an 8-bit signed binary number (using two’s complement).

4) Convert 1010 0010 to an unsigned decimal number.

5) Convert 1010 0010 to a signed decimal number (assume two’s complement).

6) Write a program that asks the user to input a number between 0 and 255. Print this number as an 8-bit binary number (of the form #### ####). Don’t use any bitwise operators.

Hint: Use method 2. Assume the largest power of 2 is 128.

Hint: Write a function to test whether your input number is greater than some power of 2. If so, print ‘1’ and return your number minus the power of 2.

**Quiz answers**

3.8 -- Bitwise operators |

Index |

3.6 -- Logical operators |

Good evening!

I have a couple questions - some are recurring and I’m finally getting around to ask them, and some are specific to the quiz in this lesson.

1: Quite often we are asking the user for a console input and assigning their input to a variable. Which of these should I use for this type of task? I’ve reviewed prior lessons discussing initialization and the concept must just not be clicking. I want to squash this point of confusion while I’m still early in learning.

1a. Copy Initialization

1b. Direct Initialization

1c. Uniform Initialization

2. At least twice now (once when writing the program to calculate a ball being dropped from a tower and now for this program to convert a decimal value to binary) the programs in the solutions to the quiz have had int main() call a function that performs two actions. I thought this was a hard “no-no” which really threw me for a loop (pun intended) when I was brainstorming my own solutions to the quizzes. For example, in this lesson’s quiz, the following was my original “outline” before I decided it wasn’t going to work, or that I didn’t know how to make it work:

Essentially I was going to try to send the input to the convert() function, write the binary number in decimal format (eg: convert 200 to 11001000), return it back to int main() where it would be sent to the print() function to be written in the console. After struggling with this for a while, I decided to peak and see if I was on the right track where I was again surprised to see functions with ‘and’ in them performing multiple actions (math and printing, in this example).

When is it ok to have a function that performs multiple tasks?

After being nudged in the proper direction, I came up with this:

3. If I'm writing a program such as the one above, is there really any need to use

since I'm not using that variable anywhere else? It's being used exactly once, when it is passed as an argument in the

function call. Is it just good habit?

4. Lastly, how do I get ‘n’ to properly copy in code brackets here on this forum? As seen in my code above, the backslashes aren’t displaying.

Hi Matt!

1:

Uniform initialization whenever possible. Enable all warnings in your compiler and you’ll get warnings when you try to initialize variables with types that would be casted.

2:

> When is it ok to have a function that performs multiple tasks?

Whenever the tasks are strongly related to each other and you don’t need the results of the called function outside of the wrapper.

The commenting system messed up your code (@Alex pls fix), from what I can tell it looks good, it could be improved with content covered in upcoming lessons.

> Is this std::endl; necessary here? What happens if our program finishes with a n?

Use std::endl whenever you don’t need high-speed bulk text output. std::endl will make sure your text gets printed right away while \n might wait a little longer.

3:

> is there really any need to use

No there’s not. const is generally only used for references and pointers (Covered in later lessons). You can use it for ‘normal’ variables like you did but other than having the compiler remind you that you’re trying to write to a variable you marked as read only there are no benefits.

4:

I’ll leave this one for @Alex, it appears to work for some people and for others it doesn’t (Especially when using the edit function).

Don’t hesitate to ask questions, that’s what the comment section is for.

Hi Alex,

Thank you for this wonderful tutorial series. You are very generous in sharing your time, energy and expertise. I would like to share my answer to question 6. I tried to incorporate as much as what I could from the previous sections. I also read ahead and incorporated a while loop. My program appears to work. I would appreciate any comments regarding good programming practice or areas where I could improve it. Thanks in advance.

in file conversion.h

#ifndef CONVERSION_H

#define CONVERSION_H

int getUserInput();

void convertDecimalToBinary(int, int, int *);

void printResult(int *);

#endif

in file conversion.cpp

/*

* This program asks the user to input an integer between 0 and 255, inclusive. The program then converts the number to base 2 and

* displays the result as an 8 bit binary number using the following format: #### ####

*/

#include <iostream>

#include "conversion.h"

int getUserInput()

{

std::cout << "Enter an integer between 0 and 255, inclusive: ";

int userInput;

std::cin >> userInput;

return userInput;

}

void convertDecimalToBinary(int numBase_10, int powerOf_2, int *ptr)

/*

* To convert to binary we will examine numBase_10 to see if it's greater than or equal to the given powerOf_2.

* If numBase_10 is greater than or equal to the given powerOf_2 then the corresponding bit must be a "1" else it is a "0."

* If numBase_10 is greater than or equal to the given powerOf_2 that we must subtract that power of 2 to get the value that must

* still be accounted for by the remaining bits; this will decrease numBase_10 until it is 0.

*/

{

while(powerOf_2 >= 1)

{

int bit = (numBase_10 >= powerOf_2) ? 1 : 0;

numBase_10 = (numBase_10 >= powerOf_2) ? (numBase_10 - powerOf_2) : numBase_10; // This will give the remaining value.

powerOf_2 = powerOf_2 / 2; // This will perform integer division and assign the next power of 2 to powerOf_2 (i.e. 128, 64, 32...

*ptr = bit;

++ptr;

}

}

void printResult(int *ptr)

{

int index{0}; // index is used to step through our array, which has 8 elements.

while(index < 8)

{

if(index != 4)

std::cout << *ptr;

else

std::cout << " " << *ptr;

++index;

++ptr;

}

std::cout << std::endl;

}

in file main.cpp

/*

* This program asks the user to input an integer between 0 and 255, inclusive. The program then converts the number to base 2 and

* displays the result as an 8 bit binary number using the following format: #### ####

*/

#include <iostream>

#include "conversion.h"

int main()

{

int numBase_10 = getUserInput();

int powerOf_2{0x80}; // 0x80 is 128 in hexadecimal, which is the largest "power of 2" in an 8-bit byte.

int *ptr, outputBits[8];

ptr = &outputBits[0];

convertDecimalToBinary(numBase_10, powerOf_2, ptr);

printResult(ptr);

return 0;

}

I think it works well but you may be overcomplicating this. I get it if you’re trying to do so for like learning purposes but really I think you should stick to getting programs to be as efficient as possible. I’ll try to provide input on what you have, and I’ll share the code I came up with so that it’s not just me telling you "Hey my way’s right and yours is wrong" because I’m in no position (or knowledge) to do so. Anyways,

1) For getUserInput(), you should at least verify that the number entered is in the desired range. There’s the extra (and recommended step) which I didn’t take either that is to verify that the input stream is not in a failed state, as well as to "remove" any of the trailing input after the integer that’s read in. Here’s how I would "replace" your getUserInput() function:

2) The conversion function looks good, just one line where you could’ve shortened it up a bit:

3) For the print function, I’d write it like this since the line to print the dereferenced ptr is shared on both (assuming you want to split it after the 4th bit, which I guess is your intent). This should make it more compact and a bit more readable (at least imo):

Other than that it looks good ! I am getting back into stuff too so there’s a chance that I didn’t pick on some other things that you could improve, or even worse, made mistakes myself. I tried to double check my answer as the most I could, so I apologize if I misled you in any way.

Good luck with future coding endeavors =)

Here’s the fixed version of how my code looked:

Thanks, Matias, for your comments and recommendations. I can definitely see the wisdom in checking to see if the input is in the desired range. I still have a lot more reading (and learning) to do. Your code seems very sophisticated!

Thanks again,

If there’s something I don’t have it’s definitely wisdom lol. I’ve been programming off and on for about 6 years now (I’m 20 now, and I started playing around with programming when I was 14). If there’s one solid piece of advice I can give you it’s this: take it easy, and enjoy what you do. Don’t try to push these lessons to be "over as fast as they can" because there’s a lot of good content in them. Also, try to find applications of each lesson and try to come up with problems to solve with code yourself, and do tiny programs that you can improve/grow in size as you progress across the tutorials. Also, don’t be afraid of checking back on sections, there’s no shame in revisiting a concept, specially in a language like c++!

Also for next time, if you want to put code, use "

" tags, or you can put a Pastebin/Ideone link to your source code.

Good luck !

In general, your coding style is great -- lots of comments, well named variables, correct use of multiple files and function prototypes in a header, etc… Good work! A few nitpicks:

1) getUserInput() should probably be in main.cpp, not conversion.cpp.

2) in main(), powerOf_2 should be const.

3) You don’t need the variable int *ptr. Just pass outputBits, and the fixed array will decay into a pointer:

Here’s my solution to the last problem, I think there’s a couple of concepts that are not familiar in the series so far, but perhaps someone can make good use of this.

Thanks for the tutorials, love them !

( I realize I could’ve added a print8BitBinary function but oh well, I guess I play into the lazy programmer stereotype :p)

There was a tiny error in the conversion method, where i was starting to evaluate at 2^8 instead of 2^7. oopsie.

Here’s the fixed version of the code above, with the print function.

Dear Alex, I think on "Converting integers to binary two’s complement" section, it can be better by showing examples from negative values to positive with step by step guide, as you just showed only positive to negative conversion. I ended up confused on "Converting binary two’s complement to integers" section, which converts negative to positive via positive to negative method. It would be easier to read in my opinion.

Thank you.

"pow" is much more powerful and efficient than eight "if/then" statements!! But I also learned some more about displaying current values in finding my ererz… DUH!!!

The main difference in my code was a test for out of range input. WOW! :blush:

Here is a way you can write the program, using a for loop:

Suggestion: this page could use a summary

Hey, why do you use 2 if statements here:

?

Isn’t that just inefficient and not so nice code?

looks much nicer to me.

Yes, I agree, but I hadn’t covered blocks at this point in the tutorials.

Ah! I knew something like this had to be possible. So far, every time I’ve thought to myself, "Self, there must be an easier way to do this thing I’m trying to do," I’ve been right. Then, a lesson or two later you show me what that easier way is. This tutorial series has been incredible.

This seems to work for me. What do you think about it? 🙂

Seems okay, but your code really could use some comments. Inside function binNum, what are b and p, and what is the for loop doing?