J'ai écrit un article sur les opérations bit à bit

Reccement, j’ai écrit un article (en anglais) sur les opérations bit à bit, que j’ai posté sur mon compte GitHub en tant que dépôt (que vous pouvez trouver ici). Je voulais également poster le texte ici. J’espère que vous l’apprécierez !

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Bitwise Operations in C

Introduction

The goal of this repository is to acquaint you with bitwise operations, explaining what they are, how they work, and what they can be used for.

Chapter 1: It’s All Binary

In C (and most other high-level languages), our variables have types. These types are indicative of a few things. Of course, a variable of type int will store an integer value, but the key to understanding these bitwise operations is to know that under the hood, all types are stored in memory (anywhere, stack, heap, wherever) as binary. Here’s an example of what happens when you store a simple integer value on the stack in C:

int main(int argc, char** argv) {
    int x = 2;
    return 0;
}

After compiling to assembly, the code might look like this (I’m using ARM assembly here, and I’ve annotated the code using comments):

.section .text
.global main

main:
    ; Set up the stack for the function
    stp x29, x30 [sp, -16]! ; Save previous frame pointer & link register
    mov x29, sp ; Setup a new frame pointer
    
    ; Initialize x with 2
    ; IN C: int x = 2;
    mov w0, 2 ; Move 2 into the w0 register
    str w0, [sp, 16] ; Store the contents of w0 (2) at a 16-byte offset from the stack pointer
    ; Essentially, the above line stores 2 on the stack.
    
    mov w0, 0 ; Move 0 into w0, prepare for return
    
    ; Clear stack
    ldp x29, x30, [sp], 32 ; Restore frame pointer and link register
    ret ; Return

Note that most compilers would not actually store a variable like the one I showed on the stack, as it is unused. However, if it is used multiple times, it would be stored on the stack something like the above.

If we looked at the location where our variable was stored on the stack (while it is there, of course), we would see something like:

Memory Address	Value Stored (Hex)	Value Stored (Binary)
0x1000	0x02	00000010
0x1001	0x00	00000000
0x1002	0x00	00000000
0x1003	0x00	00000000

This assumes that your system is little-endian. I won’t go into endianness here, but you can read more about it here.

The key thing I’d like you to notice about the table above is that even though our integer is only 2 bits long, it takes up 4 bytes (32 bits) of memory. Now, don’t freak out—this is normal and expected. One of the many things that C and your compiler do is set standards for the types you invoke. So when I create an int variable, the compiler knows to allocate 4 bytes (again, 32 bits) of memory. We can even test this using the sizeof() operator in C.

The sizeof() Operator

sizeof() is not an actual C function. Instead, at compile time, the compiler replaces the expression with the size of the given data type. You can even use this with your own types, like typedefs and/or structs:

#include <stdio.h> 

typedef struct {
  char name[64];
  int age;
} Person;

int main(int argc, char** argv) {
  printf("A Person is %lu bytes long.\n", sizeof(Person));
  return 0;
}

One other thing you might be asking is how negative numbers are stored. Excellent question. Numbers can be signed or unsigned, but by default, they’re signed. If an integer is signed, it sacrifices its most significant bit to be the “ sign bit.” If the sign bit is 1, the number is negative; otherwise it’s positive. An astute reader might realize that the change that happens here is in the range of possible numbers. Consider 8-bit numbers. There are 256 possible numbers to represent (given by 2^8). With an unsigned 8-bit integer, the values 0–255 can be represented; with a signed 8-bit int, -128–127 can be represented.

To get the inverse of a binary number, use two’s compliment. Let’s find -5 in binary.

1. Start with 5. In binary, 5 is 0101. The leading 0 is the sign.
2. Invert each bit. 0101 → 1010.
3. Add 1 to this number (ignoring any possible overflow). 1010 + 0001 = 1011.

Your Turn!

1. Confirm that -5 is 1011 in binary by performing two’s compliment on it to get 5, or 0101.
2. Write a C program that prints the size of an int in both bytes and bits. Use the code above as a starting point. Hint: to convert from bytes to bits, how many bits are in a byte?
3. Fill in the following table with sizes of different types, modifying your program to check them.

Type	Size (bytes)	Size (bits)
int
int64_t
int8_t
char
bool (you’ll need to #include <stdbool.h>)
long int
short int
long long int
double
double

Les reponses aux questions ci-dessus et l’article complete peuvent être trouvées dans le dépôt GitHub ici.