why it shows like this in hexadecimal notation?

Question

I'm trying to do bit arithmetic with variables a and b. When I reverse a, which has the value 0xAF, the result shows as 8 digits. Unlike the others that show as 2 digits.

I don't know why it happens, but guess that it is relevant to %x's showing way and little endian?

Here is my code:

#include <stdio.h>
int main()
{
    int a = 0xAF; // 10101111
    int b = 0xB5; // 10110101

    printf("%x \n", a & b); // a & b = 10100101
    printf("%x \n", a | b); // a | b = 10111111
    printf("%x \n", a ^ b); // a ^ b = 00011010
    printf("%x \n", ~a); // ~a = 1....1 01010000
    printf("%x \n", a << 2);// a << 2 = 1010111100 
    printf("%x \n", b >> 3); // b >> 3 = 00010110 

    return 0;
}

Answer 1

Considering your int a is most likely 32 bit in size, your a actually looks like this:

int a = 0xAF; // 0000 0000 0000 0000 0000 0000 1010 1111

So if you flip all the bits on that, you have

1111 1111 1111 1111 1111 1111 0101 0000

Or, in hex

0xFFFFFF50

Which is exactly what you're getting. The others show only 2 digits because trailing zeroes are omitted when printing hex, and your other bit operations do not in fact change any of the leading zeroes.

---- Credit to @ chqrlie for this ---- If you really only want to see 8 bits of the result, you can do

printf("%hhx \n", ~a); // ~a = 1....1 01010000 --> Output : 50

Which restricts the printed value to unsigned char (8 bit [on modern os, not guaranteed, but very likely for your purposes]) length.