Why are the bits reversed?

Question

In the code bellow, I am trying to store a 2 byte int in 2 char s. Then, I try to show on the screen the number I stored. Those parts work pretty well.

If I try to see the binary form of the number I stored on the other hand I don't get what I expect. 256 gives me 00000000 1 while the correct is 10000000 0

unsigned int num = 256;
unsigned int pos = 0;

unsigned char a[2] = {num << pos, ((num << pos) & 0xFF00) >> 8};
//we store the number in 2 bytes

cout << (((unsigned int)a[0] + ((unsigned int)a[1] << 8)) >> pos) << endl;
//we check if the number we stored is the num

for(int i = 0; i < 2; i++)//now we display the binary version of the number
{
    for(int j = 0; j < 8; j++)
        cout << ((a[i] >> j)&1);
    cout << " ";
}

Can someone please explain what I am doing wrong?

Answer 1

Change:

cout << ((a[i] >> j)&1);

To:

cout << ((a[i] >> (8-j-1))&1);

Or change:

for(int j = 0; j < 8; j++)

To:

for(int j = 8-1; j >= 0; j--)

Answer 2

unsigned char a[2] = {num << pos, ((num << pos) & 0xFF00) >> 8};

means that you store the low bits in a[0] and high bits in a[1].

256 = 1 00000000
a[0] = 00000000
a[1] = 00000001

And for(int j = 0; j < 8; j++) ( least significant bits first )should be for(int j = 7; j >=0; j--) ( most significant bits first)

Answer 3

Numbers are stored in memory with least significant bits first.

Think of it this way: the number 456 has 6 in the 10^0 place, 5 in 10^1 place, etc. It makes some sense to store it as "654" (where the i'th character of the string corresponds to the 10^i'th digit).

The computer stores numbers the same way: the first bit of a number represents the 2^0th place, and the i'th bit to the 2^i'th place.

This rule is actually half-broken with Big Endian, which is a way of storing hexadecimal digits in memory from biggest to smallest (human-readable way).