What's up with this reversing bit order function?

Question

I'm rather ashamed to admit that I don't know as much about bits and bit manipulation as I probably should. I tried to fix that this weekend by writing some 'reverse the order of bits' and 'count the ON bits' functions. I took an example from here but when I implemented it as below, I found I had to be looping while < 29. If I loop while < 32 (as in the example) Then when I try to print the integer (using a printBits function i've written) I seem to be missing the first 3 bits. This makes no sense to me, can someone help me out?

Thanks for everyone's help, I've added comments to show changes I've made.

int reverse(int n)
{
    int r = 0;
    int i = 0;
    for(i = 0; i < 29; i++) //Should be i < 32
    {
        r = (r << 1) + (n & 1); //| instead of + to make it obvious I'm handling bits
        n >>=1;
    }

    return r;
}

Here is my printBits function:

void printBits(int n)
{
    int mask = 0X10000000; //unsigned int mask = 0X80000000;
    while (mask)
    {
        if (mask & n)
        {
            printf("1");
        }
        else
        {
            printf("0");
        }
        mask >>= 1;
    }
    printf("\n");
}

And a working? reverse function

int reverse2(int n)
{
    int r = n;
    int s = sizeof(n) * 7; // int s = (sizeof(n) * 8) -1

    for (n >>= 1; n; n >>=1)
    {
        r <<=1;
        r |= n & 1;
        s--;


    r <<= s;
    return r;
}

Answer 1

Instead of +, you should use | (bitwise or). And you should use < 32.

Answer 2

As written, this will reverse the lower 29 bits of n into r. The top three bits of n will be left in n (shifted down 29 bits) and not returned.

I would suspect a problem with your printBits function if you see something else.

edit

Your printBits function prints the lower 29 bits of n, so it all makes sense.

Answer 3

Print Bits is wrong, its 0x80000000 not 0x10000000.

>>> bin (0x80000000)  
'0b10000000000000000000000000000000'  
>>> bin (0x10000000)  
'0b10000000000000000000000000000'

See 0x1... doesnt set the highest bit.

Answer 4

int mask = 0X10000000;

puts a 1 in bit 28. You want 0X80000000.

Answer 5

You have:

int mask = 0x10000000;

There are two problems here. You don't have the high bit set, and if you did, it still (probably) wouldn't work, as your compiler would be using arithmetic shift on a signed int.

You want to change your mask to:

unsigned int mask = 0x80000000;

For arithmetic shift, shifting 0x80000000 right will never become zero, as the sign bit will be magically extended into the other bits. See here for more details on arithmetic shift.