Bit reversal of an integer, ignoring integer size and endianness

Question

Given an integer typedef:

typedef unsigned int TYPE;

or

typedef unsigned long TYPE;

I have the following

Answer 1

@ΤΖΩΤΖΙΟΥ

In reply to ΤΖΩΤΖΙΟΥ 's comments, I present modified version of above which depends on a upper limit for bit width.

#include 
#include 
typedef int32_t TYPE;
TYPE reverse(TYPE x, int bits)
{
    TYPE m=~0;
    switch(bits)
    {
        case 64:
            x = (x&0xFFFFFFFF00000000&m)>>16 | (x&0x00000000FFFFFFFF&m)<<16;
        case 32:
            x = (x&0xFFFF0000FFFF0000&m)>>16 | (x&0x0000FFFF0000FFFF&m)<<16;
        case 16:
            x = (x&0xFF00FF00FF00FF00&m)>>8 | (x&0x00FF00FF00FF00FF&m)<<8;
        case 8:
            x = (x&0xF0F0F0F0F0F0F0F0&m)>>4 | (x&0x0F0F0F0F0F0F0F0F&m)<<4;
            x = (x&0xCCCCCCCCCCCCCCCC&m)>>2 | (x&0x3333333333333333&m)<<2;
            x = (x&0xAAAAAAAAAAAAAAAA&m)>>1 | (x&0x5555555555555555&m)<<1;
    }
    return x;
}

int main(int argc, char**argv)
{
    TYPE x;
    TYPE b = (TYPE)-1;
    int bits;
    if ( argc != 2 ) 
    {
        printf("Usage: %s hexadecimal\n", argv[0]);
        return 1;
    }
    for(bits=1;b;b<<=1,bits++);
    --bits;
    printf("TYPE has %d bits\n", bits);
    sscanf(argv[1],"%x", &x);

    printf("0x%x\n",reverse(x, bits));
    return 0;
}

Notes:

• gcc will warn on the 64bit constants
• the printfs will generate warnings too
• If you need more than 64bit, the code should be simple enough to extend

I apologise in advance for the coding crimes I committed above - mercy good sir!