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

typedef unsigned long TYPE;

TYPE reverser(TYPE n)
{
    TYPE k = 1, nrev = 0, i, nrevbit1, nrevbit2;
    int count;

    for(i = 0; !i || (1 << i && (1 << i) != 1); i+=2)
    {
        /*In each iteration, we  swap one bit 
            on the 'right half' of the number with another 
            on the left half*/

        k = 1<>= count;

        nrev |= nrevbit1;
        nrev |= nrevbit2;
    }
    return nrev;
}

This works fine in gcc under Windows, but I'm not sure if it's completely platform independent. A few places of concern are:

• the condition in the for loop - it assumes that when you left shift 1 beyond the leftmost bit, you get either a 0 with the 1 'falling out' (what I'd expect and what good old Turbo C gives iirc), or the 1 circles around and you get a 1 (what seems to be gcc's behaviour).

• the condition in the inner while loop: see above. But there's a strange thing happening here: in this case, gcc seems to let the 1 fall out and not circle around!

The code might prove cryptic: if you're interested and need an explanation please don't hesitate to ask - I'll put it up someplace.