Signed right shift = strange result?

Question

I was helping someone with their homework and ran into this strange issue. The problem is to write a function that reverses the order of bytes of a signed integer(That's how the function was specified anyway), and this is the solution I came up with:

int reverse(int x)
{
    int reversed = 0;

    reversed = (x & (0xFF << 24)) >> 24;
    reversed |= (x & (0xFF << 16)) >> 8;
    reversed |= (x & (0xFF << 8)) << 8;
    reversed |= (x & 0xFF) << 24;

    return reversed;
}

If you pass 0xFF000000 to this function, the first assignment will result in 0xFFFFFFFF. I don't really understand what is going on, but I know it has something to do with conversions back and forth between signed and unsigned, or something like that.

If I either append ul to 0xFF it works fine, which I assume is because it's forced to unsigned then converted to signed or something in that direction. The resulting code also changes; without the ul specifier it uses sar(shift arithmetic right), but as unsigned it uses shr as intended.

I would really appreciate it if someone could shed some light on this for me. I'm supposed to know this stuff, and I thought I did, but I'm really not sure what's going on here.

Thanks in advance!

Answer 1

From your results we can deduce that you are on a 32-bit machine.

(x & (0xFF << 24)) >> 24

In this expression 0xFF is an int, so 0xFF << 24 is also an int, as is x.

When you perform the bitwise & between two int, the result is also an int and in this case the value is 0xFF000000 which on a 32-bit machine means that the sign bit is set, so you have a negative number.

The result of performing a right-shift on an object of signed type with a negative value is implementation-defined. In your case, as sign-preserving arithmetic shift right is performed.

If you right-shift an unsigned type, then you would get the results that you were expecting for a byte reversal function. You could achieve this by making either operand of the bitwise & operand an unsigned type forcing conversion of both operands to the unsigned type. (This is true on any implementation where an signed int can't hold all the possible range of positive values of an unsigned int which is nearly all implementations.)

Answer 2

Since x is a signed quantity, the result of (x & (0xFF << 24)) is 0xFF000000 which is also signed and thus a negative number since the top (sign) bit is set. The >> operator on int (a signed value) performs sign extension (Edit: though this behaviour is undefined and implementation-specific) and propagates the sign bit value of 1 as the value is shifted to the right.

You should rewrite the function as follows to work exclusively on unsigned values:

unsigned reverse(unsigned x)
{
    unsigned int reversed = 0;

    reversed = (x & (0xFF << 24)) >> 24;
    reversed |= (x & (0xFF << 16)) >> 8;
    reversed |= (x & (0xFF << 8)) << 8;
    reversed |= (x & 0xFF) << 24;

    return reversed;
}