Signed right shift: which compiler use logical shift

Question

I tested right shift with Visual Studio, Ubuntu's GCC, Intel compiler, MinGW. All shift in the sign bit. I guess Xcode's GCC does the same.

I know that the behavior is implementation specific, but it looks like that all major desktop/server compilers implement arithmetic shift. Are there any widely used compiler that doesn't shift in the sign bit?

Thank you.

Answer 1

C runs on a lot of different architectures. I mean a lot of different architectures. You can get C code running on an embedded DSP and on a Cray supercomputer.

Most of the "implementation-defined" parts of the C standard that people take for granted really only do break on obscure architectures. For example, there are DSPs and Cray supercomputers where CHAR_BIT is something huge like 32 or 64. So if you try out your code on an x86, and maybe if you're generous a PowerPC, ARM, or SPARC, you're not likely to run into any of the really weird cases. And that's okay. Most code these days will always run on a byte-oriented architecture with twos-complement integers and arithmetic shifts. I have no doubt that any new CPU architectures in the foreseeable future will be the same.

But let's look at the two most common representations for integers: twos-complement and ones complement:

switch ((-1) >> 1) {
case 0:
case -0:
    puts("Hello, one's complement world!");
    // Possibly sign-magnitude.
    break;
case -1:
    puts("Hello, two's complement world!");
    break;
default:
    puts("Hello, computer without arithmetic shift");
    break;
}

Don't sweat it. Just stick to / when you want to divide, and >> when you need to shift. Even bad compilers are good at optimizing these operations. (And remember that x/2 != x>>1 if x is negative, unless you're on a one's complement machine, which is almost certainly not true.)

The standard does guarantee that if (int) x is not negative, then (int) x >> n == (unsigned) x >> n , so there is not a lot of room for a compiler to do something completely unexpected.

Answer 2

Cray C compiler does logical right shift on signed values by default, but there's an option to do arithmetic shift instead.

Generally, it's safe to assume that signed right shift is arithmetic.

Answer 3

Generally it depends more on the target architecture that the compiler used. If the arch has both arithmetic (signed) and logical (unsigned) shift instructions, then C compilers for that arch will use whichever is appropriate. On the other hand, if it has only logical shifts, the C compiler will just use that, even though it doesn't 'do the right thing' for negative values, as the C spec allows the compiler to do anything.

Answer 4

As far as I can tell, the >> operator does arithmetic shift. There is however a difference between how the shift is performed for signed and unsigned integers - signed will extend the MSB (which usually is a sign bit) and unsigned will not (they are always non-negative so sign bit is always zero).

EDIT: apply "usually" to everything I wrote above ;).