[spc476]

Odd. The code:

    int main(int argc,char *argv[])
    {
      int x = strtol(argv[1],NULL,10);
      printf("result is %d\n",x/16);
      return 0;
    }

The resulting assembly contained no division instructions, nor a call to a divide routine, but there is one instruction that does an arithmetic right shift by 4. Changing x to unsigned changed the instruction to a logical right shift by 4.

I think it really depends upon the compiler.

[mierle]

Strangely, I can't reproduce the effect with a divide by 16. With GCC 4.5.2 -O2, I see a "shrl $4, %eax" for unsigned, and "sarl $4, %eax" for signed. However, if I divide by 2, the results are different; the function:

  int div_by_2(int x) {return x/2;}

compiles to:

  div_by_2:
    movl	%edi, %eax
    shrl	$31, %eax
    addl	%edi, %eax
    sarl	%eax
    ret

Meanwhile, the signed variant:

  unsigned int div_by_2(unsigned int x) {return x/2;}

compiles to:

  div_by_2:
    movl	%edi, %eax
    shrl	%eax
    ret

My conclusion remains that for hot loops, you should check the compilers assembly to make sure the compiler is applying the strength reduction you expect.

[to3m]

I got the following (-O6 -fstrength-reduce). A single arithmetic shift is no good, so the code adjusts the value first.

    movl	%eax, %esi
    sarl	$31, %esi  ; ESI = FFFFFFFF (-) / 00000000 (+)
    shrl	$28, %esi  ; ESI = 0000000F (-) / 00000000 (+)
    addl	%eax, %esi ; offset to cancel invalid values
    sarl	$4, %esi   ; shift

It first generates an offset value in ESI: 15 if the value to divide was negative, or 0 if it was positive.

Then it adds this to the original value. This is the cunning part - well I think so anyway. It ensures that for the values where x>>4 would give the wrong value (i.e., -15<=x<0), x is made positive and smaller than 16, so that x>>4 gives the correct value of zero. For all other values, the bottom 4 bits don't affect the result of the division, so it's fine to offset them as well.

(Don't think I've ever seen VC++ generate anything like this, but I'd pretty much always have an explicit shift rather than a divide by 16 so maybe it does and I'd just never noticed. The choice of 16 usually comes from having some 4-bit field somewhere, and when dealing with bitfields you're best off using bitwise instructions.)