x86 assembly abs() implementation?

Question

I need to get the difference of 2 signed integers. Is there an ABS() function in x86 assembly language so I can do this. Any help would be greatly appreciated.

Answer 1

This is how the C library function abs() does it in assembly without branching:

   abs(x) = (x XOR y) - y 

where y = x >>> 31 (assuming 32-bit input), and >>> is arithmetic right shift operator.

Explanation of the above formula: We want to generate 2's complement of negative x only.

y = 0xFFFFFFFF, if x is negative     0x00000000, if x is positive 

So when x is positive x XOR 0x00000000 is equal to x . And when x is negative x XOR 0xFFFFFFFF is equal to 1's complement of x. Now we just need to add 1 to get its 2's complement which is what expression -y is doing . Because 0xFFFFFFFF is -1 in decimal.

Let's look at assembly generated for following code by gcc (4.6.3 on my machine):

C code:

main() {   int x;   int output = abs(x); } 

gcc 4.6.3 generated assembly snippet (AT&T syntax), with my comments:

  movl  -8(%rbp), %eax    # -8(%rbp) is memory for x on stack   sarl  $31, %eax         #  shift arithmetic right: x >>> 31, eax now represents y   movl  %eax, %edx        #     xorl  -8(%rbp), %edx    #  %edx = x XOR y   movl  %edx, -4(%rbp)    # -4(%rbp) is memory for output on stack   subl  %eax, -4(%rbp)    # (x XOR y) - y 

BONUS (from Hacker's Delight): If you have a fast multiply by +1 and -1, the following will give you abs(x):

      ((x >>> 30) | 1) * x 

Answer 2

Old thread but if I surfed in here late you might have too... abs is a brilliant example so this should be here.

; abs(eax), with no branches. ; intel syntax (dest, src)  mov ebx, eax ;store eax in ebx neg eax cmovl eax, ebx ;if eax is now negative, restore its saved value