x86 - Divide Signed Integer By 2 compiles to complex assembly output, not just a shift

Question

Consider this C function;

int triangle(int width, int height)
{
    return (width * height) / 2;
}

When compiled with gcc (gcc -m32 -g -c test.c) produces following assembly (objdump -d -M intel -S test.o).

test.o:     file format elf32-i386


Disassembly of section .text:

00000000 <triangle>:

int triangle(int width, int height)
{
   0:   55                      push   ebp
   1:   89 e5                   mov    ebp,esp
    return (width * height) / 2;
   3:   8b 45 08                mov    eax,DWORD PTR [ebp+0x8]
   6:   0f af 45 0c             imul   eax,DWORD PTR [ebp+0xc]
   a:   89 c2                   mov    edx,eax
   c:   c1 ea 1f                shr    edx,0x1f
   f:   01 d0                   add    eax,edx
  11:   d1 f8                   sar    eax,1
  13:   5d                      pop    ebp
  14:   c3                      ret    

I already know that shifting an integer n bit to right divides it by 2^n. However, according to above output, signed integers seems to be treated differently (which, of course, make sense). If I am reading assembly output correctly, sign bit of the integer added to itself before it is shifted.

What is the purpose of adding sign bit of the integer to itself before right shifting?

Answer 1

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