将IA32汇编转换为C代码

Question

I'm having some difficulty translating IA32 Assembly code back to its C code counterpart. I'm 99% of the way there, but the byte offsets and register storage are confusing me.

The assembly code in question, movl %edx, %eax , seems to set the value stored in %eax equal to the value in %edx , but wouldn't that mean sub = result ?

I'm new to this, so your guidance is appreciated!

int d(int x, int y, int z)     // x at %ebp+8, y at %ebp+12, z at %ebp+16
{
    int sub, result;

    sub = z - y;               // movl 12(%ebp), %edx
    result = sub;              // subl 16(%ebp), %edx

    ???????????                // movl %edx, %eax

    result <<= 31;             // sall $31, %eax
    result >>= 31;             // sarl $31, %eax

    result = sub * result;     // imull 8(%ebp), %edx

    sub ^= x;                  // xorl %edx, %eax

    return result;
}

Answer 1

The first two lines of asm are actually the first line of C but reversed and performed in two parts:

sub = y;                   // movl 12(%ebp), %edx
sub -= z;                  // subl 16(%ebp), %edx

You seem to have slight trouble with the fact that at&t syntax (that this is) puts the destination operand on the right. As such the movl %edx, %eax is indeed the result = sub as written in the code. Also, the imull 8(%ebp), %edx clearly writes into edx so that's sub = x * result (the eax operand is implicit). Finally xorl %edx, %eax is of course result ^= sub . x , which is 8(%ebp) , is not even mentioned on that line.