Finding missing C code, given assembly code?

Question

The code is

int f(int x, int y, int z) {
  if (/* missing code here */)
    return z;
  else
    return -z;
}

And the assembly is

    pushl %ebp
    movl %esp, %ebp
    movl 8(%ebp), %eax
    cmpl 12(%ebp), %eax
    jge .L2
    movl 16(%ebp), %eax
    jmp .L3
.L2:
    movl 16(%ebp), %eax
    negl %eax
.L3:
    popl %ebp
    ret

And the question asks for me to find what the missing test expression must be to yield the assembly code given. Okay, easy enough. There is an obvious comparison going on between x and y. The jge operator is going to preform the jump into the body of the loop if 12(%ebp) > %eax.

The possible choices are

x<=y x>=y x>y x<y

My answer was x<=y, since 12(%ebp) is a reference to y, and it is the destination. But this answer was wrong, and I do not see how. Any hints? Thank you so much.

Answer 1

Here is the annotated x86 assembly:

pushl %ebp ; save the old stack movl %esp, %ebp ; set up your local, new stack movl 8(%ebp), %eax ; take the first function argument and store it into eax cmpl 12(%ebp), %eax ; compare the 2nd function arg with the 1st (in eax)

After this, there's a jge which means, essentially, "jump if greater than or equal", which you can do after the cmp instruction.

This means that it jumps if the first argument is greater than the second argument, and thus, x >= y.

However, this jump (to L2) will actually negate z, and then return z. What you actually want is the jump to L3, which would happen if x < y, which should be the end result.