Can someone convert this assembly to C - c

Question

I have this class assignment that i can't seem to figure out.
the point is to convert this assembly to C (Code assembled with GNU assembler AT&T syntax):
.section .rdata,"dr"
LC0:
.ascii "%d\12\0"
.text
.globl _main
_main:
pushl %ebp
movl %esp, %ebp
andl $-16, %esp
subl $32, %esp
movl 12(%ebp), %eax
addl $4, %eax
movl (%eax), %eax
movl %eax, (%esp)
call _atoi
movl %eax, 24(%esp)
cmpl $4, 24(%esp)
je L2
cmpl $6, 24(%esp)
jle L3
cmpl $9, 24(%esp)
jg L3
L2:
movl 24(%esp), %eax
addl $20, %eax
movl %eax, 28(%esp)
jmp L4
L3:
cmpl $0, 24(%esp)
jne L5
movl $44, 28(%esp)
jmp L4
L5:
cmpl $-1, 24(%esp)
jne L6
movl $-44, 28(%esp)
jmp L4
L6:
movl $99, 28(%esp)
L4:
movl 28(%esp), %eax
movl %eax, 4(%esp)
movl $LC0, (%esp)
call _printf
movl $0, %eax
leave
ret
however I always get lost on L2. it seems that no matter if we get into the first if statement or not we will always execute the label L2, which doesn't make any sense. I tried to make sense of it all and the closest I got was this C code:
#include <stdio.h>
int main(int argc, char *argv[]){
int y = 0;
int x = atoi(*++argv);
if (x != 4){
if (x > 6 && x <= 9){
y = 20 + x;
}
else if (x == 0){
y = 44;
}
else if (x == -1){
y = -44;
}
else {
y = 99;
}
}
else{
y = 20 + x;
}
printf("%d %d", y, x);
}
can someone please help with this confusing issue i'm having,
thanks

Answer 1

In decoding compiler output it often helps to rewrite the assembly code in a denser, but still very low-level form (pseudo code), using only elementary, low-level transformations. The point is to group/combine small numbers of instructions in a way that is difficult to get wrong but that exposes the inner logic of the code fragment better. The next step is to eliminate redundant stores and temporaries (like EAX used for storing x + 20 to y). I'm skipping the first step here but it can be dangerous to do that in more complicated code.
The sequence of conditionals then becomes:
x equ [esp + 24]
y equ [esp + 28]
if (x == 4) goto L2;
if (x <= 6) goto L3;
if (x > 9) goto L3;
L2: y = x + 20; goto L4;
L3: if (x != 0) goto L5;
y = 44; goto L4;
L5: if (x != -1) goto L6;
y = -44; goto L4;
L6: y = 99;
L4: printf("%d\f", y);
The first three conditionals form a conspicuous pattern employed by compilers to evaluate complex conditions. The compiler inverted the second and third terms of that conditional to use its 'jump around' solution fragment; inverting again allows you to code the original condition (jumps to L2/'then' are ORs, jumps to L3/'else' are AND NOTs):
if (x == 4 || !(x <= 6) && !(x > 9))
->
if (x == 4 || (x > 6) && (x <= 9))
and Bob's your uncle. The other conditionals could have been the result of a chained if or of a switch statement, it's difficult to tell. But that hardly matters. Hence your decompilation was almost perfect already, you just missed a tiny beat. In an intermediate step the C-ified conditionals look like this:
if (x == 4 || x > 6 && x <= 9)
{
y = x + 20;
}
else // L3
{
if (x == 0)
{
y = 44;
}
else // L5
{
if (x == -1)
{
y = -44;
}
else // L6
{
y = 99;
}
}
}
This can then be tightened to:
if (x == 4 || x > 6 && x <= 9)
{
y = x + 20;
}
else if (x == 0)
{
y = 44;
}
else if (x == -1)
{
y = -44;
}
else
{
y = 99;
}
P.S.: the value of (argv + 1) is not stored back to argv, just dereferenced. Hence it's atoi(*(argv + 1)) or atoi(argv[1]).