What are the results of this sbb instruction?

Question

I'm doing a crackme to learn some reversing, and I stumbled upon this code generated by C++ MFC:

sbb eax, eax
sbb eax, -1
test eax, eax
jz exit

Before that code a comparison is done, such as cmp al, bl where al and bl hold some value read from the serial

The thing that confused me, is I figured that the cmp and two sbb instructions are equivalent to this pseudocode:

cmp a,b
eax=-1 if b>a
eax=1 otherwise

However this confused me because eax can never be 0, so the zero flag will always be set. Therefore, I figure the chunk of code

test eax, eax
jz exit

is useless because it does nothing- but how can this be? I don't think their C++ compiler would generate useless code like that

Where am I wrong here?

Answer 1

are you sure it was compiler generated ?
may be it was hand crafted or deliberately coded like wise may be it checks the sign flag further down in the jz path may be red herring ?

well whatever it requires more info based on the info provided only thing that pops up is a check for sign flag further down the path

here is a small compilable c code that shows what flags and results are for the operations in the query

#include <stdio.h>
int helper (unsigned char a , unsigned char b )
{
    unsigned char res = 0;
    unsigned char flag = 0;
    __asm
    {
        xor eax,eax
        xor ecx,ecx
        mov al,a        ;          = 0  1  2  3  4  5  6  7
        mov cl,b        ;          = 4  4  4  4  4  4  4  4
        cmp al,cl       ;cf        = 1  1  1  1  0  0  0  0
        sbb eax,eax     ;x-x-cf    =-1 -1 -1 -1  0  0  0  0
        sbb eax,-1      ;x-(-1)-cf =-1 -1 -1 -1  1  1  1  1
        mov res,al      ;          = ---------""-----------
        pushfd
        xor eax,eax
        mov eax,dword ptr ss:[esp]
        popfd
        lahf
        mov flag , al
    }
    bool zf = ((flag & 64)==64);
    bool sf = ((flag & 128)==128);
    bool cf = ((flag & 1)==1);
    printf("%2x %2x %2x %2x %2x %2x\n" ,a,b,res,zf,sf,cf);
}
int main (void)
{
    printf(" a  b  r  z  s  c\n");
    for (unsigned char i = 0; i < 8 ;i++){
        helper(i,4);
    }
    return 0;
}

result

:\>dir /b & cl /nologo runasm.cpp & runasm.exe
runasm.cpp
runasm.cpp
 a  b  r  z  s  c
 0  4 ff  0  1  1
 1  4 ff  0  1  1
 2  4 ff  0  1  1
 3  4 ff  0  1  1
 4  4  1  0  0  1
 5  4  1  0  0  1
 6  4  1  0  0  1
 7  4  1  0  0  1