re-construction of c code: movsx with test instruction

Question

I just started learning about reconstructing C code from assembler instructions.

I have the following piece of assembler code:

mov eax dword ptr [ebp+8]
add eax, dword ptr[ebp-4]
movsx ecx, byte ptr [eax]
test ecx, ecx
jne XXXXX
comp dword ptr [ebp-4], ffh
jle XXXX

I want to reconstruct the condition of these lines but I have some problems or things I am not sure about: I re-constructed the following:

Let,ebp+8= param1 and ebp-4 = i

if(param1[i]!=’\0’ || i<=0xff){
…
}

But I am wondering about this line:

movsx ecx, byte ptr [eax]

As I understand movsx it moves a signed value into a register and sign-extends it with 1. Thus, ecx shouble look something like 0xFFFFFF<eax>. Because only the lower bytes (byte ptr eax) of eax are moved and ecx is 1 extended. Where am I wrong?

Answer 1

You may want to familiarize yourself with two's complement notation. It will only extend the sign bit (msb) into the larger register. So, the msb will be extended into all the additional bits -- sxxxxxxx -> sssssssssssssssssssssssssxxxxxxx

For example, if the byte value pointed to by eax was 1 (0b00000001), ecx would be 1 (0b00000000000000000000000000000001). Similarly, if it was the most positive number possible in a two's complement byte, 127 (0b01111111), ecx would be 127 as well (0b00000000000000000000000001111111).

Where the sign extension is needed is to maintain the value for negative numbers. If the value were -1 (0b11111111), then if you just simply zero-extended into a 32-bit value you would get 255 (0b00000000000000000000000011111111) in ecx instead of -1 (0b11111111111111111111111111111111).