CODEWITHSUNDEEP
assemblyx86
Sebastian Hoffmann
Sebastian Hoffmann

Reputation: 11502

Add 0xFFFFFFFF in x86 assembly

I'm currently reversing disassembly and stumbled upon a chain of instructions I don't understand:

Given an object pointer in esi.

.text:00C20263                 cmp     dword ptr [esi+80h], 0
.text:00C2026A                 jnz     short loc_C2027D

As you can see if the member +0x80 is not 0 (the member is an integer) the code jumps to 00C2027D:

.text:00C2027D                 add     dword ptr [esi+80h], 0FFFFFFFFh
.text:00C20284                 jnz     short loc_C20291

These two instructions are those I don't really understand. First of all, the member is incremented by 0xFFFFFFFF; but since the member is not 0, wouldn't this exceeds the max value of an 32-bit integer? And when does the jnz instruction jumps?

Could one maybe point out what the purpose of these two instructions is?

Upvotes: 5

Views: 2163

Answers (1)

500 - Internal Server Error
500 - Internal Server Error

Reputation: 28839

For a signed variable, 0FFFFFFFFh is the same as -1, so this is subtracting one from the value and checking if that made it zero. Compilers will often emit "add negative value" rather than a sub instruction, presumably because it allows for reuse of compiler logic for both addition and subtraction.

Upvotes: 16

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