CODEWITHSUNDEEP
functionvariablesassemblyx86
user1502256
user1502256

Reputation:

Modifying given variables in ASM

This is the ASM code:

__declspec(naked) void foo(something* t)
{
__asm {
push    ebp
mov ebp, esp
mov eax, dword ptr [t]
mov dword ptr [eax], 0
mov dword ptr [eax+4], 0
mov esp, ebp
pop ebp
}

This would be the C version of the code:

struct something {
_int64 a;
_int64 b;
_int64 c;
};

void foo(struct* something) {
something->a = 0;
}

Now, I am wonder if I could do the same thing without storing t in eax. And just use ebp instead. But I am not sure where "a" would be (ebp+28 or ebp), and if it is even possible. This doesn't seem to work. Does anyone if this is possible, and how?

mov     dword ptr [ebp+28], 0
mov     dword ptr [ebp+24], 0

Upvotes: 0

Views: 46

Answers (1)

Seva Alekseyev
Seva Alekseyev

Reputation: 61351

Arbitrary nesting of expressions is not possible in assembly. That's what high level languages were invented for. In other words, yes, you have to load the value of t into a register if you want to dereference it. Assembly does not support constructs like 

mov     dword ptr [[ebp+28]], 0

which is what you're aiming for. ebp+28 is not the address of t->a; it's the address of t, which is the address of t->a.

Also, the assembly snippet zeros out both t->a and t->b while the C one only does a. They're not equivalent.

Upvotes: 2

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