Hot answers tagged

1

I am unsure what you mean by "solving" this, but the meaning of the code is rather obvious and even more so in the screenshot you provided. Simplified version: call $+5 add [esp], 5 ret xor eax, eax ret And with annotations: _main: call $+5 ; call address of next instruction, placing return address on stack (esp) add [esp], 5 ; add 5 bytes to the ...


Only top voted, non community-wiki answers of a minimum length are eligible