What is the size of each asm instruction? Every instruction takes how many bytes? 8 bytes? Four for the opcode and Four for the argument? What happens when you have one opcode and 2 arguments, in mov, for example? Do they have a fixed size in memory or do they vary? Does EIP have anything to do with this, of its value is always incremented by one, being totally independent of what kind of instruction it is passing by?
I ask this as when I was reading http://en.wikibooks.org/wiki/X86_Disassembly/Functions_and_Stack_Frames , I stumbled across the fact that it seems a call instruction is equivalent to a push and jmp instruction.
call MYFUNCTION
mov my_var, eax
being the same as
push [eip + 2];
jmp MYFUNCTION;
mov my_var, eax
When we’re pushing [eip + 2] on the stack, to what value are we pointing then? To the line right next to “jmp MYFUNCTION”, mov my_var eax, right?
ps: MSVC++ flags an error on the first line, as it says eip is undefined. It works for eax, esp, ebp, etc. What am I doing wrong?
The size of a machine instruction depends on the processor architecture – there are architectures with fixed size instruction, but you are obviously refering to the IA-32 and Intel 64 and they have strongly varing instruction lengths. The instruction pointer is of course always incremented by the length of the processed instruction.
You can download the IA-32 and Intel 64 manuals from Intel – they contain almost everything you can know about the architecture. You can find an opcode map and instruction set format in
Intel® 64 and IA-32 Architectures Software Developer’s Manual Volume 2B: Instruction Set Reference, N-Z on pages 623 to 768.