Why does C not push a pointer on the stack when calling a assembly function?

问题

I am currently trying to get some experience with calling assembly functions from C. Therefore, I created a little program which calculates the sum of all array elements.

The C Code looks like this:

#include <stdio.h>
#include <stdint.h>

extern int32_t arrsum(int32_t* arr,int32_t length);

int main()
{
    int32_t test[] = {1,2,3};
    int32_t length = 3;
    int32_t sum = arrsum(test,length);
    printf("Sum of arr: %d\n",sum);
    return 0;
}

And the assembly function looks like this:

.text
.global arrsum
arrsum:

    pushq %rbp
    movq %rsp, %rbp

    pushq %rdi
    pushq %rcx

    movq 24(%rbp),%rcx
    #movq 16(%rbp),%rdi

    xorq %rax,%rax

    start_loop:
    addl (%rdi),%eax
    addq $4,%rdi
    loop start_loop

    popq %rcx
    popq %rdi

    movq %rbp , %rsp
    popq %rbp
    ret

I assumed that C obeys the calling convention and pushes all arguments on the stack. And indeed, at position 24(%rbp) I am able to find the length of the array. I expected to find the pointer to the array at 16(%rbp), but instead I just found 0x0. After some debugging I found that C didn't push the pointer at all but instead moved the whole pointer into the %rdi register.

Why does this happen? I couldn't find any information about this behavior.

回答1:

The calling convention the C compiler will use depends on your system, metadata you pass to the compiler and flags. It sounds like your compiler is using the System V AMD64 calling convention detailed here: https://en.m.wikipedia.org/wiki/X86_calling_conventions (implying that you're using a Unix-like OS on a 64 bit x86 chip). Basically, in this convention most arguments go into registers because it's faster and the 64 bit x86 systems have enough registers to make this work (usually).

回答2:

I assumed that C obeys the calling convention and pushes all arguments on the stack.

There is no "the" calling convention. Passing arguments via the stack is only one possible calling convention (of many). This strategy is commonly used on 32-bit systems, but even there, it is not the only way that parameters are passed.

Most 64-bit calling conventions pass the first 4–6 arguments in registers, which is generally more efficient than passing them on the stack.

Exactly which calling convention is at play here is system-dependent; your question doesn't give much of a clue whether you're using Windows or *nix, but I'm guessing that you're using *nix since the parameter is being passed in the rdi register. In that case, the compiler would be following the System V AMD64 ABI.

In the System V AMD64 calling convention, the first six integer-sized arguments (which can also be pointers) are passed in the registers RDI, RSI, RDX, RCX, R8, and R9, in that order. Each register is dedicated to a parameter, thus parameter 1 always goes into RDI, parameter 2 always goes into RSI, and so on. Floating-point parameters are instead passed via the vector registers, XMM0-XMM7. Additional parameters are passed on the stack in reverse order.

More information about this and other common calling conventions is available in the x86 tag wiki.

来源：https://stackoverflow.com/questions/41535864/why-does-c-not-push-a-pointer-on-the-stack-when-calling-a-assembly-function

标签

pointers

assembly

x86

x86-64