分析生成的汇编代码以操作命令行参数

Question

#include <stdio.h>
int main(int argc, char * argv[])
{
 argv[1][2] = 'A';
 return 0;
}

Here is the corresponding assembly code from GCC for a 32-bit Intel architecture. I can't totally understand what is going on.

main:
        leal    4(%esp), %ecx  - Add 4 to esp and store the address in ecx
        andl    $-16, %esp  - Store first 28 bits from esp's address into esp??
        pushl   -4(%ecx)  - Push the old esp on stack
        pushl   %ebp         - Preamble
        movl    %esp, %ebp
        pushl   %ecx          - push old esp + 4 on stack
        movl    4(%ecx), %eax   - move ecx + 4 to eax. this is the address of argv. argc stored at (%ecx).
        addl    $4, %eax - argv[1]
        movl    (%eax), %eax - argv[1][0]
        addl    $2, %eax  - argv[1][2]
        movb    $65, (%eax) - move 'A'
        movl    $0, %eax - move return value (0)
        popl    %ecx - get old value of ecx
        leave
        leal    -4(%ecx), %esp  - restore esp
        ret

What is going on in the beginning of the code before the preamble? Where is argv store according to the following code? On the stack?

Answer 1

The funny code (the first two lines) that you are seeing is the alignment of the stack to 16 bytes ( -16 is the same as ~15 , and x & ~15 rounds x to a multiple of 16).

argv would be stored at ESP + 8 when entering the function, what leal 4(%esp), %ecx does is create a pointer to a pseudo-struct containing argc and argv , then it proceeds to access them from there. movl 4(%ecx), %eax access argv from this pseudo-struct.

Answer 2

argv是“main（）”的参数，因此在许多ABI中，它确实会在堆栈上传递。