从C语言中的函数调用过程理解计算机执行原理

本文是Mooc 课程第一次作业.
姓名:石维康
转载请注明出处.
经过简单的数字修改,需要编译的C语言代码如下:
int g(int x)
{
return x + 2;
}

int f(int x)
{
return g(x);
}

int main(void)
{
return f(5) + 4;
}
编译后生成的汇编代码如下所示:

Screen Shot 2016-02-27 at 8.02.50 PM.png

生成的完整汇编代码如下:

g:
  pushl   %ebp
  movl    %esp, %ebp
  movl    8(%ebp), %eax
  addl    $2, %eax
  popl    %ebp
  ret
f:
  pushl   %ebp
  movl    %esp, %ebp
  subl    $4, %esp
  movl    8(%ebp), %eax
  movl    %eax, (%esp)
  call    g
  leave
  ret
main:
  pushl   %ebp
  movl    %esp, %ebp
  subl    $4, %esp
  movl    $5, (%esp)
  call    f
  addl    $4, %eax
  leave
  ret

完整截图(对应汇编代码函数):

汇编代码

与视频中采用相同约定,假设初始esp= ebp =0.
先进入main函数执行到call f后,esp=3,ebp=1,堆栈目前的情况
              编号   内容
              | --- | --- |
ebp           |  1 |   0  |
              | --- | --- |
              | --- | --- |
              |  2  |  5  |
              | --- | --- |
              | --- | --- |
esp           |  3 |   23 |
              | --- | --- |
              | --- | --- |
              |  4 |      |
              | --- | --- |


进入f函数,执行到call g后,堆栈如下

ebp=4


              | --- | --- |
              |  1 |   0  |
              | --- | --- |
              | --- | --- |
              |  2  |  5  |
              | --- | --- |
              | --- | --- |
              |  3 |   23 |
              | --- | --- |
              | --- | --- |
 ebp->        |  4 |   1  |
              | --- | --- |
              | --- | --- |
              |  5 |   5  |
              | --- | --- |
              | --- | --- |
esp->         |  6 |   14 |
              | --- | --- |


 进入g函数执行运算,执行到addl  $2, %eax 这句. eax=5+2=7



              | --- | --- |
              |  1 |   0  |
              | --- | --- |
              | --- | --- |
              |  2  |  5  |
              | --- | --- |
              | --- | --- |
              |  3 |   23 |
              | --- | --- |
              | --- | --- |
              |  4 |   1  |
              | --- | --- |
              | --- | --- |
              |  5 |   5  |
              | --- | --- |
              | --- | --- |
              |  6 |   15 |
              | --- | --- |
              | --- | --- |
esp,ebp->     |  7  |  4  |
              | --- | --- |

在g函数值中,在eip指向ret语句时(即执行完popl %ebp),堆栈如下:
              | --- | --- |
              |  1 |   0  |
              | --- | --- |
              | --- | --- |
              |  2  |  5  |
              | --- | --- |
              | --- | --- |
              |  3 |   23 |
              | --- | --- |
              | --- | --- |
ebp->         |  4 |   1  |
              | --- | --- |
              | --- | --- |
              |  5 |   5  |
              | --- | --- |
              | --- | --- |
esp->         |  6 |   15 |
              | --- | --- |

ret 后 eip=15,堆栈变成如下:

              | --- | --- |
              |  1 |   0  |
              | --- | --- |
              | --- | --- |
              |  2  |  5  |
              | --- | --- |
              | --- | --- |
              |  3 |   23 |
              | --- | --- |
              | --- | --- |
ebp->         |  4 |   1  |
              | --- | --- |
              | --- | --- |
esp->         |  5 |   5  |
              | --- | --- |

f中的leave执行完之后:

              | --- | --- |
ebp->         |  1 |   0  |
              | --- | --- |
              | --- | --- |
              |  2  |  5  |
              | --- | --- |
              | --- | --- |
esp->         |  3 |   23 |
              | --- | --- |

执行f中的ret后,eip=23,eax=7:
              | --- | --- |
ebp->         |  1 |   0  |
              | --- | --- |
              | --- | --- |
esp->         |  2  |  5  |
              | --- | --- |
执行main中的leave后,堆栈如下,eax=11,esp=ebp=0:
ebp,esp->     | --- | --- |
              |  1 |   0  |
              | --- | --- |

回到初始态.

我对计算机执行程序的理解:
计算机只是机械的从PC指针指向的位置取指令并解析执行.而数据与指令都是以二进制形式进行存储.
在C语言层面上的函数调用与返回的语义,在x86层面上是通过call指令与ret指令来完成.