linux内核分析作业：以一简单C程序为例，分析汇编代码理解计算机如何工作

 1 int g(int x)
 2 {
 3   return x + 3;
 4 }
 5 int f(int x)
 6 {
 7   return g(x);
 8 }
 9 int main(void)
10 {
11   return f(8) + 1;
12  }

 1 g:
 2     pushl    %ebp
 3     movl    %esp, %ebp
 4     movl    8(%ebp), %eax
 5     addl    $4, %eax
 6     popl    %ebp
 7     ret
 8 f:
 9     pushl    %ebp
10     movl    %esp, %ebp
11     subl    $4, %esp
12     movl    8(%ebp), %eax
13     movl    %eax, (%esp)
14     call    g
15     leave
16     ret
17 main:
18     pushl    %ebp
19     movl    %esp, %ebp
20     subl    $4, %esp
21     movl    $2, (%esp)
22     call    f
23     addl    $3, %eax
24     leave
25     ret

 1 g:
 2     pushl    %ebp      
 3     movl    %esp, %ebp  　
 4     movl    8(%ebp) ,%eax 
 5     addl    $4, %eax ;eax + 4 = 6

 6     popl    %ebp ;
 7     ret
 8 f:
 9     pushl    %ebp ;ebp1入栈 ebp指向1 esp指向4
10     movl    %esp, %ebp ;ebp = esp =4
11     subl    $4, %esp ;
12     movl    8(%ebp), %eax
13     movl    %eax, (%esp)
14     call    g
15     leave
16     ret
17 main:
18     pushl    %ebp ;ebp0入栈 ebp指向0 esp指向1
19     movl    %esp, %ebp ;ebp = esp =1
20     subl    $4, %esp ; esp-4 esp指向2
21     movl    $2, (%esp) ;2入栈
22     call    f ;调用f函数
23     addl    $3, %eax ;eax+3
24     leave ; esp = ebp = 1 esp0出栈 ebp = 0 ebp-4 = 0
25     ret