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//32位系统
#include<stdio.h>
struct
{
    int i;    
    char c1;  
    char c2;  
}x1;

struct{
    char c1;  
    int i;    
    char c2;  
}x2;

struct{
    char c1;  
    char c2; 
    int i;    
}x3;

int main()
{
    printf("%d\n",sizeof(x1));  // 输出8
    printf("%d\n",sizeof(x2));  // 输出12
    printf("%d\n",sizeof(x3));  // 输出8
    return 0;
}

command

stack frame

rbp: stack frame bottom;
rsp: stack frame top, chaning as stack grow;

functon call

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add(int, int):
   0x080483b4 <+0>:	push   %ebp
   0x080483b5 <+1>:	mov    %esp,%ebp
   0x080483b7 <+3>:	sub    $0x4,%esp
   0x080483ba <+6>:	mov    0xc(%ebp),%eax
   0x080483bd <+9>:	mov    0x8(%ebp),%edx
   0x080483c0 <+12>:	add    %edx,%eax
   0x080483c2 <+14>:	mov    %eax,-0x4(%ebp)
   0x080483c5 <+17>:	mov    -0x4(%ebp),%eax
   0x080483c8 <+20>:	leave  
   0x080483c9 <+21>:	ret   
main:
   0x080483ca <+0>:	push   %ebp
   0x080483cb <+1>:	mov    %esp,%ebp
   0x080483cd <+3>:	sub    $0xc,%esp
   0x080483d0 <+6>:	movl   $0x2,0x4(%esp)
   0x080483d8 <+14>:	movl   $0x28,(%esp)
   0x080483df <+21>:	call   0x80483b4 <add>
   0x080483e4 <+26>:	mov    %eax,-0x4(%ebp)
   0x080483e7 <+29>:	leave  
   0x080483e8 <+30>:	ret    

2. stack

stackFrame: a stack segment; [return address, lastBasePointer,local variable,args]

push Next address: 将该方法执行完之后需要继续执行的指令地址压入栈中；
jump:

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int add(int a, int b)
{
   int result = a + b;
   return result;
}

int main(int argc)
{
   int answer;
   answer = add(40, 2);
}



```c 
main;  
   push %ebp; #老的栈底存储起来
   move %esp %ebp;  #更新新的栈底值
   sub  $12 %esp; #栈空间增加 12byte；
   mov  $2 4(%esp); #2入栈
   mov  $40 %esp; #40入栈
   call add;
   mov   %eax,-0x4(%ebp)
   ret


add:
   push %ebp; #老栈顶存起来
   move %esp %ebp;  #更新新的栈底值
   sub   $0x4,%esp
   move 12(%ebp) %eax;
   move 8(%ebp) %edx;
   add %edx %eax; # 计算写入 %eax
   move %eax (-4)%ebp 
   move (-4)%ebp  %eax;
   leave 
   ret


leave; 
leave;
ret; #program counter
# 栈顶值更新为当前栈底；
# 栈底更新为老的；
# program counter 更新为新的指令；


move %eax  (-4)%ebp #将结果写入answer

leave;
leave;
ret;