汇编语言
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寄存器::

    rbp: register base pointer 栈基址寄存器（栈帧指针），指向当前栈帧的(栈底)地址
        rbp is the base pointer, which points to the base of the current stack frame
    rsp: register stack pointer 栈顶寄存器（栈指针），指向当前栈帧的(栈顶)地址
        rsp is the stack pointer, which points to the top of the current stack frame.





实例-函数调用
=============

C 语言::

    // function_example.c
    #include <stdio.h>
    int static add(int a, int b)
    {
        return a+b;
    }
    int main()
    {
        int x = 5;
        int y = 10;
        int u = add(x, y);
    }


执行命令::

    $ gcc -g -c function_example.c
    $ objdump -d -M intel -S function_example.o 

汇编代码-add::

    int static add(int a, int b)
    {
       0:   55                      push   rbp
       1:   48 89 e5                mov    rbp,rsp
       4:   89 7d fc                mov    DWORD PTR [rbp-0x4],edi
       7:   89 75 f8                mov    DWORD PTR [rbp-0x8],esi
        return a+b;
       a:   8b 55 fc                mov    edx,DWORD PTR [rbp-0x4]
       d:   8b 45 f8                mov    eax,DWORD PTR [rbp-0x8]
      10:   01 d0                   add    eax,edx
    }
      12:   5d                      pop    rbp
      13:   c3                      ret    

说明::
    
    push   rbp: 
        把之前调用函数，也就是 main 函数的栈帧的栈底地址，压栈
    mov    rbp,rsp: 
        把 rsp 的值复制到 rbp 里，而 rsp 始终会指向栈顶
        这意味着：rbp 和 rsp 都指向 main 函数的栈顶
        也就是 add 函数的栈帧的栈底地址了

汇编代码-main::

    0000000000000014 <main>:
    int main()
    {
      14:   55                      push   rbp
      15:   48 89 e5                mov    rbp,rsp
      18:   48 83 ec 10             sub    rsp,0x10
        int x = 5;
      1c:   c7 45 fc 05 00 00 00    mov    DWORD PTR [rbp-0x4],0x5
        int y = 10;
      23:   c7 45 f8 0a 00 00 00    mov    DWORD PTR [rbp-0x8],0xa
        int u = add(x, y);
      2a:   8b 55 f8                mov    edx,DWORD PTR [rbp-0x8]
      2d:   8b 45 fc                mov    eax,DWORD PTR [rbp-0x4]
      30:   89 d6                   mov    esi,edx
      32:   89 c7                   mov    edi,eax
      34:   e8 c7 ff ff ff          call   0 <add>
      39:   89 45 f4                mov    DWORD PTR [rbp-0xc],eax
      3c:   b8 00 00 00 00          mov    eax,0x0
    }
      41:   c9                      leave  
      42:   c3                      ret    


内联优化编译::

    $ gcc -g -c -O function_example_inline.c
    $ objdump -d -M intel -S function_example_inline.o