GDB - 什么是神秘的汇编代码？

Question

Dump of assembler code for function main:
   0x0804833e <+0>:     push   %ebp
   0x0804833f <+1>:     mov    %esp,%ebp
   0x08048341 <+3>:     sub    $0x8,%esp
   0x08048344 <+6>:     and    $0xfffffff0,%esp
   0x08048347 <+9>:     mov    $0x0,%eax
   0x0804834c <+14>:    add    $0xf,%eax
   0x0804834f <+17>:    add    $0xf,%eax
   0x08048352 <+20>:    shr    $0x4,%eax
   0x08048355 <+23>:    shl    $0x4,%eax
   0x08048358 <+26>:    sub    %eax,%esp
=> 0x0804835a <+28>:    movl   $0x10,-0x4(%ebp)
   0x08048361 <+35>:    movl   $0x0,-0x8(%ebp)
   0x08048368 <+42>:    pushl  -0x4(%ebp)
   0x0804836b <+45>:    call   0x8048334 <myfunc1 at test.c:4>
   0x08048370 <+50>:    add    $0x4,%esp
   0x08048373 <+53>:    pushl  -0x8(%ebp)
   0x08048376 <+56>:    call   0x8048339 <myfunc2 at test.c:8>
   0x0804837b <+61>:    add    $0x4,%esp
   0x0804837e <+64>:    mov    $0x0,%eax
   0x08048383 <+69>:    leave
   0x08048384 <+70>:    ret
End of assembler dump.
(gdb) info line
Line 16 of "test.c" starts at address 0x804835a <main+28 at test.c:16> and ends at 0x8048361 <main+35 at test.c:17>.------------------------------------(1)
(gdb) shell cat test.c
#include<stdio.h>

void myfunc1(int recv_arg1)
{
        /* does nothing */
}
void myfunc2(int recv_arg1)
{
   /* does nothing */
}

int main(int argc,char **argv)
{
        int var1;
        int var2;
        var1 = 16;
        var2 = 0;
        myfunc1(var1);
        myfunc2(var2);
        return 0;
}

注意在(1)中main的asm代码在该范围内!! 和此范围之前的asm代码是用于其他什么？什么 ？肯定是神秘的东西!!

Answer 1

请允许我为您发表评论.

   0x0804833e <+0>:     push   %ebp                    ; Establish standard
   0x0804833f <+1>:     mov    %esp,%ebp               ; stack frame record
   0x08048341 <+3>:     sub    $0x8,%esp               ; Make room for locals
   0x08048344 <+6>:     and    $0xfffffff0,%esp        ; Align esp to 16-byte memory
   0x08048347 <+9>:     mov    $0x0,%eax               ; eax=0
   0x0804834c <+14>:    add    $0xf,%eax               ; eax=f
   0x0804834f <+17>:    add    $0xf,%eax               ; eax=  (eax + 0xf)
   0x08048352 <+20>:    shr    $0x4,%eax               ;      (             >> 4)
   0x08048355 <+23>:    shl    $0x4,%eax               ;     (                    << 4)
   ;The above math rounds up eax as set by 0x0804834c to the next 16-byte boundary
   ;In this case, eax will be 0x10, rounded up from 0x0f.  You compiled without
   ;optimizations?  This could be a "probe" checking whether the upcoming call 
   ;will fail?

   0x08048358 <+26>:    sub    %eax,%esp               ; Make room for "0x10 more mystery bytes"
   0x0804835a <+28>:    movl   $0x10,-0x4(%ebp)        ; var1 = 16
   0x08048361 <+35>:    movl   $0x0,-0x8(%ebp)         ; var2 = 0
   0x08048368 <+42>:    pushl  -0x4(%ebp)              ; push           var1
   0x0804836b <+45>:    call   0x8048334 <myfunc1 at test.c:4> ;myfunc1(    );
  0x08048370 <+50>:    add    $0x4,%esp                ; pop (var1)
   0x08048373 <+53>:    pushl  -0x8(%ebp)              ; push           var2
   0x08048376 <+56>:    call   0x8048339 <myfunc2 at test.c:8> ;myfunc2(    );
   0x0804837b <+61>:    add    $0x4,%esp               ; pop (var2)
   0x0804837e <+64>:    mov    $0x0,%eax               ; return 0;
   0x08048383 <+69>:    leave                          ; undo standard stack frame
   0x08048384 <+70>:    ret                            ; actual return

我认为这是一个很好的问题,为什么最终执行0x08048358哪个分配看似无用的空间.我怀疑这是在执行调用之前检查esp超出范围异常.如果你指定你正在使用的处理器,我想知道它是否会"消失" - 它闻起来可能是特定芯片的勘误表.

Answer 2

来自up 0x0804833e <+0>(包括)的代码0x08048358 <+26>正在设置所谓的堆栈帧.

前四个陈述非常标准.首先保存旧的基指针(在维基百科文章中称为帧指针).然后,使用堆栈指针的当前值设置新的基指针.

接下来,您递减堆栈指针以为您的局部变量腾出空间(注意您减去0x8,这足以让您获得两个整数).最后,它确保堆栈指针与16位地址对齐.

下一组线(从0x08048347 <+9>到0x08048358 <+26>)有点奇怪.效果是增加堆栈,但我无法解释为什么它使用5条指令来计算值(因为没有变量,它应该能够在编译时执行此操作)或为什么需要更多地增加堆栈.