您将如何解释此拆卸清单? [英] How would you explain this disassembly listing?

问题描述

我有一个简单的C语言函数，位于单独的文件字符串中.c:

I have a simple function in C language, in separate file string.c:

void var_init(){
    char *hello = "Hello";
}

编译为:

gcc -ffreestanding -c string.c -o string.o

然后我使用命令

objdump -d string.o

查看反汇编清单.我得到的是:

to see disassemble listing. What I got is:

string.o:     file format elf64-x86-64


Disassembly of section .text:

0000000000000000 <var_init>:
   0:   55                      push   %rbp
   1:   48 89 e5                mov    %rsp,%rbp
   4:   48 8d 05 00 00 00 00    lea    0x0(%rip),%rax        # b <var_init+0xb>
   b:   48 89 45 f8             mov    %rax,-0x8(%rbp)
   f:   90                      nop
  10:   5d                      pop    %rbp
  11:   c3                      retq

我迷失了对这个清单的了解.《从头开始写操作系统》一书讲述了一些关于旧的反汇编的内容，并稍微揭开了神秘的面纱，但是它们的清单完全不同，而且我什至没有看到数据像作者所说的那样被解释为我的代码.

I lost in understanding this listing. The book "Writing OS from scratch" says something about old disassembly and slightly uncover the mistery, but their listing is completely different and I even not see that data interpreted as code in mine as author says.

推荐答案

此命令

lea    0x0(%rip),%rax

将字符串文字的地址存储在寄存器 rax 中.

stores the address of the string literal in the register rax.

这条命令

mov    %rax,-0x8(%rbp)

将地址从寄存器 rax 复制到分配的堆栈存储器中.从堆栈 -0x8 中的偏移量可以看到，该地址占用8个字节.

copies the address from the register rax into the allocated stack memory. The address occupies 8 bytes as it is seen from the offset in the stack -0x8.

此存储仅发生是因为您在调试模式下进行了编译；通常可以将其优化掉.接下来发生的事情是本地变量(在在堆栈指针下方的红色区域)将被有效地丢弃，因为该函数会破坏其堆栈框架并返回.

This store only happens at all because you compiled in debug mode; it would normally be optimized away. The next thing that happens is that the local vars (in the red-zone below the stack pointer) are effectively discarded as the function tears down its stack frame and returns.

您正在查看的材料可能包括 sub $ 16，％rsp 或类似的东西，用于为低于RBP的本地人分配空间，然后稍后再分配该空间；x86-64 System V ABI在叶函数(不需要调用任何其他函数)中不需要它；他们可以只使用读取区域.(另请参见 x86-64上的红色区域到底在哪里?).或使用 gcc -mno-red-zone 进行编译，无论如何您可能都想要独立代码:

The material you're looking at probably included a sub $16, %rsp or similar to allocate space for locals below RBP, then deallocating that space later; the x86-64 System V ABI doesn't need that in leaf functions (that don't call any other functions); they can just use the read zone. (See also Where exactly is the red zone on x86-64?). Or compile with gcc -mno-red-zone, which you probably want anyway for freestanding code: Why can't kernel code use a Red Zone

然后，它恢复调用者RBP的保存值(之前已将其设置为帧指针；请注意，相对于RBP寻址了本地人的空间).

Then it restores the saved value of the caller's RBP (which was earlier set up as a frame pointer; notice that space for locals was addressed relative to RBP).

pop    %rbp

并退出，有效地将返回地址弹出到RIP

and exits, effectively popping the return address into RIP

retq

这篇关于您将如何解释此拆卸清单?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！