gdb'x'命令是做什么的？ [英] What does the gdb 'x' command do?

问题描述

为了更好地了解计算机，我偶然发现了一本关于反汇编，x86汇编语言以及C和x86汇编之间关系的书。现在我一直在阅读这个 GDB 命令，但我无法完全理解它。

该命令与其结果如下：

pre $ code $（gdb）x / 32xw $ esp
0xbffff7e0：0xb8000ce0 0x08048510 0xbffff848 0xb7eafebc
0xbffff7f0：0x00000002 0xbffff874 0xbffff880 0xb8001898
0xbffff800：00000000 00000001 00000001 00000000
0xbffff810：0xb7fd6ff4 0xb8000ce0 00000000 0xbffff848
0xbffff820：0x40f5f7f0 0x48e0fe81 00000000 00000000
0xbffff830：00000000 0xb7ff9300 0xb7eafded 0xb8000ff4
0xbffff840：0x00000002 0x08048350 0x00000000 0x08048371
0xbffff850：0x08048474 0x00000002 0xbffff874 0x08048510

现在，我明白了，我发出的命令告诉调试器：

• x（第一个）：检查内存


• 32：得到32后面的内容


• x：启用十六进制表示形式


• w：sh请注意：** 字尺码数据
我知道我问过esp寄存器，但我不完全理解$在它前面做什么。当我尝试不使用它时，我得到一个丢失的符号错误，所以我得到它与引用/去引用有关系吗？

一直困扰着我的是我如何找到所有这些字节？由于我正在检查一个寄存器，谁的大小是32位，我不应该只得到32位或4个字节（只有上面的1行）？如果我对我的假设是正确的，那么我们是否可以找到其余的数据？它是否需要对堆栈和特定的堆栈框架进行操作，目前我还没有意识到这一点？

我很感谢您的意见，以便我能够澄清我的想法。

解决方案

这给你32字的内存，其中 esp 寄存器正在指向（显然这个寄存器包含地址0xbffff7e0）。

In my quest to learn more about the computer in general, I stumbled upon a book which has some chapters about disassembling, the x86 assembly language, and the relationship between C and x86 assembly. Now I have been reading this GDB command but I am unable to fully understand it.

The command, along with its results, follows:

(gdb) x/32xw $esp
0xbffff7e0:    0xb8000ce0 0x08048510 0xbffff848 0xb7eafebc
0xbffff7f0:    0x00000002 0xbffff874 0xbffff880 0xb8001898
0xbffff800:    0x00000000 0x00000001 0x00000001 0x00000000
0xbffff810:    0xb7fd6ff4 0xb8000ce0 0x00000000 0xbffff848
0xbffff820:    0x40f5f7f0 0x48e0fe81 0x00000000 0x00000000
0xbffff830:    0x00000000 0xb7ff9300 0xb7eafded 0xb8000ff4
0xbffff840:    0x00000002 0x08048350 0x00000000 0x08048371
0xbffff850:    0x08048474 0x00000002 0xbffff874 0x08048510

Now, from what I understand, the command that I issue tells the debugger to:

• x (first one): examine the memory
• 32: get 32 of what follows
• x: enable hexadecimal representation
• w: show me Word size data.
• **Note:** I know that I ask about the esp register, but I don't quite fully understand what $ is doing in front of it. When I try not to use it, I get a missing symbol error, so I get it has something to do with reference/de-reference?

What has been bugging me is how did I find all those bytes? Since I am examining a register, who's size is 32 bit, shouldn't I get only 32 bits, or 4 bytes (only 1 row of the above)? If I am correct with my assumption, then were did we find the rest of the data? Does it have to do something with the stack, and a particular stack frame, which I currently am unaware of?

I would appreciate your input so that I can clarify things in my mind.

解决方案

It's giving you 32 words of memory where the esp register is pointing (apparently that register contains the address 0xbffff7e0).

这篇关于gdb'x'命令是做什么的？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！