在gdb中翻转字节序? [英] Flipping endianness in gdb?

问题描述

在gdb中，我是一个困惑的困惑者，为什么有时它会从左到右，有时是从右到左打印字节.这是一个示例，其中没有指令在它们之间运行:

In gdb I'm a bix perplexed as to why sometimes it prints bytes left-to-right and sometimes right-to-left. Here is an example where no instruction has been run in between:

>>> x/4b $rbp-4
0x7fffffffe43c: 0x04    0x00    0x03    0x16
>>> x/2h $rbp-4
0x7fffffffe43c: 0x0004  0x1603

为什么在gdb中完成此操作?我认为它将始终将它们打印为:

Why is this done in gdb? I would think it would always print them as:

04 00 03 16

推荐答案

GDB使用目标计算机的本机字节序(默认为 ¹ )将您请求的大小的块解释为整数，地址增加在大块之间从左到右. x86是低端字节序.

GDB uses the target machine's native endiannes (by default¹) to interpret chunks of the size you requested as integers, with addresses increasing from left to right between chunks. x86 is little-endian.

在一个块中，04 00解释为16位小尾数整数是 0x0004.例如，这使GDB以您期望的方式转储short数组.您要求GDB提供16位整数块，所以它按照您的指示进行操作，使用标准的位置值表示法显示整数 value ，其中最左边的数字最高.它并不是要单独打印字节，因为您要求输入半字.

Within a chunk, 04 00 interpreted as a 16-bit little-endian integer is 0x0004. This makes GDB dump an array of short the way you'd expect, for example. You asked GDB for 16-bit integer chunks, so it does what you told it, showing the integer value using standard place-value notation with the leftmost digit the most significant. It's not trying to print the bytes separately because you asked for half-words.

如果要按内存顺序排列字节，请使用b.那就是它的目的.

If you want bytes in memory order, use b. That's what it's for.

脚注1 :

您可以更改GDB的字节序设置； show endian显示当前设置.但是，set endian big会导致问题，损坏寄存器值.例如在_start停下的时间:

You can change GDB's endianness setting; show endian shows the current setting. However, set endian big causes problems, corrupting register values. e.g. when stopped at _start:

(gdb) p /x $rsp
$1 = 0x7fffffffe6d0       # this is normal for x86-64
(gdb) set endian big
The target is assumed to be big endian
(gdb) x /16xw $rsp
0xd0e6ffffff7f0000:     Cannot access memory at address 0xd0e6ffffff7f0000
(gdb) p /x $rsp
$2 = 0xd0e6ffffff7f0000   # this is obviously broken, byte-reversed

寄存器没有字节序，并且在扩展其值时将其翻转，因为另一条命令的地址被完全破坏了.

Registers don't have an endianness, and flipping them when expanding their value as an address for another command is just totally broken.

不完全相关的重复项:

• GDB是否正确解释了内存地址?
• 为什么存储在内存中的数据被反转了?
• Big Endian和Little Endian有点困惑

• Is GDB interpreting the memory address correctly?
• Why is data stored in memory reversed?
• Big Endian and Little endian little confusion

这篇关于在gdb中翻转字节序?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！