为什么gdb告诉我x86-64上的指针是4个字节? [英] Why does gdb tell me a pointer is 4 bytes on x86-64?

问题描述

在openSUSE，CentOS，Fedora和Ubuntu上与gdb一起观看:

Seen with gdb on openSUSE, CentOS, Fedora, and Ubuntu:

此gdb被配置为"x86_64-unknown-linux-gnu".

This gdb was configured as "x86_64-unknown-linux-gnu".

(gdb)p sizeof(void *)

(gdb) p sizeof(void *)

$ 1 = 4

(gdb)p sizeof(long)

(gdb) p sizeof(long)

$ 2 = 4

为什么gdb在我所有的64位系统上给我错误的答案?

Why is gdb giving me the wrong answers on all of my 64-bit systems?

推荐答案

当您不调试任何特定代码时，似乎gdb选择了一些令人惊讶的默认值.如果您按以下方式加载64位可执行文件: gdb/bin/sh ，您会得到意想不到的结果:

It seems like gdb chooses some surprising defaults when you're not debugging any particular piece of code. If you load up a 64-bit executable as in: gdb /bin/sh you get a less-surprising result:

(gdb) p sizeof(void *)
$1 = 8

您还可以明确告诉gdb该怎么做:

You can also specifically tell gdb what to do:

(gdb) show architecture
The target architecture is set automatically (currently i386)
(gdb) p sizeof(void *)
$1 = 4
(gdb) set architecture
Requires an argument. Valid arguments are i386, i386:x86-64, i8086, i386:intel, i386:x86-64:intel, auto.
(gdb) set architecture i386:x86-64
The target architecture is assumed to be i386:x86-64
(gdb) p sizeof(void *)
$2 = 8

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