使用 GDB 打印 C++ vtables
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Print C++ vtables using GDB
提问by Nathaniel Flath
I'm trying to print an object's vtable using gdb; I found the
我正在尝试使用 gdb 打印对象的 vtable；我找到了
show print vt bl on
setting, but I still don't actually know how to print the vtable - p *object still doesn't print it out.
设置，但我实际上仍然不知道如何打印 vtable - p *object 仍然没有打印出来。
How do I print the vtable?
如何打印 vtable？
采纳答案by Will Bradley
(gdb) set $i = 0
(gdb) while $i < 10
>print $i
>p /a (*(void ***)obj)[$i]
>set $i = $i + 1
>end
Where "obj" is the object whose vtable you'd like to print, and 10 is the number of methods.
其中“obj”是您要打印其 vtable 的对象，10 是方法的数量。
回答by Jasancos
A more compact solution:
更紧凑的解决方案：
p /a (*(void ***)obj)[0]@10
回答by pnkfelix
If you have a sufficiently new version of gdb, you may want to look at the "info vtbl" command.
如果您有足够新版本的 gdb，您可能需要查看“info vtbl”命令。
I only noticed the feature when googling for an answer to this question and I noticed posts to the gdb mailing list circa 2012, notably this one from March 2012:
我只在谷歌搜索这个问题的答案时注意到了这个功能，我注意到大约在 2012 年的 gdb 邮件列表中的帖子，特别是 2012 年 3 月的这个帖子：
回答by Klaus
In the actual gdb 7.5.1 the command is notinfo vtable!
在实际的 gdb 7.5.1 中，命令不是info vtable！
Use info vtbl
使用信息 vtbl
回答by smh
For the example at http://en.cppreference.com/w/cpp/language/virtual
例如http://en.cppreference.com/w/cpp/language/virtual
Without using 'info vtbl'
不使用'info vtbl'
(gdb) p b
= {_vptr.Base = 0x400a60 <vtable for Base+16>}
(gdb) x/16x 0x400a60
0x400a60 <_ZTV4Base+16>: 0x0040094c 0x00000000 0x72654437 0x64657669
(gdb) x/16x 0x0040094c
0x40094c <Base::f()>: 0xe5894855 0x10ec8348 0xf87d8948 0x400a15be
0x40095c <Base::f()+16>: 0x10c0bf00 0xf9e80060 0xc9fffffd 0x485590c3
0x40096c <Derived::f()+2>: 0x8348e589 0x894810ec 0x1bbef87d 0xbf00400a
0x40097c <Derived::f()+18>: 0x006010c0 0xfffddbe8 0x66c3c9ff 0x00841f0f
