显示没有方法名称的调用堆栈 [英] Display callstack without method names
问题描述
在WinDbg中,我可以使用 k
命令获取调用堆栈。对于没有符号的DLL,它会显示错误的方法名称和较大的偏移量,例如
In WinDbg, I can get the callstack using the k
command. For DLLs without symbols, it displays an incorrect method name and a large offset, e.g.
0018f9f0 77641148 syncSourceDll_x86!CreateTimerSyncBridge+0xc76a
由于我没有符号,因此我必须将此信息提供给DLL的开发人员。我不知道谁会处理该错误以及他有多少调试知识。我想避免开发人员认为问题出在CreateTimerSyncBridge()方法中。
Since I don't have symbols, I have to give this information to the developer of the DLL. I don't know who will work on the bug and how much debugging knowledge he has. I want to avoid that the developer thinks the problem is in the CreateTimerSyncBridge() method.
有没有一种方法可以在没有方法名称的情况下获取调用堆栈,只需使用
此刻,我正在使用以下解决方法:
At the moment I'm using the following workaround:
0:000> ? syncSourceDll_x86!CreateTimerSyncBridge+0xc76a
Evaluate expression: 1834469050 = 6d57c6ba
0:000> ? syncSourceDll_x86
Evaluate expression: 1834287104 = 6d550000
0:000> ? 6d57c6ba-6d550000
Evaluate expression: 181946 = 0002c6ba
所以我可以手动将调用堆栈修改为
So I can modify the callstack manually to
0018f9f0 77641148 syncSourceDll_x86!+0x2c6ba
但是对于很多线程中的许多帧来说,确实很难做到。
But that's really hard to do for a lot of frames in a lot of threads.
推荐答案
您可以使用更严格的评估来指定符号必须完全匹配,方法是使用命令行参数 -ses
启动windbg或发出命令:
You can specify that the symbols must match exactly using a stricter evaluation, either by starting windbg with command line parameter -ses
or issuing the command:
.symopt + 0x400
如果调试器默认为false希望重设此选项,然后删除选项:
The default is false for the debugger, if you wish to reset this then just remove the option:
.symopt -0x400
See the msdn docs: https://msdn.microsoft.com/en-us/library/windows/hardware/ff558827(v=vs.85).aspx#symopt_exact_symbols
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