当试图获取某个进程的线程的调用堆栈时,我总是得到一个相同的帧,尽管它肯定有更多(至少 5 帧)。
StackWalk64() 总是在第一次调用时成功 - 返回一个帧:
AddrPC.Offset = 18446744072850558156
但是,在第二次调用时,它立即失败,错误 id 为 998-ERROR_NOACCESS(可能是这个错误不是因为这个调用,正如 MSDN 所说)。
此外,尝试使用 SymFromAddr() 将此地址解析为其符号名称失败,并出现错误 126-ERROR_MOD_NOT_FOUND(在成功调用 SymInitialize(m_processHandler,NULL,TRUE) 之后)。
这是代码:
#ifdef _M_IX86
//
// Disable global optimization and ignore /GS waning caused by
// inline assembly.
//
#pragma optimize( "g", off )
#pragma warning( push )
#pragma warning( disable : 4748 )
#endif
bool EchoProfiler::getThreadStackTrace(__in HANDLE h_thread, __out vector<DWORD64> &framesVec)
{
CONTEXT threadContext;
if (GetThreadContext(h_thread, &threadContext) == 0)
{
cout << "Error: GetThreadContext() failed with error ID " << GetLastError() << endl;
return false;
}
//initialize stack frame
DWORD MachineType;
STACKFRAME64 StackFrame;
ZeroMemory( &StackFrame, sizeof( STACKFRAME64 ) );
MachineType = IMAGE_FILE_MACHINE_I386;
StackFrame.AddrPC.Offset = threadContext.Eip;
StackFrame.AddrPC.Mode = AddrModeFlat;
StackFrame.AddrFrame.Offset = threadContext.Ebp;
StackFrame.AddrFrame.Mode = AddrModeFlat;
StackFrame.AddrStack.Offset = threadContext.Esp;
StackFrame.AddrStack.Mode = AddrModeFlat;
PVOID contextRec = (MachineType == IMAGE_FILE_MACHINE_I386) ? NULL : &threadContext;
int i=0;
// enumerate all the frames in the stack
for (i=1 ; ; i++)
{
if (StackWalk64( MachineType, targetProcessHandler, h_thread, &StackFrame,
contextRec, NULL, SymFunctionTableAccess64, SymGetModuleBase64, NULL ) == false)
{
// in case it failed or we have finished walking the stack.
cout << "Error: StackWalk64() failed with error ID " << GetLastError() << endl;
i--;
break;
// return false;
}
if ( StackFrame.AddrPC.Offset != 0 )
{
// Valid frame.
cout << "Frame #" << i << " address - " << StackFrame.AddrPC.Offset << endl;
framesVec.push_back(StackFrame.AddrPC.Offset);
}
else
{
// Base reached.
break;
}
}
//cout << "StackWalk64 found " << i << " stack frames:" << endl;
//i = 1;
//for (FramesConstItr itr=framesVec.begin() ; itr != framesVec.end() ; itr++ , i++)
// cout << i << " - " << *itr << endl;
return true;
}
#ifdef _M_IX86
#pragma warning( pop )
#pragma optimize( "g", on )
#endif
会是什么呢?