今天與一個朋友討論死鎖的時候,我說,我簡單寫一個程序,驗證一下.
一測不要緊,本應該死鎖的程序,怎么不死鎖了呢?
我們來看代碼吧
?1?
//?testMutex.cpp?:?Defines?the?entry?point?for?the?console?application.
?2?
//
?3?
?4?
#include?"stdafx.h"
?5?
#include?"ACE/Thread_Mutex.h"
?6?
#include?"ACE/Log_Msg.h"
?7?
#include?"ACE/Guard_T.h"
?8?
typedef?ACE_Thread_Mutex?MUTEX;
?9?
class?Logger
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{
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public:
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????void?log(void)
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????{
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????????ACE_GUARD(MUTEX,mon,mutex_);
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????????ACE_DEBUG((LM_DEBUG,"(%t)?進入互斥體1\n"));
16?
????????logCritical();
17?
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????}
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????void?logCritical()
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????{
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????????ACE_GUARD(MUTEX,mon,mutex_);
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????????ACE_DEBUG((LM_DEBUG,"(%t)?為什么還能再進入進入互斥體1,為什么不在這里死鎖\n"));
23?
????}
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private:
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????MUTEX?mutex_;
26?
};
27?
28?
29?
int?ACE_TMAIN(int?argc,?_TCHAR*?argv[])
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{
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????ACE_DEBUG((LM_DEBUG,"(%t)?主線程\n"));
32?
33?
????Logger?l;
34?
????l.log();??//影響不在這里死鎖呢
35?
36?
???ACE_DEBUG((LM_DEBUG,"(%t)?主線程2\n"));
37?
????return?0;
38?
}
39?
40?
屏幕輸出的結果是
??(4492)?主線程
??(4492)?進入互斥體1
??(4492)?為什么還能再進入進入互斥體1,為什么不在這里死鎖
??(4492)?主線程2