c++ - LeakSanitizer : get run time leak reports?

Question

我继承了一些遗留代码，似乎某处有内存泄漏。我的第一直觉是用

编译

-faddress=sanitize -fno-omit-frame-pointer

让 Address Sanitizer 的工具系列帮我找到漏洞。然而，我非常失望。我希望出现某种运行时错误消息(类似于地址清理程序在您不应该读取或写入内存时的错误)。在程序成功完成之前，泄漏 sanitizer 似乎不会进行任何泄漏检查分析。我的问题是我继承的代码有多个线程，它并不是为了将所有线程都加入到软着陆准备中而设计的。

我用一个简单的例子简化了我的问题:

#include <thread>                                           
#include <chrono>                                           
#include <iostream>                                         

bool exit_thread = false;                                   

void threadFunc()                                           
{                                                           
   while(!exit_thread)                                      
   {                                                        
      char* leak = new char[256];                           
      std::this_thread::sleep_for(std::chrono::seconds{1}); 
   }                                                        
}                                                           

int main() {                                                
   std::thread t(threadFunc);                               
   std::cout << "Waiting\n";                                
   std::this_thread::sleep_for(std::chrono::seconds{5});    
   exit_thread = true;                                      
   std::cout << "Exiting\n";                                
   //Without joining here I do not get the leak report.     
   t.join();                                                
   return 0;                                                
}    

我编译这个

clang++ leaker.cpp -fsanitize=address -fno-omit-frame-pointer -g -O0 -std=c++1y -o leaker      

然后运行

ASAN_OPTIONS='detect_leaks=1' LSAN_OPTIONS='exitcode=55:report_objects=true:log_threads=true:log_pointers=true' ./leaker                                                

(我在这里对“LSAN_OPTIONS”有点疯狂，因为我在玩……没有一个选项能达到我想要的效果，但是在得知泄漏后退出)。

如代码中所述，如果我加入线程然后退出程序，我会收到一份漂亮的泄漏报告。否则我什么也得不到。正如您想象的那样，在遗留代码库中跟踪 10-100 个线程并让它们全部正常运行是很笨拙的。

几年前我记得玩过Visual Leak Detector并且祝你好运，因为它会生成包含所有潜在内存泄漏的报告(而且我不记得必须很好地删除所有内容)。问题是此工具仅适用于 Windows，而我的代码仅适用于 Linux。我可以让 LeakSanitizer 工具做类似的事情吗？

Answer 1

LeakSanitizer 的公共(public)接口(interface) ( sanitizer/lsan_interface.h ) 具有可满足您需求的各种功能。函数 __lsan_do_leak_check() 执行检查并在发现泄漏时终止。还有__lsan_do_recoverable_leak_check不会终止，可以多次调用。

考虑对您的程序进行此修改:

#include <thread>
#include <chrono>
#include <iostream>
#include <sanitizer/lsan_interface.h>

bool exit_thread = false;

void threadFunc()
{
   while(!exit_thread)
   {
      char* leak = new char[256];
      std::this_thread::sleep_for(std::chrono::seconds{1});
   }
}

int main() {
   std::thread t(threadFunc);
   std::cout << "Waiting\n";
   std::this_thread::sleep_for(std::chrono::seconds{5});
   exit_thread = true;
   std::cout << "Exiting\n";
   //Without joining here I do not get the leak report.
   //t.join();
   __lsan_do_recoverable_leak_check();
   std::cout << "Done\n";
   return 0;
}

输出:

Waiting
Exiting

=================================================================
==29240==ERROR: LeakSanitizer: detected memory leaks

Direct leak of 1024 byte(s) in 4 object(s) allocated from:
    #0 0x4d9a30 in operator new[](unsigned long) (leaker+0x4d9a30)
    #1 0x4dc663 in threadFunc() leaker.cpp:12:20
    #2 0x4dffe3 in void std::_Bind_simple<void (*())()>::_M_invoke<>(std::_Index_tuple<>) /usr/bin/../lib64/gcc/x86_64-unknown-linux-gnu/5.2.0/../../../../include/c++/5.2.0/functional:1530:18
    #3 0x4dff94 in std::_Bind_simple<void (*())()>::operator()() /usr/bin/../lib64/gcc/x86_64-unknown-linux-gnu/5.2.0/../../../../include/c++/5.2.0/functional:1520:16
    #4 0x4dfcc8 in std::thread::_Impl<std::_Bind_simple<void (*())()> >::_M_run() /usr/bin/../lib64/gcc/x86_64-unknown-linux-gnu/5.2.0/../../../../include/c++/5.2.0/thread:115:13
    #5 0x7f0a9664034f in execute_native_thread_routine /build/gcc-multilib/src/gcc-5.2.0/libstdc++-v3/src/c++11/thread.cc:84

SUMMARY: AddressSanitizer: 1024 byte(s) leaked in 4 allocation(s).
Done
terminate called without an active exception
Aborted