Question

为什么valgrind会说这可能会泄漏。

#include <stdexcept>

int main(int argc, char const *argv[])
{
    throw std::runtime_error("");
    return 0;
}

用valgrind说（用全面检查更新）

    ==26803== Using Valgrind-3.7.0 and LibVEX; rerun with -h for copyright info
==26803== Command: ./test
==26803== 
terminate called after throwing an instance of 'std::runtime_error'
  what():  
==26803== 
==26803== HEAP SUMMARY:
==26803==     in use at exit: 144 bytes in 1 blocks
==26803==   total heap usage: 2 allocs, 1 frees, 176 bytes allocated
==26803== 
==26803== 144 bytes in 1 blocks are possibly lost in loss record 1 of 1
==26803==    at 0x4C2B6CD: malloc (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==26803==    by 0x4E8FE42: __cxa_allocate_exception (in /usr/lib/x86_64-linux-gnu/libstdc++.so.6.0.20)
==26803==    by 0x400A07: main (in /home/arynaq/HioA/DAVE3605/oppgaver/uke910/test)
==26803== 
==26803== LEAK SUMMARY:
==26803==    definitely lost: 0 bytes in 0 blocks
==26803==    indirectly lost: 0 bytes in 0 blocks
==26803==      possibly lost: 144 bytes in 1 blocks
==26803==    still reachable: 0 bytes in 0 blocks
==26803==         suppressed: 0 bytes in 0 blocks
==26803== 
==26803== For counts of detected and suppressed errors, rerun with: -v
==26803== ERROR SUMMARY: 1 errors from 1 contexts (suppressed: 2 from 2)
Aborted (core dumped)

这144个字节是什么？

Answer 1

std::runtime_error的构造函数分配一些内存来保持它的内部结构，但是析构函数（由于堆栈展开称为终止，它应该释放它更新）。这些valgrind力学只是强迫你捕捉主要的一切（好的做法恕我直言）。

int main() {
  try {
    /* all of your code goes here */
  } catch (std::exception& e) {
    std::err << "error: " << e.what() << std::endl;
    return 1;
  } catch(...) {
    std::err << "error: unknown exceptions" << std::endl;
    return 1;
  }
}

这种方法允许您捕获一些异常以便以其他方式跟踪它们。

异常可能泄漏？

1 个答案: