为什么 Valgrind 在这个实现中会报告内存泄漏？

Question

我有以下一段 C++ 代码，我正在尝试执行一个例程：

#include <thread>
#include <unistd.h>
#include <sys/wait.h>
#include <memory>

using namespace std;

static void upload(const string url, const string path){ int sync_status;}

int main(){
    bool flag = true;
    for(int worker_id = 0; worker_id < 1; worker_id++){
        int worker_pid = fork();
        if (worker_pid == 0){
            while (true){
                std::unique_ptr<std::thread> uploader(nullptr);
                if (flag){       // This block only occurs once inside while loop
                    flag = false;
                    string path = "l", url = "k";
                    // Valgrind reports the next line
                    uploader = std::make_unique<std::thread>(upload, url, path);
                }
                int child_id = fork();
                if (!child_id){
                    // Do something
                    exit(0);
                }
                waitpid(child_id, NULL, 0);
                if(uploader && uploader->joinable()){
                    uploader->join();
                }
            }
            // Do something
            exit(0);
        }
    }
    return 0;
}

但 valgrind 似乎总是将这段代码报告如下：

==16424== 80 bytes in 1 blocks are definitely lost in loss record 2 of 2
==16424==    at 0x4C3017F: operator new(unsigned long) (in /usr/lib/valgrind/vgpreload_memcheck-amd64-linux.so)
==16424==    by 0x10AA39: std::unique_ptr<std::thread::_State, std::default_delete<std::thread::_State> > std::thread::_S_make_state<std::thread::_Invoker<std::tuple<void (*)(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >), std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > > > >(std::thread::_Invoker<std::tuple<void (*)(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >), std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > > >&&) (thread:197)
==16424==    by 0x10A04A: std::thread::thread<void (&)(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >), std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&>(void (&)(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >), std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&) (thread:126)
==16424==    by 0x109A64: std::_MakeUniq<std::thread>::__single_object std::make_unique<std::thread, void (&)(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >), std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&>(void (&)(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >), std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&) (unique_ptr.h:825)
==16424==    by 0x109514: main (main.cpp:21)

这次泄漏背后的原因是什么？我认为这是由于exit() 通话，但即使在通话之前加入unique_ptr 也无济于事。

Answer 1

你可以在[basic.start.main]找到这句话：

在不离开当前块的情况下终止程序（例如，通过调用函数std​::​exit(int) ([support.start.term])）不会破坏任何具有自动存储持续时间的对象（[class.dtor]）。

因此，您的 unique_ptr 的分叉副本将永远不会被销毁。这是 valgrind 报告的泄漏。

尝试销毁unique_ptr 的两个副本似乎也不是一个好主意。所以这个例子的设计可能存在更深层次的缺陷。