Question

我是新手，尝试在单独的线程中实现自由函数，静态函数和成员函数。它在调试模式下运行良好，但在发布模式下崩溃。通常它意味着未初始化的数组或值，但我找不到问题..

class test {
public:
    static void func2() {
        cout<< "function2"<<endl;
    }

    void func3(string msg) {
        cout<<msg<<endl;
    }

    void operate() {
        // Constructs the new thread and runs it. Does not block execution. 
        thread t2(&test::func2);   // static function               
        thread t3(boost::bind(&test::func3,this,"function3"));  // member function

        //Makes the main thread wait for the new thread to finish execution, therefore blocks its own execution.
        t2.join();
        t3.join();
    }
};

void func1(string msg) {
    cout<<msg<<endl;
}

int main() {
    test example;
    example.operate();
    thread t1(&func1,"function1"); // free function
    t1.join();
    return 0;
}

Answer 1

一个简单的解决方法是使用互斥锁来保证只使用一次cout。

std::mutex mut;

void log(const std::string& ref)
{
    std::lock_guard<std::mutex> lock(mut);
    std::cout<<ref<<std::endl;
}

然后你的代码看起来就像那样。

class test {
public:
    static void func2() {
        log("function2");
    }

    void func3(const std::string & msg) {
        log(msg);
    }

    void operate() {
        // Constructs the new thread and runs it. Does not block execution. 
        std::thread t2(&test::func2);   // static function               
        std::thread t3(&test::func3,this,"function3");  // member function

        //Makes the main thread wait for the new thread to finish execution, therefore blocks its own execution.
        t2.join();
        t3.join();
    }
};

需要注意三点：

我正在通过const ref
我不再使用boost bind了
您仍然可以直接使用cout，因此如果您想要阻止编译时间，您需要在单独的单元（.h + .cpp）中声明日志功能并删除主程序的#include <iostream> < / LI>

希望有所帮助，

在发布模式下提升线程崩溃

1 个答案: