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Boost Thread - 在不加入(a)的情况下建立地下室
原标题:Boost Thread - Create a Thread without join()

我的申请创造了一个线索,我将听取即将到来的联系。 主要线索将做其他事情。

boost::mutex mutex;
void
ThreadFunction(int port, int(*callbackFunc)(int, int))
{
    mutex.lock();
    std::cout << "Cannot get to this point" << std::endl;
    mutex.unlock();
    Application app;
    app.run(port, callbackFunc);
}

void 
Init(int port, int(*callbackFunc)(int, int))
{
    std::cout << callbackFunc(1,1) << std::endl;
    boost::thread t(boost::bind(&ThreadFunction, port, callbackFunc));
}

int
main(){
    int port = 2340;
    Init(port, *callbackfunction);
    return 0;
}

我面临的问题是,它从来都无法获得<代码>std:cout << “Cannot go to this point” << end:endl; 然而,如果我呼吁(......)在我打造了read子之后加入进来,那么它只是做的是罚款,而是阻碍申请。

我需要做些什么去做“read”?

最佳回答

你的申请在校方有机会完成工作之前即告终止。 一旦你实施“接通”循环,问题就会得到解决。 因此,不需要做任何事情。

问题回答

暂无回答




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