Question

我已经做了一个简单的线程安全的hmmscan_fun () { local file=$1 local marker_profiles=$2 local n_threads=$3 local out_dir=$4 fname=$(echo $file | rev | cut -d'/' -f1 | rev) echo 'filename' echo $out_dir$fname".txt" echo 'n threads' echo $n_threads echo 'marker profiles' echo $marker_profiles echo $out_dir$fname".txt" >> $out_dir"out.txt" hmmscan -o $out_dir$fname".txt" --tblout $out_dir$fname".hmm" -E 1e-10 --cpu $n_threads $marker_profiles $file }实现，创建了10个线程在Buffer队列上工作，以随机推送和弹出一些数字。我的实现应让等待弹出的线程仅等待3秒，然后终止。发生这种情况时，我会显示超时消息。

问题是仅打印一条超时消息，然后主线程将加入所有线程并返回。为什么？

这是代码main.cpp

buffer

Buffer.h

#include <thread>
#include <vector>
#include <iostream>
#include <sstream>
#include "Buffer.h"

int main() {

    std::vector<std::thread> workers;
    Buffer<std::string> buffer(3);

    srandom(time(NULL));

    for (int i = 0; i < 10; i++) {
        workers.emplace_back([&buffer]{
            long num = random();
            if(num%2==0) {
                std::stringstream msg;
                msg << std::this_thread::get_id() << " pushing " << num << std::endl;
                std::cout << msg.str();
                buffer.push(std::to_string(num));
            } else {
                std::stringstream msg1;
                msg1 << std::this_thread::get_id() << " waiting to pop" << std::endl;
                std::cout << msg1.str();
                std::string popped_string = buffer.pop();
                std::stringstream msg2;
                msg2 << std::this_thread::get_id() << " popped " << popped_string << std::endl;
                std::cout << msg2.str();
            }
        });
    }

    for (auto &w: workers) {
        if (w.joinable()) w.join();
    }

    return 0;
}

wait_until超时导致仅一个线程终止，无法捕获所有线程超时

0 个答案: