C++ 11线程简单的例子

Question

我是c ++的新手,我正在研究一些c ++跨平台线程教程.我正在研究这个问题:http://solarianprogrammer.com/2011/12/16/cpp-11-thread-tutorial/

并试图执行以下代码:

#include <iostream>
#include <thread>

static const int num_threads = 10;

//This function will be called from a thread

void call_from_thread(int tid) {
    std::cout << "Launched by thread " << tid << std::endl;
}

int main() {
    std::thread t[num_threads];

    //Launch a group of threads
    for (int i = 0; i < num_threads; ++i) {
        t[i] = std::thread(call_from_thread, i);
    }

    std::cout << "Launched from the main\n";

    //Join the threads with the main thread
    for (int i = 0; i < num_threads; ++i) {
        t[i].join();
    }

    return 0;
}

我得到的输出如下,我无法理解为什么:

syd@syd-HP-Compaq-dx7500-Microtower:~/Desktop$ ./ref
Launched by thread Launched by thread Launched by thread Launched by thread Launched by thread 201
Launched by thread 5

Launched by thread 6
4
Launched by thread 7
3

Launched by thread 8
Launched from the main
Launched by thread 9

我知道这些数字每次都是随机的,但有时我没有显示数字,我想知道为什么？

Answer 1

他们都在那里.由于控制台输出以模糊的随机顺序发生,因此它们只是被破坏了.

特别是看看第一行输出的结尾.

Answer 2

您需要做的就是添加一个互斥锁并将其锁定在正确的位置:

std::mutex mtx;

-

void call_from_thread(int tid) {
    mtx.lock();
-----------------------------------------------------------
    std::cout << "Launched by thread " << tid << std::endl;
-----------------------------------------------------------
    mtx.unlock();
}

-

mtx.lock();
-----------------------------------------------------------
std::cout << "Launched from the main\n";
-----------------------------------------------------------
mtx.unlock();

参考这个