c++ - 带线程的后台函数c++

Question

这是我的代码:

#include <wiringPi.h>

#include <stdio.h>
#include <iostream>
#include <string>
#include <unistd.h>
#include <thread>

using namespace std;

bool running;
const int pin = 1; // Pin 18 = Pin 1 in WiringPi

void stayAwake() {
    while(running) {
        digitalWrite(pin, 1);
        usleep(1000);
        digitalWrite(pin, 0);
        usleep(1000);
    }
}

int main() {
    if (wiringPiSetup() == -1)
        return 1;

    pinMode(pin, OUTPUT);

    running = true;
    thread t(stayAwake);

    while(1) {
        int input = 0;
        cout << "put in a freq between 1000 and 2000:";
        while(input < 1000 || input > 2000) cin >> input;

        running = false;
        t.join();
        for(int i=0; i<=1000; i++) {
            digitalWrite(pin, 1);
            usleep(input);
            digitalWrite(pin, 0);
            usleep(1000);
        }

        running = true;
        thread t(stayAwake);
    }
}

我需要“stayAwake”函数在后台一直运行，直到用户输入有效数字，然后它应该停止并在 for 循环完成后立即再次启动。这只要我想要它。 (= 直到 Ctrl-C)

但是程序只是中断了:

terminate called without an active exception

Answer 1

您有两个 thread t，第二个位于 while 循环的底部，并立即超出范围并在线程未完成的情况下死亡，从而触发错误。

这应该会为您提供您想要的行为，或者类似的行为。

int main() {
    if (wiringPiSetup() == -1) {
        return 1;
    }
    pinMode(pin, OUTPUT);

    while(1) {
        running = true;
        thread t(stayAwake);
        int input = 0;
        cout << "put in a freq between 1000 and 2000:";
        while(input < 1000 || input > 2000) cin >> input;

        running = false;
        t.join();
        for(int i=0; i<=1000; i++) {
            digitalWrite(pin, 1);
            usleep(input);
            digitalWrite(pin, 0);
            usleep(1000);
        }
    }
}