Question

我正在构建一个多线程应用程序，但我遇到了问题。在应用程序中，我使用变量在线程之间进行通信，但它不起作用。因此，当您键入 exit 时，应用程序不会停止执行。

代码：

//Program that controls an cadrocopter with the use of a
//Raspberry Pi, MPU 6050, an ultrasonic sensor, an HMC5883L and Arduino Leonardo
//to compile use "g++ ./quad.cpp -o ./quad -std=c++0x -pthread"
//
//Copyright Jan Hendrik Farr

#include <iostream>             //used to input data
#include <string>               //used to procces the userdata
#include <stdlib.h>             //used to convert strings into floats
#include "./serial/serial.h"    //used to communicate with the Arduino
#include <thread>               //used to do multithreating

using namespace std;

//userinterface thread
void userInterface(int cmdPos1, float cmdPos[]){
    string cmd = "";

    cout << "************************" << endl;
    cout << "* Quadrocopter control *" << endl;
    cout << "*       software       *" << endl;
    cout << "*                      *" << endl;
    cout << "*     version 0.1      *" << endl;
    cout << "*                      *" << endl;
    cout << "* Copyright J.H. Farr  *" << endl;
    cout << "************************" << endl << endl << endl;

    while(cmdPos1 != 4){
        cin >> cmd;

        if(cmd == "move"){
            cin >> cmdPos[0] >> cmdPos[1] >> cmdPos[2] >> cmdPos[3];
            cmdPos1 = 1;
            cout << endl << endl;
        } else if(cmd == "move+"){
            cin >> cmdPos[0] >> cmdPos[1] >> cmdPos[2] >> cmdPos[3];
            cmdPos1 = 2;
            cout << endl << endl;
        } else if(cmd == "land"){
            cmdPos1 = 3;
            cout << endl << endl;
        } else if(cmd == "exit"){
            cmdPos1 = 4;
            cout << endl;
        } else {
            cout << "invalid argument!!" << endl << endl;
        }
    }
}

//algorithm
void algorithm(float tele[], int cmdPos1, float cmdPos[]){
    while(cmdPos1 != 4){
        switch (cmdPos1){
            case 2:
            cout << "works!!";
            break;

            case 1:

            break;

            case 3:

            break;
        }
    }
}

//gets telemetrydata from mpu
void gettelempu(float tele[], int cmdPos1){
    while(cmdPos1 != 4){

    }
}

//gets height from ultrasonic sensor
void getheight(float tele[], int cmdPos1){
    while(cmdPos1 != 4){

    }
}

//main function
int main(){
    //telemetry data
    float tele[12] = {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0};
    //what to do
    int cmdPos1 = 0;
    //where to go
    float cmdPos[4] = {0, 0, 0, 0};

    thread t1(userInterface, cmdPos1, cmdPos);
    thread t2(algorithm, tele, cmdPos1, cmdPos);
    thread t3(gettelempu, tele, cmdPos1);
    thread t4(getheight, tele, cmdPos1);

    t1.join();
    t2.join();
    t3.join();
    t4.join();
    return 0;
}

Answer 1

void gettelempu(float tele[], int cmdPos1)

此函数获取自己的cmdPos副本;在调用它时，传递给它的值是0，函数永远不会改变该值，因此循环永远不会终止。 algorithm和getheight中也是如此。

为了在一个地方更改此参数的值并让其他函数看到更改，函数必须通过引用获取值：

void gettelempu(float tele[], int& cmdPos1)

并且在创建线程时，您必须传递引用：

thread t3(gettelempu, tele, std::ref(cmdPos1));

但等等，还有更多！无法保证在一个线程中对cmdPos1所做的更改对其他线程可见。为了确保这一点，让它成为原子。在main中，更改

int cmdPos1 = 0;

到

std::atomic<int> cmdPos1 = 0;

并更改功能签名以取std::atomic<int>而不是int：

void gettelempu(float tele[], std::atomic<int>& cmdPos1)

std :: thread通信问题

1 个答案: