不能取值对象 - C++

Question

我在获取 "p1.media" 的值时遇到问题。值是 media: 6.91026e-310 并且必须像 1000、5000...。我尝试了太多的解决方案，但任何人都有效。这是代码：

Code of Calculos.h

#ifndef CALCULOS_H
#define CALCULOS_H
//includes
#define N 100
using namespace std;

class Calculos {
public:
    Calculos(double T[], int op);  
    Calculos(double T[], int op, double media);
    Calculos(); //constructor por defecto
    void run();

    int op;
    double desvtipica, media, *T;
};
#endif

Code of Calculos.cpp

//includes
#include "Calculos.h"
using namespace std;

Calculos::Calculos(double T[], int op) { 
    this->T = T;
    this->op = op;
    desvtipica = 0.0;
};

Calculos::Calculos() { 
}
Calculos::Calculos(double T[], int op, double media) {
    this->T = T;
    this->op = op;
    this->media=media;

};
void Calculos::run() {
    if(op == 1) {   //calcular media
        double suma = 0.0;
        for(int i = 0; i < N; i++) {
            suma = suma + T[i];
        }   
        media = (double)(suma/N);
    }
    else {  //op=3 calcular desviacion tipica
        desvtipica = 3.0; //partially
    }
};

Code of Ejercicio.cpp

//includes
#include "Calculos.h"
#include <iostream>
#include <cstdlib>
#include <thread>

using namespace std;

double fRand(double fMin, double fMax)
{
    double f = (double)rand() / RAND_MAX;
    return fMin + f * (fMax - fMin);
};

int main() {
    cout << "#datos: " << N << endl;
    double T[N];
    for(int i = 0 ; i < N; i++) {
        T[i] = fRand(1.0, 1000.0);
    }

    thread P[2];

    Calculos p1(T, 1);
    P[0] = thread(&Calculos::run, p1);
    P[0].join();
    double m = p1.media;   // Here is the problem. p1.media return bad values
    cout << "media: " << m << endl;
    Calculos p3(T, 3, m);
    P[1] = thread(&Calculos::run, p3);
    P[1].join();

    cout << "Fin\n";
    return 0;
}

Answer 1

您将参数传递给 thread 构造函数的方式存在问题。

变化：

   P[0] = thread(&Calculos::run, p1); // creates a copy of p1

至：

   P[0] = thread(&Calculos::run, std::ref(p1)); // pass p1 by reference

现在输出是：

#datos: 100
media: 547.278
Fin

Notes

The arguments to the thread function are moved or copied by value. If a reference argument needs to be passed to the thread function, it has to be wrapped (e.g. with std::ref or std::cref).

Source: http://en.cppreference.com/w/cpp/thread/thread/thread (emphasis mine)

在您的代码中，p1.media 是 0，因为您没有对其进行任何操作。您已经使用了 p1.

的临时副本

Answer 2

问题是您使用了 std::thread 的构造函数。看它的第三个定义here。 P[0] = thread(&Calculos::run, p1); 行正在创建 p1 的 copy，然后对其进行操作。尝试使用简单的引用捕获：

P[0] = thread([&p1](){ p1.run(); });

或者，我相信你也可以尝试像这样传入 p1 的地址：

P[0] = thread(&Calculos::run, &p1);

我知道该语法适用于 std::bind，但我不确定 std::thread。

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