Mar*_*ras 2 c++ syntax-error c++11
我在获取"p1.media"的价值方面遇到了问题.价值是媒体:6.91026e-310,必须像1000,5000,....我尝试了太多的解决方案,但任何人的工作.这是代码:
Calculos.h代码
#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
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Calculos.cpp代码
//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
}
};
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Ejercicio.cpp代码
//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;
}
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将参数传递给thread构造函数的方式存在问题.
更改:
P[0] = thread(&Calculos::run, p1); // creates a copy of p1
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至:
P[0] = thread(&Calculos::run, std::ref(p1)); // pass p1 by reference
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现在的输出是:
#datos: 100
media: 547.278
Fin
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笔记
线程函数的参数按值移动或复制.如果需要将引用参数传递给线程函数,则必须将其包装(例如,使用
std::ref或std::cref).来源:http://en.cppreference.com/w/cpp/thread/thread/thread (强调我的)
在你的代码中,p1.media是0因为你没有对它做任何事情.您使用过的临时副本p1.
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