【发布时间】:2020-07-22 04:30:54
【问题描述】:
我想在作为模板类的派生类成员的虚函数中返回一个指向指针的数组。详细来说,我的类定义是:
Sampler.h
#ifndef SAMPLER_H
#define SAMPLER_H
template <class T>
class Sampler
{
public:
virtual T getnumber()=0;
virtual T* simulation(int n)=0;
};
class UniformSampler:public Sampler<double>
{
public:
virtual double getnumber();
virtual double* simulation(int n);
UniformSampler(double a=0.0, double b=1.0);
private:
double low_bound;
double up_bound;
};
#endif
类 Sampler 是一个模板类,以便以后能够派生另一个带有向量的采样器。实现是:
Sampler.cpp
#include "Sampler.h"
#include<iostream>
#include<cstdlib>
#include<cmath>
using namespace std;
//Uniform
UniformSampler::UniformSampler(double a, double b)
{
low_bound=a;
up_bound=b;
}
double UniformSampler::getnumber()
{
int myrand=rand();
while((myrand==0)||(myrand==RAND_MAX)){myrand = rand(); } //We want a number in (0, RAND_MAX).
double myuni = myrand/static_cast<double>(RAND_MAX); //Create a number in (0,1).
return low_bound + myuni*(up_bound-low_bound);
}
double* UniformSampler::simulation(int n){
double simulations[n];
for(int i=0; i<n; i++){
simulations[i] = this->getnumber();
}
return simulations;
}
我的问题是,当我尝试在main() 中调用这个程序时,看起来指针的分配不起作用。这是我的main.cpp:
#include <iostream>
#include <math.h>
#include <cstdlib>
#include <time.h>
using namespace std;
#include "Sampler.h"
int main(){
srand(time(0));
int n=10;
double *unif = new double[n];
UniformSampler uni;
unif = uni.simulation(n);
for ( int i = 0; i < n; i++ ) {
cout << "*(p + " << i << ") : ";
cout << *(unif + i) << endl;
}
delete[] unif;
return 0;
}
当我运行它时,它不会打印unif 指向的任何元素。我不明白那里有什么问题。
【问题讨论】:
标签: c++ templates inheritance virtual