有没有办法在非模板类中定义模板成员？ ＃2 [英] Is there a way to define a template member in a non-template class? #2

问题描述

我有一个模板类 class_A ：

// class_A.h
#pragma once

#include <iostream>
#include <tuple>

template <class T>
class class_A
{
public:
    class_A(){}

    T Function_A(T parameter)
    {
        return parameter;
    }
};

我正尝试在非模板类 class_B 作为私有成员：

that I am trying to use in a non-template class class_B as private member:

// class_B.h
#pragma once

#include <tuple>
#include <iostream>

#include "class_A.h"

class class_B
{
public:
    class_B();

    template <typename T> T Evaluate(T parameter);

private:
    std::tuple<class_A<double>, class_A<char> > As;
};

和

// class_B.cc
#include "class_B.h"

class_B::class_B(){}

template <typename T>
T class_B::Evaluate(T parameter)
{
    return std::get<class_A<T>>(As).Function_A(parameter); //This is causing error
    //return parameter // This works
}

template double class_B::Evaluate(double parameter);
template char class_B::Evaluate(char parameter);

和我的 main.cc 是：

// main.cc
#include<iostream>
#include <string>
#include "class_B.h"

using namespace std;


int main()
{
    class_B B;

    std::cout<< B.Evaluate(5.2) <<std::endl;
    std::cout << B.Evaluate('h') << std::endl;

    return 0;
}

我收到以下错误：

src/class_B.cc:8:12: error: no matching function for call to 'get'
    return std::get<class_A<T>>(As).Evaluate(parameter);
.
.
.
etc.

这是对答案的尝试：https://stackoverflow.com/a/55357742/9203360 无效，但如果可行，则非常理想。

This is a trial of the answer: https://stackoverflow.com/a/55357742/9203360 that didn't work, but would be ideal if it did.

推荐答案

正如我在评论中提到的， std :: get 以类型作为模板参数并返回该类型的元组元素仅在C ++ 14起可用。

As I mentioned in the comments, the overload of std::get taking a type as template argument and returning the tuple element of that type is available only since C++14.

根据您的评论，您并未针对C +进行编译+14或更高版本，因此没有匹配类型为第一个模板参数的 std :: get 。

Per your comment you were not compiling against C++14 or later and so there was no match for std::get with a type as first template argument.

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