Question

有人可以解释一下我做错了什么吗？这是我从编译器得到的错误。

非常感谢

1>------ Build started: Project: Ch16, Configuration: Release Win32 ------
1>  p643_inclusion.cpp
1>  p643_inclusion_main.cpp
1>  p643_print.cpp
1>p643_print.cpp(5): error C2065: 'T1' : undeclared identifier
1>p643_print.cpp(5): error C2065: 'T2' : undeclared identifier
1>p643_print.cpp(6): warning C4552: '<<' : operator has no effect; expected operator with side-effect
1>p643_print.cpp(7): warning C4552: '<<' : operator has no effect; expected operator with side-effect
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

p643_inclusion.h

#ifndef P643H
#define P643H

template< class T1, class T2> class Car { 

    friend void  print (const Car<T1, T2> &c1);
    private:
        T1 Wheels;
        T2 DriversName;
    public:
        Car(): Wheels(4), DriversName("None") {}
        Car(T1, T2);
};



template <class T1, class T2> class Driver {
    private:
        T1 Name;
        T2 Surname;
    public:
        Driver(): Name("None"), Surname("None") {}
};


#include "p643_inclusion.cpp"

#endif

p643_inclusion.cpp

# ifndef P643CC
#define P643CC

#include <iostream>
#include <string>
using std::string;
using std::cout;
using std::endl;
#include "p643_inclusion.h"

template<class T1, class T2>
Car<T1, T2>::Car(T1 w, T2 d) {
    Wheels = w;
    DriversName = d;
}

#endif

p643_print.cpp

#include "p643_inclusion.h"
template< class T1, class T2> class Car;

void  print (const Car<T1, T2> &c1) {
    cout << c1.Wheels << endl;
    cout << c1.DriversName << endl;
    }

主

#include "p643_inclusion.h"
#include<iostream>
#include<string>
using namespace std;


int main()
{

    Car<int, string> myCar;
    Driver<string, string> myDriver;

    print(myCar);

    return 0;
}

Answer 1

您的功能实际上不是非模板功能。

void  print (const Car<T1, T2> &c1) {
    cout << c1.Wheels << endl;
    cout << c1.DriversName << endl;
}

这是错误的。你能问问自己究竟是什么T1？和T2？

您应该将其实现为：

template<class T1, class T2>
void  print (const Car<T1, T2> &c1) {
    cout << c1.Wheels << endl;
    cout << c1.DriversName << endl;
}

你应该把它friend作为：

template< class T1, class T2> class Car { 

    //choose different name for type params, because enclosing class 
    //already using T1, and T2
    template<class U, class V>
    friend void  print (const Car<U, V> &c1);

    //...

Answer 2

混合模板和友谊并不总是像看起来那么简单。我的建议是你在类定义中定义了友好函数，然后你的问题基本上会消失：

template <typename T1, typename T2>
struct test {
   friend void print( test const & t ) { ... };
};

对于模板test的每个实例化，它将声明并定义一个（非模板化的）自由函数，该函数使用触发模板实例化的相同模板参数实例化test对象。

其他可用选项（如果可能，我会清除这些选项）：

您可以将print作为模板，并将该模板声明为您的班级模板的朋友（整个模板）：

template <typename T1, typename T2>
struct test {
   template <typename U, typename V>
   friend void foo( test<U,V> const & ); // befriend template, all instantiations
};
template <typename T1, typename T2>
void foo( test<X,Y> const & x ) {...}

这将打开您的内部模板的所有潜在实例化，包括可能的特化，您可能不想这样做。如果你想只与该模板的特定实例建立联系，你可以这样做，但它变得更加麻烦：

template <typename T1, typename T2> struct test;  // forward declaration

template <typename T1, typename T2>
void foo( test<T1,T2> const & );                  // forward declaration

template <typename T1, typename T2>
struct test {
   friend void foo<T1,T2>( test<T1,T2> const & ); // befriend specific instantiation
};

template <typename T1, typename T2>
void foo( test<T1,T2> const & x ) { ... }         // implement

有关进一步说明，您可以查看答案here

模板类的非模板功能朋友

2 个答案: