没有模板函数继承的 C++ 接口(interface)

Question

有两个不同的类具有交叉的方法集:

class A {
public:
    int Value1() { return 100; }
    char Value2() { return "a"; }
    void actionA() { };
}

class B {
public:
    int Value1() { return 200; }
    char Value2() { return "b"; }
    void actionB() { };
}

类接口(interface)的公共(public)部分可以这样描述:

class GenericPart {
public:
  virtual int Value1();
  virtual char Value2();
}

请注意，类 A 和 B 来自某个库，因此不能从 GenericPart 继承。

有一个模板函数可以与实现GenericPart中描述的方法的对象一起使用:

template <typename T>
void function(T record) {
    std::cout << record.Value1() << " " << record.Value2() << std::endl;
}

是否可以专门化此模板函数以使其仅接收与 GenericPart 匹配的对象？

Answer 1

您可以使用 C++20 功能:concepts & constraints ，但更简单的解决方案可能涉及 static_assert。解释见function中的注释

#include <type_traits>
#include <iostream>

class A {
public:
    int Value1() { return 100; }
    char Value2() { return 'a'; }
    void actionA() { };
};

class B {
public:
    int Value1() { return 200; }
    char Value2() { return 'b'; }
    void actionB() { };
};

class C  { // Has the required functions but with different return types
public:
    double Value1() { return 0.0; }
    double Value2() { return 1.0; }
};

class D  { // Has only one required function
public:
    int Value1() { return 300; }
};


template <typename T>
void function(T record) {
    // Check statically that T contains a `Value1()` that returns an int
    static_assert(std::is_same_v<int, decltype(record.Value1())>, "int Value1() required!");

    // Check statically that T contains a `Value2()` that returns an char
    static_assert(std::is_same_v<char, decltype(record.Value2())>, "char Value2() required!");

    std::cout << record.Value1() << " " << record.Value2() << std::endl;
}

int main()
{
    A a;
    B b;
    C c;
    D d;

    function(a); // Ok
    function(b); // Ok
    function(c); // causes static_assert to fail
    function(d); // causes static_assert to fail
}