c++ - 继承成员函数的模板特化

Question

我们想专门化基类的成员函数。但是，它不编译。有人知道可以编译的替代方案吗？

举个例子

struct Base
{
    template<typename T>
    void Foo()
    {
        throw "Foo() is not defined for this type";
    }
};

struct Derived : public Base
{
    template<>
    void Foo<int>() { cout << "Foo<int>()" << endl; } // compile error (cannot specialize members from a base class)

    template<>
    void Foo<double>() { cout << "Foo<double>()" << endl; }  // compile error (cannot specialize members from a base class)
};

Answer 1

最终，我们使用重载解决了它。

这是基类的样子

struct Base
{
    template<typename T>
    class OfType {}

    template<typename T>
    void Foo(OfType<T>) { static_assert(false, "Foo is not implemented for this type. Please look in the compiler error for more details."); }
};

struct Derived : public Base
{
    using Base::Foo;

    void Foo(OfType<int>) { // here comes logic for ints }
    void Foo(OfType<double>) { // here comes logic for doubles }
};

这是一个使用 Foo()

的客户端代码示例

template<typename S>
class ClassThatUsesFoo
{
    private: S s;

    template<typename T>
    void Bar(T item) 
    {  
        s.Foo(Base::OfType<T>());  // this is the code that uses Foo
        DoSomeStuffWithItem(item); 
    } 
};

void main()
{
    ClassThatUsesFoo<Derived> baz;
    baz.Bar(12); // this will internally use Foo for ints
    baz.Bar(12.0); // this will use Foo for doubles
    baz.Bar("hello world"); // this will give a verbose compile error
}