如何在 C++ 中的模板化类中为单个方法创建特化? [英] How to create specialization for a single method in a templated class in C++?

问题描述

已经提出了许多问题，它们与我将在这里提出的问题相似，但我认为它们并不相同.

Many questions have been asked and they are similar to the one I am going to ask here, but they are not the same I think.

我有一个模板类:

namespace app {
template <typename T>
class MyCLass {
public:
  void dosome();
  void doother();
}
} /*ns*/

和实现:

template <typename T>
app::MyClass<T>::dosome() {}
template <typename T>
app::MyClass<T>::doother() {}

当我有一个该类的实例时，将 char 作为模板参数提供给它，我希望函数 dosome() 以完全不同的方式运行.但我只是希望该函数的行为有所不同，其他一切都必须保持不变.

When I have an instance of that class to which a char is provided as template parameter, I want function dosome() to behave in a totally different way. But I just want that function to behave differently, everything else must still act the same.

我尝试输入:

template<>
app::MyCLass<char>::dosome() {
}

但是编译器告诉我我正在尝试在不同的命名空间中创建一个专业化.

But the compiler tells me that I am trying to create a specialization in a different namespace.

所以当我有这样的代码时:

So when I have a code like this:

app::MyCLass<int> a;
app::MyCLass<char> b;
a.dosome(); // This must call the generic template definition
b.dosome(); // This must call the specialization
a.doother(); // This must call the generic template definition
b.doother(); // This must call the generic template definition

在其他问题中，我看到人们为整个班级创造了完全不同的专业.但我只想要一种方法的特化.

In other questions I saw people creating totally different specialization of the entire class. But I only want a specialization of a single method.

推荐答案

你可以为所欲为:http://ideone.com/oKTFPC

//标题

namespace My
{

template <typename T>
class MyClass {
public:
    void dosome();
    void doother();
};

template <typename T> void MyClass<T>::dosome() {}
template <typename T> void MyClass<T>::doother() {}

template<> void MyClass<char>::dosome();

}

//cpp 或在标题中

// cpp or in header

template<>
void My::MyClass<char>::dosome() {
    std::cout << "specialization" << std::endl;
}

或使用替代符号

namespace My {
template<>
void MyClass<char>::dosome() {
    std::cout << "specialization" << std::endl;
}
}

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