英文:
how to create diffrent class implemantations based off template
问题
以下是您提供的代码的翻译部分：
在我的animal.hpp文件中，我有这个模板类声明：
enum class ANIMAL_TYPE {DOG,CAT,};template <ANIMAL_TYPE T>class animal {public:void makeSound();};
这是一个经典的示例，用于解释接口和继承，但我希望它在编译时知道，并且不是虚函数，因此我在animal.cpp文件中实现如下：
template <>class Animal<ANIMAL_TYPE::DOG> {public:void makeSound() {std::cout << "woof";}};template <>class Animal<ANIMAL_TYPE::CAT> {public:void makeSound() {std::cout << "MEOW";}};template class Animal<ANIMAL_TYPE::DOG>;template class Animal<ANIMAL_TYPE::CAT>;
然后，在主文件中，我这样使用它：
#include <animal.hpp>int main() {Animal<ANIMAL_TYPE::DOG> doggy;doggy.makeSound();return 0;}
但是，当我编译和链接时，我得到了一个链接错误："undefined reference to Animal<(ANIMAL_TYPE)0>::makeSound"。
英文:
Let's say I have this template class declared in my animal.hpp file:
enum class ANIMAL_TYPE{DOG,CAT,};template <ANIMAL_TYPE T>class animal{public:void makeSound();};
Pretty classic example used in explanations of interfaces and inheritance, but I want it to be known at compile time, and not be a virtual function, so I implemented it like this in the animal.cpp file:
template <>class Animal<ANIMAL_TYPE::DOG>{public:void makeSound(){std::cout << "woof";}};template <>class Animal<ANIMAL_TYPE::CAT>{public:void makeSound(){std::cout << "MEOW";}};template class Animal<ANIMAL_TYPE::DOG>;template class Animal<ANIMAL_TYPE::CAT>;
And this in the main file:
#include <animal.hpp>int main(){Animal<ANIMAL_TYPE::DOG> doggy;doggy.makeSound();return 0;}
But when I compile and link this, I get an error during linking "undefined reference to Animal<(ANIMAL_TYPE)0>::makeSound"
答案1
得分: 0
以下是翻译的内容：
而不是实例化类本身，您应该实例化方法。换句话说，不要这样做：
template <typename>class Animal<ANIMAL_TYPE::DOG>{public:void makeSound(){std::cout << "woof";}};template <typename>class Animal<ANIMAL_TYPE::CAT>{public:void makeSound(){std::cout << "MEOW";}};
您应该这样做：
template <>void Animal<ANIMAL_TYPE::DOG>::makeSound(){std::cout << "woof";}template <>void Animal<ANIMAL_TYPE::CAT>::makeSound(){std::cout << "MEOW";}
此外，我注意到您在animal.cpp底部有以下内容：
template class Animal<ANIMAL_TYPE::DOG>;template class Animal<ANIMAL_TYPE::CAT>;
在这种情况下，它们没有任何效果，您可以将它们删除。
编辑： 好的，如果您想要有两个模板参数，一个是类型，一个不是类型，您需要在头文件中直接实例化类，因为它仍然是一个模板，而不是一个完整的实例化。
例如，假设之前的Animal类接受一个类型参数。
template <ANIMAL_TYPE T, typename R>class Animal{public:void makeSound(R foo);};
您将不得不像这样实例化它（请记住将其放在头文件中，因为它仍然是一个模板，而不是一个完整的实例化）。
template <typename R>class Animal<ANIMAL_TYPE::DOG, R>{public:void makeSound(R foo){std::cout << "woof";}};template <typename R>class Animal<ANIMAL_TYPE::CAT, R>{public:void makeSound(R foo){std::cout << "MEOW";}};
希望这对您有所帮助。
英文:
Instead of instantiating the classes themselves, you should be instantiating the methods. In other words, instead of doing this:
template <>class Animal<ANIMAL_TYPE::DOG>{public:void makeSound(){std::cout << "woof";}};template <>class Animal<ANIMAL_TYPE::CAT>{public:void makeSound(){std::cout << "MEOW";}};
You should be doing this:
template <>void Animal<ANIMAL_TYPE::DOG>::makeSound(){std::cout << "woof";}template <>void Animal<ANIMAL_TYPE::CAT>::makeSound(){std::cout << "MEOW";}
Also, I've noticed that you have this at the bottom of animal.cpp.
template class Animal<ANIMAL_TYPE::DOG>;template class Animal<ANIMAL_TYPE::CAT>;
Those have no effect in this case, you can just remove them.
EDIT: Okay, so if you want to have two template arguments, one of which is a type, and one that is not a type, you will need to directly instantiate the class, in the header because it's still a template, not a complete instantiation.
For example, let's say that the Animal class from before takes in a type argument.
template <ANIMAL_TYPE T, typename R>class Animal{public:void makeSound(R foo);};
You will have to instantiate it like this (Remember to put this in the header file, as it is still a template and not a complete instantiation).
template <typename R>class Animal<ANIMAL_TYPE::DOG, R>{public:void makeSound(R foo){std::cout << "woof";}};template <typename R>class Animal<ANIMAL_TYPE::CAT, R>{public:void makeSound(R foo){std::cout << "MEOW";}};
Hopefully, this is helpful.
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