如何强制模板化匹配基类？ [英] How do you force a templatization to match a base class?

问题描述

我有一个模板函数显式实例化 Base 类，但不是 Derived 类。如何强制传递 Derived 类（或其他派生类）的用法与 Base 类匹配？

头文件：

  class Base {
} ; 
 class Derived：public Base {
}; 
 class Derived2：public Base {
}; 
 
 template< typename Example> void function（Example& arg）; 
  

执行文件：

  //显式实例化基类：
 template void function< Base>（Base& arg）; 
 
 //定义模板函数：
 template< typename Example> void function（Example& arg）{
 //做某事。因为我没有为显式实例化函数> Derived <$ c $ b> 
  


$ b 或 Derived2 ，我得到未定义的引用，但是我想绑定到 Base

如何强制模板使用C ++ - 03从Base派生到Base的所有对象解析为Base类？

我可以通过将 Derived 类专门化到 Base 类定义

解决方案

如何：

 code> template<> void function（Derived& arg）
 {
 function< Base>（arg）; 
} 
  

编辑：您也可以使用函数重载来执行此操作， aschepler 已建议：

  void function（Derived& arg）
 {
 function& Base>（arg）; 
} 
  

这在概念上是一样的， p>

I have a template function which is explicitly instantiated for Base class, but not for Derived class. How can I force the uses that pass a Derived class (or other derived classes) to match against the Base class?

Header file:

class Base {
};
class Derived : public Base {
};
class Derived2 : public Base {
};

template <typename Example> void function(Example &arg);

Implementation file:

// Explicitly instantiate Base class:
template void function<Base>(Base &arg);

// Define the template function:
template <typename Example> void function(Example &arg) {
  // Do something.
}

Because I have not explicitly instantiated function for Derived or Derived2, I get undefined references, however, I would like to bind against the Base class which is explicitly defined.

How can I force the template to resolve to the Base class for all objects derived from Base using C++-03?

Can I do it somehow with a specialization of the Derived class to the Base class definition?

解决方案

How about:

template <> void function(Derived &arg)
{
     function<Base>( arg );
}

EDIT: You can also do this with function overloading, as aschepler has suggested:

void function(Derived &arg)
{
     function<Base>( arg );
}

It's conceptually the same, although, I agree, slightly better :)

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