带有一个专门用于 C++ 的方法的模板类 [英] template class with a single method specialized in C++

问题描述

我只有一个用 C++ 编写的学校作业的 hpp 文件(我不允许添加 cpp 文件，声明和实现都应该写在文件中).

I only have an hpp file for a school assignment in C++ (I am not allowed to add a cpp file, declaration and implementation should be both written in the file).

我在里面写了这段代码:

I wrote this code inside it:

template<class T>
class Matrix
{
   void foo()
   {
       //do something for a T variable.
   }
};

我想添加另一个 foo 方法，但是这个 foo() 将专门用于一个 .我在某些地方读到过，我需要声明一个新的专业化类才能使其工作.但我想要的是专门的 foo 将位于原始 foo 的正下方，所以它看起来像这样:

I would like to add another foo method, but this foo() will be specialized for only an <int>. I have read in some places that I need to declare a new specialization class for this to work. But what I want is that the specialized foo will lie just beneath the original foo, so it will look like this:

template<class T>
class Matrix
{
   void foo(T x)
   {
       //do something for a T variable.
   }
   template<> void foo<int>(int x)
   {
       //do something for an int variable.
   }
};

• 为什么我收到此语法的错误('<' 标记之前的预期未限定 ID")?
• 为什么这不可能?
• 如何在不声明新的专用类的情况下解决此问题?

谢谢

推荐答案

foo 不是模板.它是模板的成员函数.因此 foo 是没有意义的.(此外，必须在命名空间范围内声明显式特化.)

foo isn't a template. It's a member function of a template. Thus foo<int> is meaningless. (Also, explicit specializations must be declared at namespace scope.)

您可以显式特化类模板的特定隐式实例化的成员函数:

You can explicitly specialize a member function of a particular implicit instantiation of a class template:

template<class T>
class Matrix
{
   void foo(T x)
   {
       //do something for a T variable.
   }
};

// must mark this inline to avoid ODR violations
// when it's defined in a header
template<> inline void Matrix<int>::foo(int x)
{
     //do something for an int variable.
}

这篇关于带有一个专门用于 C++ 的方法的模板类的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！