如果我只想在模板中专门化一种方法，我该怎么

问题描述

假设我有一个模板类

template <typename T> struct Node
{
    // general method split
    void split()
    {
        // ... actual code here (not empty)
    }
};

需要在 Triangle 类的情况下专门化这个......类似

Need to specialise this in the Triangle class case.. something like

template <>
struct Node <Triangle*>
{
    // specialise the split method
    void split() {}
} ;

但我不想想重新重写整个模板！唯一需要更改的是 split() 方法，仅此而已.

but I don't want to rewrite the entire template over again! The only thing that needs to change is the split() method, nothing more.

推荐答案

您可以在类声明之外仅为该函数提供专门化.

You can provide a specialization for only that function outside the class declaration.

template <typename T> struct Node
{
    // general method split
    void split()
    {
        // implementation here or somewhere else in header
    }
};

//cpp中声明的函数原型void splitIntNode( Node & node );

// prototype of function declared in cpp void splitIntNode( Node & node );

template <>
void Node<int>::split()
{
     splitIntNode( this ); // which can be implemented
}

int main(int argc, char* argv[])
{
   Node <char> x;
   x.split(); //will call original method
   Node <int> k;
   k.split(); //will call the method for the int version
}

如果 splitIntNode 需要访问私有成员，您可以将这些成员传递给函数而不是整个节点.

If splitIntNode needs access to private members, you can just pass those members into the function rather than the whole Node.

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