带模板功能的Typedef [英] Typedef with template functions

问题描述

说我在命名空间A中有一个模板函数。我还有另一个命名空间B。在命名空间A中声明了一个模板函数，定义为

Say I have a template function in namespace A. I also have another namespace B. There is a template function declared in namespace A, which is defined as

template<typename T, typename U>
void f(T a, U b);

现在在命名空间B中，我想声明一种特殊类型的模板函数。我在想是否可以 typedef 模板函数，以便在命名空间B中将其声明为

Now in namespace B, I would want to declare a specialized type of the template function. I was thinking if I could typedef the template function so it is declared in namespace B as

void f(int a, double b);

实际上没有实现调用模板函数的函数。由于有一种方法可以使用特定的模板参数声明新的类型名，难道不应该有一种方法也可以通过函数来​​做到这一点吗？
我尝试了不同的方法来实现它，但是并没有完全解决问题。

without actually implementing the function calling the template function. As there is a way to declare new typenames with specific template parameters, shouldn't there be a way to do that with functions aswell? I tried different methods to achieve it, but it didn't quite work out.

所以C ++中已经存在一种用给定的函数重新声明函数的方法。模板参数而没有实际实现新功能？如果不是这样，它可以在C ++ 11中实现吗？

So is there already a way in C++ to redeclare the function with given template parameters without actually implementing a new function? If not, is it somehow achievable in C++11?

这将是一个简单的功能，因为它将使功能的目的更加清楚，并且语法上更好：）

It would be a neat feature to have since it would make the purpose of the function more clear and would be syntactically better :)

编辑：所以可以这样写：

So one could write:

using A::f<int, double>;

在B名称空间中，该函数将显示这些模板参数

in B namespace and the function would show up with those template parameters

推荐答案

您可以使用使用：

namespace A {
    template <typename T> void f(T);
    template <> void f<int>(int);    // specialization
}

namespace B {
    using ::A::f;
}

您无法区分专业化 （因为使用仅与名称有关），但是应该足以使所需的专业化可见。

You can't distinguish between the specializations like that (since using is only about names), but it should be enough to make the desired specialization visible.

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