模板等效还是模板功能等效? [英] Template equivalence or template functional equivalence?

问题描述

在C++标准[temp.over.link]中，解释了函数模板等价性的确定不应涉及编译器的英勇努力".

In the C++ standard [temp.over.link], it is explained that the determination of function template equivalence should not involve "heroic efforts" of the compiler.

举个例子，C++ 标准提出了这个:

As an example, the C++ standard propose this:

// guaranteed to be the same
template <int I> void f(A<I>, A<I+10>);
template <int I> void f(A<I>, A<I+10>);
// guaranteed to be different
template <int I> void f(A<I>, A<I+10>);
template <int I> void f(A<I>, A<I+11>);
// ill-formed, no diagnostic required
template <int I> void f(A<I>, A<I+10>);
template <int I> void f(A<I>, A<I+1+2+3+4>);

此规则是否也适用于涉及元编程的情况，如下例所示?

Is this rule also applies to cases involving meta programming, as in the example below?

template<class T>
struct t_{
   using type = T;
   };
//ill-formed or different?
template<class T> T f(T);
template<class T> typename t_<T>::type f(T);

推荐答案

让我们从简单的案例开始:

Let's start with the simple case:

template <typename T> using id = T;

template<class T> T f(T);
template<class T> id<T> f(T);

根据 相关规则.这两个声明在功能上是等效的，但不是简单地重命名模板参数(并且没有任何依赖名称需要考虑).

This to me is clearly ill-formed, no diagnostic required, per the relevant rule. The two declarations are functionally equivalent, but not in a way that's simply renaming template parameters (and there aren't any dependent names to consider).

更复杂的情况:

template <typename T> struct id_t { using type = T; };
template <typename T> using id = typename id_t<T>::type;

template<class T> T f(T);
template<class T> id<T> f(T);

我认为这可能不是格式错误，因为它们在功能上并不真正等效.可能存在 id_t 的特化，使得这两个声明实际上是不同的 - 所以这些实际上是不同的声明.

I think this is probably not ill-formed, because they aren't truly functionally equivalent. There could be specializations of id_t such that these two declarations are actually different - so these are actually different declarations.

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