C++11:如何给函数取别名? [英] C++11: How to alias a function?

问题描述

如果我在命名空间栏中有一个 Foo 类:

If I have a class Foo in namespace bar:

namespace bar
{
    class Foo { ... }
};

然后我可以:

using Baz = bar::Foo;

现在就像我在命名空间中定义了名为 Baz 的类一样.

and now it is just like I defined the class in my namespace with the name Baz.

是否可以对函数做同样的事情?

Is it possible to do the same for functions?

namespace bar
{
    void f();
}

然后:

using g = bar::f; // error: ‘f’ in namespace ‘bar’ does not name a type

最干净的方法是什么?

该解决方案也应适用于模板函数.

The solution should also hold for template functions.

定义:如果某个实体 B 是 A 的别名，那么如果 A 的任何或所有用法(当然不是声明​​或定义)在源代码比(剥离)生成的代码保持不变.例如 typedef A B 是一个别名.#define B A 是一个别名(至少).<代码>T&B = A 不是别名，B 可以有效地实现为间接指针，而未别名"的 A 可以使用立即语义".

Definition: If some entity B is an alias of A, than if any or all usages (not declarations or definitions of course) of A are replaced by B in the source code than the (stripped) generated code remains the same. For example typedef A B is an alias. #define B A is an alias (at least). T& B = A is not an alias, B can effectively implemented as an indirect pointer, wheres an "unaliased" A can use "immediate semantics".

推荐答案

您可以使用完美转发定义函数别名(需要一些工作):

You can define a function alias (with some work) using perfect forwarding:

template <typename... Args>
auto g(Args&&... args) -> decltype(f(std::forward<Args>(args)...)) {
  return f(std::forward<Args>(args)...);
}

即使 f 重载和/或函数模板，此解决方案也适用.

This solution does apply even if f is overloaded and/or a function template.

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