如何使用模板推导类型获得右值和左值引用的不同重载? [英] How to get different overloads for rvalue and lvalue references with a template-deduced type?

问题描述

我有一个函数 foo 通过引用获取参数，我希望它对右值和左值引用有不同的作用.(我还应该提到 foo() 尊重常量性；它不会改变引用的值.)我知道如果我写:

I have a function foo taking a parameter by reference, and I want it to act differently on rvalue and on lvalue references. (I should also mention foo() respects constness; it doesn't change the refered-to value.) I know that if I write:

template <typename T> foo(T&& x);

我已声明转发参考，而不是右值引用，意思是这样:

I've declared a forwarding reference, rather than an rvalue reference, meaning that this way:

template <typename T> foo(const T& x);
template <typename T> foo(T&& x);

可能得不到我想要的.

所以，我的问题是:影响两种引用之间不同行为的正确方法是什么?

So, my question is: What's the right way to affect different behavior between the two kinds of references?

推荐答案

标签调度是最简单的解决方案.

Tag dispatching is the simplest solution.

namespace details {
  template <typename T>
  void foo(std::true_type is_lvalue, const T& x) {
    std::cout << x << " is an lvalue\n";
  }
  template <typename T>
  void foo(std::false_type is_lvalue, T&& x) {
    std::cout << x << " is an rvalue\n";
  }
}
template <typename T>
void foo(T&& t) {
  return details::foo(
    typename std::is_lvalue_reference<T>::type{},
    std::forward<T>(t)
  );
}

当您实际上不想支持为重载解析目的选择两个版本时，SFINAE 是严重的矫枉过正.

SFINAE is serious overkill when you don't actually want to support neither version being selected for overload resolution purposes.

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