如何将std :: enable_if与自推断返回类型一起使用？ [英] How do I use std::enable_if with a self-deducing return type?

问题描述

C ++ 14 将具有其推导类型可以推断出的函数根据返回值。

C++14 will have functions whose return type can be deduced based on the return value.

auto function(){
    return "hello world";
}

我可以将此行为应用于使用 enable_if 通过返回类型成语用于SFINAE吗？

Can I apply this behaviour to functions that use enable_if for the SFINAE by return type idiom?

例如，让我们考虑以下两个功能：

For example, let's consider the following two functons:

#include <type_traits>
#include <iostream>

//This function is chosen when an integral type is passed in
template<class T >
auto function(T t) -> typename std::enable_if<std::is_integral<T>::value>::type {
    std::cout << "integral" << std::endl;
    return;
}

//This function is chosen when a floating point type is passed in
template<class T >
auto function(T t) -> typename std::enable_if<std::is_floating_point<T>::value>::type{
    std::cout << "floating" << std::endl;
    return;
}

int main(){

  function(1);    //prints "integral"
  function(3.14); //prints "floating"

}

如您所见，使用SFINAE通过返回类型惯用法选择正确的函数。
但是，这两个都是无效函数。 enable_if 的第二个参数默认设置为 void 。将会是相同的：

As you can see, the correct function is chosen using the SFINAE by return type idiom. However, these are both void functions. The second parameter of enable_if is default set to void. This would be the same:

//This function is chosen when an integral type is passed in
template<class T >
auto function(T t) -> typename std::enable_if<std::is_integral<T>::value, void>::type {
    std::cout << "integral" << std::endl;
    return;
}

//This function is chosen when a floating point type is passed in
template<class T >
auto function(T t) -> typename std::enable_if<std::is_floating_point<T>::value, void>::type{
    std::cout << "floating" << std::endl;
    return;
}

我可以对这两个函数执行某些操作，以便它们的返回类型是由返回值推导出来的？

Is there something I can do to these two functions, so that their return type is deduced by the return value?

gcc 4.8.2（使用-std = c ++ 1y ）

gcc 4.8.2 (using --std=c++1y)

推荐答案

std :: enable_if 不必位于返回类型，从C ++ 11开始，它可以是模板参数的一部分。

std::enable_if doesn't have to be in the return type, as of C++11 it can be part of the template parameters.

因此，您的等效函数可以是（或者，可以达到这种效果）：

So your equivalent functions can be (or, well, something to this effect):

enum class enabler_t {};

template<typename T>
using EnableIf = typename std::enable_if<T::value, enabler_t>::type;

//This function is chosen when an integral type is passed in
template<class T, EnableIf<std::is_integral<T>>...>
auto function(T t) {
    std::cout << "integral" << std::endl;
    return;
}

//This function is chosen when a floating point type is passed in
template<class T, EnableIf<std::is_floating_point<T>>...>
auto function(T t) {
    std::cout << "floating" << std::endl;
    return;
}

它也可以是函数中的参数：

It can also be a parameter in the function:

//This function is chosen when an integral type is passed in
template<class T>
auto function(T t, EnableIf<std::is_integral<T>>* = nullptr) {
    std::cout << "integral" << std::endl;
    return;
}

//This function is chosen when a floating point type is passed in
template<class T>
auto function(T t, EnableIf<std::is_floating_point<T>>* = nullptr) {
    std::cout << "floating" << std::endl;
    return;
}

这将保留自动类型推断和SFINAE。

This will keep the automatic type deduction and SFINAE.

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