enable_if使用constexpr bool测试不起作用 [英] enable_if using a constexpr bool test not working

问题描述

我有一个数学函数，希望能够接受双精度数或双精度数的数组/向量/容器，并且行为略有不同。

I have a maths function that I want to be able to accept either a double, or a array/vector/container of doubles, and behave slightly differently.

我正在尝试使用SFINAE并键入traits选择正确的函数。

I am attempting to use SFINAE and type traits to select the correct function.

这里是一个最小的示例：

Here is a minimal example:

#include <iostream>
#include <vector>
#include <type_traits>

template <typename T>
constexpr bool Iscontainer()
{
    if constexpr (std::is_class<T>::value && std::is_arithmetic<typename T::value_type>::value) {
        return true;
    }
    return false;
}

// Function 1 (double):
template <typename T>
typename std::enable_if<std::is_arithmetic<T>::value>::type g(T const & t)
{
    std::cout << "this is for a double" << t << std::endl;
}

// Function 2 (vec), version 1:
template <typename T>
typename std::enable_if<IsContainer<T>()>::type g(T const & t)
{
    std::cout << "this is for a container" << t[0] << std::endl;
}

int main()
{
    std::vector<double> v {1, 2};
    std::array<double, 2> a {1, 2};
    double d {0.1};

    g<>(v);
    g<>(a);
    g<>(d);  // error here
}

我得到一个编译时错误：

I get a compile time error:

../main.cpp:8:47: error: ‘double’ is not a class, struct, or union type
     if constexpr (std::is_class<T>::value && std::is_arithmetic<typename     T::value_type>::value) {
                   ~~~~~~~~~~~~~~~~~~~~~~~~^~~~~~

但是，当我更换时函数2具有：

However, when I replace function 2 with:

// Function 2 (vec), version 2:
template <typename T>
typename std::enable_if<std::is_class<T>::value && std::is_arithmetic<typename T::value_type>::value>::type
g(T const & t)
{
    std::cout << "this is for a vector" << t[0] << std::endl;
}

有效。

我的问题是我不明白为什么第一个版本不起作用。.
而且我更喜欢第一个版本的可读性。

My problem is I don't understand why the first version does not work.. And I prefer the readability of the first version.

推荐答案

失败的原因很简单。您没有调用SFINAE，并且当编译器尝试评估表达式时，它会看到：

The reason why it fails is simple. You do not invoke SFINAE, and when the compiler tries to evaluate the expressions it sees:

if constexpr (std::is_class<double>::value // this is fine it's false
   && std::is_arithmetic<typename double::value_type>::value // problem here!
)

整个语句都被求值，如果if没有短路。与您当前最接近的解决方案是显式拆分 if ，以便在 T 时丢弃有问题的部分

The whole statement is evaluated, there is no short-circuit for the if. The closest solution to what you currently have is to explicitly split the if, so that the problematic part is discarded when T is not a class and the second check is nonsensical.

#include <iostream>
#include <vector>
#include <type_traits>

template <typename T>
constexpr bool IsVector()
{
    if constexpr (std::is_class<T>::value) {
        if constexpr (std::is_arithmetic<typename T::value_type>::value) {
            return true;
        }
    }
    return false;
}

// Function 1 (double):
template <typename T>
typename std::enable_if<std::is_arithmetic<T>::value>::type g(T const & t)
{
    std::cout << "this is for a double" << t << std::endl;
}

// Function 2 (vec), version 1:
template <typename T>
typename std::enable_if<IsVector<T>()>::type g(T const & t)
{
    std::cout << "this is for a vector" << t[0] << std::endl;
}

int main()
{
    std::vector<double> v {1, 2};
    double d {0.1};

    g<>(v);
    g<>(d);  // error here
}

或者我建议使用别名：

template <typename T>
using IsVector2 = std::conjunction<typename std::is_class<T>, std::is_arithmetic<typename T::value_type> >;

template <typename T>
typename std::enable_if<IsVector2<T>::value>::type g(T const & t)
{
    std::cout << "this is for a vector" << t[0] << std::endl;
}

您也可以更好地命名。它并不会真正检查 T 是向量还是容器（在您编辑后）。您当前的定义也有点宽松。

You could also name it better. It doesn't really check whether T is a vector, or a container (after your edit). Your current definition is a bit loose as well.

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