C ++ enable

C ++ enable_if可以有默认实现吗？(Can C++ enable_if have a default implementation?)

我想编写一个函数print ，根据其参数的类型，它的行为会有所不同。

这是我的实施：

template <typename T, typename std::enable_if<std::is_array<T>::value, int>::type = 0> void print(const T &v) { std::cout << "array: "; for (const auto &e : v) { std::cout << e << ", "; } std::cout << std::endl; } template <typename T, typename std::enable_if<std::is_integral<T>::value, int>::type = 0> void print(const T &v) { std::cout << "integral: " << v << std::endl; } template <typename T, typename std::enable_if<!(std::is_array<T>::value || std::is_integral<T>::value), int>::type = 0> void print(const T &v) { std::cout << "default: " << v << std::endl; }

此代码按预期工作，但最后一个规范中的条件过于复杂。

有没有解决方案来简化最后一个？

I want to write a function print which behaves differently according to the type of its argument.

Here is my implementation:

template <typename T, typename std::enable_if<std::is_array<T>::value, int>::type = 0> void print(const T &v) { std::cout << "array: "; for (const auto &e : v) { std::cout << e << ", "; } std::cout << std::endl; } template <typename T, typename std::enable_if<std::is_integral<T>::value, int>::type = 0> void print(const T &v) { std::cout << "integral: " << v << std::endl; } template <typename T, typename std::enable_if<!(std::is_array<T>::value || std::is_integral<T>::value), int>::type = 0> void print(const T &v) { std::cout << "default: " << v << std::endl; }

This code works as expected, but the conditions in the last specification are too complicated.

Is there any solution to simplify the last one?

最满意答案

可以用于默认情况的一般方法是使用一个带有变量参数列表的函数。 只有在没有其他功能匹配时才会使用此选项。 这是一个例子：

template <typename T, typename std::enable_if<std::is_array<T>::value, int>::type = 0> void print_helper(const T &v,int) { std::cout << "array: "; for (const auto &e : v) { std::cout << e << ", "; } std::cout << std::endl; } template <typename T, typename std::enable_if<std::is_integral<T>::value, int>::type = 0> void print_helper(const T &v,int) { std::cout << "integral: " << v << std::endl; } template <typename T> void print_helper(const T &v,...) { std::cout << "default: " << v << std::endl; } template <typename T> void print(const T &v) { print_helper(v,0); }

对于只有两个重载，额外的功能可能不值得，但是当你获得更多的重载时，这个表单确实可以为默认情况付出代价。

A general approach you can use for a default case is to have a function which takes a variable argument list. This will only be used if no other function matches. Here is an example:

template <typename T, typename std::enable_if<std::is_array<T>::value, int>::type = 0> void print_helper(const T &v,int) { std::cout << "array: "; for (const auto &e : v) { std::cout << e << ", "; } std::cout << std::endl; } template <typename T, typename std::enable_if<std::is_integral<T>::value, int>::type = 0> void print_helper(const T &v,int) { std::cout << "integral: " << v << std::endl; } template <typename T> void print_helper(const T &v,...) { std::cout << "default: " << v << std::endl; } template <typename T> void print(const T &v) { print_helper(v,0); }

For only two overloads, the extra function may not be worth it, but as you get more overloads this form can really pay off for the default case.