使用 enable_if 选择类构造函数

2021-12-13 00:00:00 constructor templates c++ sfinae

考虑以下代码:

#include <iostream>
#include <type_traits>

template <typename T>
struct A {
    int val = 0;

    template <class = typename std::enable_if<T::value>::type>
    A(int n) : val(n) {};
    A(...) { }

    /* ... */
};

struct YES { constexpr static bool value = true; };
struct NO { constexpr static bool value = false; };

int main() {
    A<YES> y(10);
    A<NO> n;
    std::cout << "YES: " << y.val << std::endl
              << "NO:  " << n.val << std::endl;
}

我想有选择地定义构造函数 A::A(int) 仅用于使用 enable_if 的某些类型.对于所有其他类型，默认构造函数 A::A(...) 应该是替换失败时编译器的默认情况.然而，这对我来说很有意义，编译器(gcc 版本 4.9.0 20130714)仍在抱怨

I want to selectively define constructor A::A(int) only for some types using enable_if. For all other types there is default constructor A::A(...) which should be the default case for compiler when substitution fails. However this makes sense for me compiler (gcc version 4.9.0 20130714) is still complaining

sfinae.cpp:在结构 A"的实例化中:sfinae.cpp:19:11:
从这里需要 sfinae.cpp:9:5: 错误:
中没有名为type"的类型'struct std::enable_if'
A(int n) : val(n) {};

sfinae.cpp: In instantiation of 'struct A': sfinae.cpp:19:11:
required from here sfinae.cpp:9:5: error: no type named 'type' in
'struct std::enable_if'
A(int n) : val(n) {};

这样的事情对构造函数来说是可能的吗?这是否可以与另一个构造函数(复制构造函数和移动构造函数)一起使用?

Is something like this possible for constructor? Is this possible with another constructor(s) (copy-constructor and move-constructor)?

推荐答案

With C++20
您只需将 requires 添加到模板即可实现:

With C++20

You can achieve that simply by adding requires to the template:

template <typename U = T> requires U::value A(int n) : val(n) { }

requires 子句获取一个 常量表达式，其计算结果为 true 或 false> 因此决定是否在重载决议中考虑此方法，如果 requires 子句为真，否则忽略它.

The requires clause gets a constant expression that evaluates to true or false deciding thus whether to consider this method in the overload resolution, if the requires clause is true, or ignore it otherwise.

代码:https://godbolt.org/z/CKTDFE

相关文章