問題描述

考慮以下代碼:

#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;
}

我想有選擇地定義構造函數(shù) A::A(int) 僅用于使用 enable_if 的某些類型.對于所有其他類型，默認構造函數(shù) A::A(...) 應該是替換失敗時編譯器的默認情況.然而，這對我來說很有意義，編譯器(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:在結構 A"的實例化中:sfinae.cpp:19:11:
從這里需要 sfinae.cpp:9:5: 錯誤:
中沒有名為type"的類型'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) {};

這樣的事情對構造函數(shù)來說是可能的嗎?這是否可以與另一個構造函數(shù)(復制構造函數(shù)和移動構造函數(shù))一起使用?

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

推薦答案

With C++20

您只需將 requires 添加到模板即可實現(xiàn):

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 子句獲取一個 常量表達式，其計算結果為 true 或 false> 因此決定是否在重載決議中考慮此方法，如果 requires 子句為真，否則忽略它.

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.

代碼:https://godbolt.org/z/CKTDFE

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