問題描述

在回答這個問題時，關于嘗試構造一個可變參數(shù)轉發(fā)引用構造函數(shù)，只有在沒有其他構造函數(shù)時才應調用該構造函數(shù)有效的.也就是說，如果有:

In answering this question about trying to construct a variadic forwarding reference constructor that should only be called if no other constructor is valid. That is, if there was a:

C(const char*, size_t) { }                     // 1
template <typename... T, ???> C(T&&... ) { }   // 2

我們希望 C c1{"abc", 2}; 調用 (1)，盡管需要轉換，但是 C c2{1, 2, 3}; 調用 (2)，因為 (1) 不適用.

We'd want C c1{"abc", 2}; to call (1), despite the required conversion, but C c2{1, 2, 3}; to call (2), as (1) cannot apply.

我提出了以下解決方案:

I proposed the following solution:

template <typename... T,
          typename = std::enable_if_t<!std::is_constructible<C, T&&...>::value>
           >
C(T&&... ) { }

我的意思是說，我嘗試過并驚訝地發(fā)現(xiàn)它確實有效.它編譯并完成了我對 gcc 和 clang 的期望.但是，我無法解釋為什么它起作用，或者即使它實際上應該起作用，而且 gcc 和 clang 都特別包容.是嗎?為什么?

And by proposed, I mean, I tried it and was surprised to discover that it actually works. It compiles and does exactly what I had hoped for on both gcc and clang. However, I am at a loss to explain why it works or even if it's actually supposed to work and gcc and clang are both just being particularly accommodating. Is it? Why?

推薦答案

你的代碼的問題是我們只是在一個上下文中實例化了 is_constructible，它得到了錯誤的答案.模板代碼中的任何類型的緩存都可能導致錯誤——嘗試在調用構造函數(shù)后在相同的參數(shù)上打印 is_constructible！很可能會弄錯.

The issue with your code is that we just instantiated is_constructible in a context where it gets the answer wrong. Any kind of caching in the template code is likely to result in bugs -- try printing is_constructible on the same parameters after you call the constructor! It is likely to get it wrong.

現(xiàn)場示例，說明如何出錯.請注意，它聲稱 C 不能從 int& 構造，盡管在前一行已經(jīng)這樣做了.

Live example of how it can go wrong. Notice it claims C cannot be constructed from an int&, despite having done so on the previous line.

struct C {
  C(const char*, size_t) {}
  template <class... Ts,
    typename = std::enable_if_t<!std::is_constructible<C, Ts&&...>::value>
  >
  C(Ts&&... ) { }
};

int main() {
  int a = 0;
  C x{a};
  std::cout << std::is_constructible<C, int&>{} << '
';
}

糟糕.

我懷疑這可能是 ODR 違規(guī)——is_constructible 的兩個定義在不同的位置有不同的類型?或者也許不是.

I suspect this might be an ODR violation -- the two definitions of is_constructible have different types at different spots? Or maybe not.

沒有這個問題的原始問題的解決方案也貼出來了.

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