什么时候私有构造函数不是私有构造函数? [英] When is a private constructor not a private constructor?

问题描述

假设我有一个类型，并且想将其默认构造函数设为私有.我写以下内容:

Let's say I have a type and I want to make its default constructor private. I write the following:

class C {
    C() = default;
};

int main() {
    C c;           // error: C::C() is private within this context (g++)
                   // error: calling a private constructor of class 'C' (clang++)
                   // error C2248: 'C::C' cannot access private member declared in class 'C' (MSVC)
    auto c2 = C(); // error: as above
}

太好了.

但是随后，构造函数却不像我想象的那样私有:

But then, the constructor turns out to not be as private as I thought it was:

class C {
    C() = default;
};

int main() {
    C c{};         // OK on all compilers
    auto c2 = C{}; // OK on all compilers
}    

这让我感到非常惊奇，意外和明显不受欢迎的行为.为什么这样可以?

This strikes me as very surprising, unexpected, and explicitly undesired behavior. Why is this OK?

推荐答案

诀窍在C ++ 14 8.4.2/5 [dcl.fct.def.default]中:

The trick is in C++14 8.4.2/5 [dcl.fct.def.default]:

...如果函数是由用户声明的且未明确指定为默认值，则该函数为用户提供的或 在其第一个声明中删除. ...

... A function is user-provided if it is user-declared and not explicitly defaulted or deleted on its first declaration. ...

这意味着C的默认构造函数实际上不是不是用户提供的，因为它是在其第一个声明中明确默认的.因此，C没有用户提供的构造函数，因此根据8.5.1/1 [dcl.init.aggr]:

Which means that C's default constructor is actually not user-provided, because it was explicitly defaulted on its first declaration. As such, C has no user-provided constructors and is therefore an aggregate per 8.5.1/1 [dcl.init.aggr]:

aggregate 是一个数组或一个类(第9条)，没有用户提供的构造函数(12.1)，没有私有的或 受保护的非静态数据成员(条款11)，没有基类(条款10)和虚拟函数(10.3).

An aggregate is an array or a class (Clause 9) with no user-provided constructors (12.1), no private or protected non-static data members (Clause 11), no base classes (Clause 10), and no virtual functions (10.3).

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