可变参数构造函数应该隐藏隐式生成的构造函数吗？ [英] Are variadic constructors supposed to hide the implicitly generated ones?

问题描述

可变参数构造函数应该隐藏隐式生成的构造函数，即默认构造函数和复制构造函数？

Are variadic constructors supposed to hide the implicitly generated ones, i.e. the default constructor and the copy constructor?

struct Foo
{
    template<typename... Args> Foo(Args&&... x)
    {
        std::cout << "inside the variadic constructor\n";
    }
};

int main()
{
    Foo a;
    Foo b(a);
}

不知何故，我希望在读取这个回答，但是它在里面打印可变参数构造函数在g ++上两次4.5.5 :(这是正确的行为吗？

Somehow I was expecting this to print nothing after reading this answer, but it prints inside the variadic constructor twice on g++ 4.5.0 :( Is this behavior correct?

没有可变参数模板时发生：

It also happens without variadic templates:

struct Foo
{
    Foo()
    {
        std::cout << "inside the nullary constructor\n";
    }

    template<typename A> Foo(A&& x)
    {
        std::cout << "inside the unary constructor\n";
    }
};

int main()
{
    Foo a;
    Foo b(a);
}

推荐答案

隐含声明的拷贝构造函数的声明实际上不是被抑制的。

Declaration of the implicitly declared copy constructor is not, in fact, being suppressed. It's just not being called due to the rules of overload resolution.

隐式声明的拷贝构造函数具有 Foo（const Foo&）。这个的重要部分是它需要一个const引用。您的构造函数模板采用非const引用。

The implicitly declared copy constructor has the form Foo(const Foo&). The important part of this is that it takes a const reference. Your constructor template takes a non-const reference.

a 不是const，声明的构造函数模板比​​隐式声明的拷贝构造函数优先。要调用隐式声明的复制构造函数，可以使 a const：

a is not const, so the non-const user-declared constructor template is preferred over the implicitly-declared copy constructor. To call the implicitly-declared copy constructor, you can make a const:

const Foo a;
Foo b(a);

或者您可以使用 static_cast const引用 a ：

or you can use static_cast to obtain a const reference to a:

Foo a;
Foo b(static_cast<const Foo&>(a));

描述这种情况的重载解析规则主要发现在C ++的§13.3.3.2/ + 0x FCD。这种特殊情况，包括左值和右值引用的组合，在第303页的各种示例中进行了描述。

The overload resolution rules that describe this are found mostly in §13.3.3.2/3 of the C++0x FCD. This particular scenario, with a combination of lvalue and rvalue references, is sort of described by the various examples on page 303.

可变参数构造函数模板将抑制隐式声明的默认构造函数，因为可变参数构造函数模板是用户声明的，并且仅当没有用户声明的构造函数（C ++ 0x FCD§12.1/ 5）时才提供隐式声明的默认构造函数：

A variadic constructor template will suppress the implicitly declared default constructor because a variadic constructor template is user-declared and the implicitly declared default constructor is only provided if there are no user-declared constructors (C++0x FCD §12.1/5):

如果类 X 没有用户声明的构造函数，没有参数隐式声明为默认。

If there is no user-declared constructor for class X, a constructor having no parameters is implicitly declared as defaulted.

可变参数构造函数模板不会抑制隐式声明的拷贝构造函数，因为只有一个非模板构造函数可以是复制构造函数（C ++ 0x FCD§12.8/ 2,3和8）：

A variadic constructor template will not suppress the implicitly declared copy constructor because only a non-template constructor can be a copy constructor (C++0x FCD §12.8/2, 3, and 8):

如果第一个参数是 X& ， const X& ; ， volatile X& 或 const volatile X& 参数或其他所有参数都有默认参数。

A non-template constructor for class X is a copy constructor if its first parameter is of type X&, const X&, volatile X& or const volatile X&, and either there are no other parameters or else all other parameters have default arguments.

X 的非模板构造函数是一个move构造函数如果其第一个参数是 X&& ， const X&&& ， volatile X&& 或 const volatile X&&& ，并且没有其他参数或所有其他参数都有默认参数。

A non-template constructor for class X is a move constructor if its first parameter is of type X&&, const X&&, volatile X&&, or const volatile X&&, and either there are no other parameters or else all other parameters have default arguments.

如果类定义没有显式声明一个复制构造函数，并且没有用户声明的move构造函数，那么一个复制构造函数被隐式声明为默认。

If the class definition does not explicitly declare a copy constructor and there is no user-declared move constructor, a copy constructor is implicitly declared as defaulted.

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