标准为何不将模板构造函数视为副本构造函数? [英] Why doesn't the standard consider a template constructor as a copy constructor?

问题描述

以下是复制构造函数的定义， [class.copy.ctor/1] :

Here's the definition of copy constructor, [class.copy.ctor/1]:

如果类X的第一个参数的类型为X& ;、 const X& ;、 volatile X&类型，则它是副本构造器.或const volatile X& ;，或者没有其他参数，或者所有其他参数都有默认参数([dcl.fct.default]).

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 ([dcl.fct.default]).

为什么标准不将模板作为复制构造函数?

Why does the standard exclude templates as copy constructors?

在这个简单的示例中，两个构造函数都是副本构造函数:

In this simple example, both constructors are copy constructors:

struct Foo {
    Foo(const Foo &); // copy constructor
    Foo(Foo &);       // copy constructor
};

请参阅以下类似示例:

struct Foo {
     Foo() = default;

     template <typename T>
     Foo(T &) {
         printf("here\n");
     }
};

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

在此示例中，将在此处进行打印.因此，看来我的模板构造函数是一个复制构造函数，至少它的行为类似于一个(在通常调用复制构造函数的上下文中被调用).

In this example, here will be printed. So it seems that my template constructor is a copy constructor, at least it behaves like one (it is called in a context where copy constructors are usually called).

为什么文本中存在非模板"要求?

Why is the "non-template" requirement there in the text?

推荐答案

让我们将模板放置一秒钟.如果一个类未声明副本构造函数，则将生成一个隐式默认的构造函数.可以将其定义为已删除，但是仍然是默认设置.

Let's put templates aside for a second. If a class doesn't declare a copy constructor, an implicitly defaulted one is generated. It may be defined as deleted, but it's defaulted nonetheless.

成员模板不是成员函数.仅在需要时才从中实例化成员.

A member template is not a member function. Members are instantiated from it only when needed.

那么，编译器如何仅从类定义中知道是否需要使用 T = Foo 进行专门化?不可以但这正是需要决定如何处理对隐式默认副本构造函数(AND move构造函数)的潜在需求的基础.变得混乱了.

So how can a compiler know from the class definition alone whether or not a specialization with T = Foo will ever be needed? It can't. But it's exactly that on which it needs to base a decision of how to handle a potential need for an implicitly defaulted copy constructor (AND move constructor). That becomes messy.

最简单的方法是排除模板.无论如何，我们总会有一些拷贝构造函数，它默认会做正确的事情 ^TM ，并且由于它不是从模板中实例化的，因此将受到重载解析的青睐.

The easiest approach is to exclude templates. We'll always have some copy constructor anyway, it will do the correct thing^TM by default, and will be favored by overload resolution because it's not instantiated from a template.

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