模板复制构造函数 [英] template copy constructor
问题描述
给定下面的代码,Foo有复制构造函数吗?用STL容器使用Foo是否安全?
class Foo
{
public:
Foo(){}
template< typename T>
Foo(const T&){}
};
标准明确地说复制构造函数是一个非模板化的构造函数,它引用一个可能的相同类型的const volatile对象。在上面的代码中,您有一个转换而不是复制构造函数(即,它将用于所有
是的,隐式声明/定义的拷贝构造函数。
使用标准库容器安全使用
Foo
吗?
当前定义 Foo
是,但在一般情况下,它取决于成员 Foo
是否具有隐式定义的复制构造函数正确地管理它们。
Given the following code, does Foo have copy constructor? Is it safe to use Foo with STL containers?
class Foo
{
public:
Foo() {}
template <typename T>
Foo(const T&) {}
};
The standard explicitly says that a copy constructor is a non-templated constructor that takes a reference to a possibly const-volatile object of the same type. In the code above you have a conversion but not copy constructor (i.e. it will be used for everything but copies, where the implicitly declared constructor will be used).
Does
Foo
have a copy constructor?
Yes, the implicitly declared/defined copy constructor.
Is it safe to use
Foo
with standard library containers?
With the current definition of Foo
it is, but in the general case, it depends on what members Foo
has and whether the implicitly defined copy constructor manages those correctly.
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