复制智能指针的构造函数 [英] Copy constructor with smart pointer
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问题描述
我有一个类 std :: unique_ptr
作为类成员。我想知道如何正确定义复制构造函数,因为我得到以下编译器错误消息:错误C2248:std :: unique_ptr< _Ty> :: unique_ptr:无法访问在类中声明的私有成员'std :: unique_ptr< _Ty>
。我的类设计看起来像:
I have a class with one std::unique_ptr
as class member. I was wondering, how to correctly define the copy constructor, since I'm getting the following compiler error message: error C2248: std::unique_ptr<_Ty>::unique_ptr : cannot access private member declared in class 'std::unique_ptr<_Ty>
. My class design looks something like:
template <typename T>
class Foo{
public:
Foo(){};
Foo( Bar<T> *, int );
Foo( const Foo<T> & );
~Foo(){};
void swap( Foo<T> & );
Foo<T> operator = ( Foo<T> );
private:
std::unique_ptr<Bar> m_ptrBar;
int m_Param1;
};
template < typename T >
Foo<T>::Foo( const Foo<T> & refFoo )
:m_ptrBar(refFoo.m_ptrBar),
m_Param1(refFoo.m_Param1)
{
// error here!
}
template < typename T >
void Foo<T>::swap( Foo<T> & refFoo ){
using std::swap;
swap(m_ptrBar, refFoo.m_ptrBar);
swap(m_Param1, refFoo.m_Param1);
}
template < typename T >
Foo<T> Foo<T>::operator = ( Foo<T> Elem ){
Elem.swap(*this);
return (*this);
}
推荐答案
假设目标是复制 - 构造独特拥有的Bar,
Assuming the goal is to copy-construct the uniquely-owned Bar,
template < typename T >
Foo<T>::Foo( const Foo<T> & refFoo )
: m_ptrBar(refFoo.m_ptrBar ? new Bar(*refFoo.m_ptrBar) : nullptr),
m_Param1(refFoo.m_Param1)
{
}
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