我有一个问题,如果我有一个模板类,它又有一个模板方法,它接受该类的另一个实例的参数(具有不同的模板参数),它不能访问被传递的类的受保护或私有成员一个参数,例如：
template<typename T>class MyClass
{
T v;
public:
MyClass(T v):v(v){}
template<typename T2>void foo(MyClass<T2> obj)
{
std::cout << v << " ";
//error C2248: 'MyClass<T>::v' : cannot access private member declared in class 'MyClass<T>'
std::cout << obj.v << " ";
std::cout << v + obj.v << std::endl;
}
};
int main()
{
MyClass<int> x(5);
MyClass<double> y(12.3);
x.foo(y);
}
有没有可以说MyClass< T>中的方法有完全访问MyClass< SomeOtherT>？
解决方法
它们是不同的类型：模板从模板构建新类型.
你必须对你的班级朋友做其他的例子：
template <typename T>class MyClass
{
T v;
public:
MyClass(T v):v(v){}
template<typename T2>void foo(MyClass<T2> obj)
{
std::cout << v << " ";
std::cout << obj.v << " ";
std::cout << v + obj.v << std::endl;
}
// Any other type of MyClass is a friend.
template <typename U>
friend class MyClass;
// You can also specialize the above:
friend class MyClass<int>; // only if this is a MyClass<int> will the
// other class let us access its privates
// (that is,when you try to access v in another
// object,only if you are a MyClass<int> will
// this friend apply)
};
