问题
I have a base class, say BassClass
, with some fields, which I made them protected, and some pure virtual functions. Then the derived class, say DerivedClass
, like class DerivedClass : public BassClass
. Shouldn't DerivedClass inherit the protected fields from BassClass? When I tried to compile the DerivedClass, the compiler complains that DerivedClass does NOT have any of those fields, what is wrong here?
thanks
回答1:
If BassClass
(sic) and DerivedClass
are templates, and the BassClass
member you want to access from DerivedClass
isn't specified as a dependent name, it will not be visible.
E.g.
template <typename T> class BaseClass {
protected:
int value;
};
template <typename T> class DerivedClass : public BaseClass<T> {
public:
int get_value() {return value;} // ERROR: value is not a dependent name
};
To gain access you need to give more information. For example, you might fully specify the member's name:
int get_value() {return BaseClass<T>::value;}
Or you might make it explicit that you're referring to a class member:
int get_value() {return this->value;}
回答2:
This works:
#include <iostream>
struct Base {
virtual void print () const = 0;
protected:
int val;
};
struct Derived : Base {
void print () { std::cout << "Bases's val: " << val << std::endl; }
};
来源:https://stackoverflow.com/questions/1813671/problem-with-protected-fields-in-base-class-in-c