c++ Inheritance with templates and visibility

Question

I don't understand att all inheritance with templates..

template <typename T> 
class Mere 
{ 
protected:
    Mere();
}; 

class Fille2 : public Mere<int>
{ 
protected:
    Fille2(){
        Mere();
    }
}; 


int main(int argc, char** argv) {

    return 0;
}

Why do I have this error?

main.cpp:22:5: error: 'Mere<T>::Mere() [with T = int]' is protected
Mere();

And all works when I put "Mere()" in public? I can't have "protected" functions for my class "Mere"? Why?

this is not how you invoke constructors of a parent class in c++... see [here](http://melpon.org/wandbox/permlink/DtiZ3q60YyVytvju) on how it can be done — W.F., Dec 13 '16 at 20:37

score 3 · Accepted Answer · edited May 23 '17 at 10:30

3

Yes, you can call the base class constructor, even if it is protected. Here is the correct syntax:

class Fille2 : public Mere<int>
{ 
protected:
    Fille2(): Mere() {
    }
};

For a detailed discussion, see Why is protected constructor raising an error this this code?

edited May 23 '17 at 10:30

Community

1
1

answered Dec 13 '16 at 20:37

NPE

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c++ Inheritance with templates and visibility

1 Answers1