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Home/ Questions/Q 6693427
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T05:58:53+00:00 2026-05-26T05:58:53+00:00

The following code compiles and works on G++ 4.4.0 and MS VC2008 Express. #include

  • 0

The following code compiles and works on G++ 4.4.0 and MS VC2008 Express.

#include <iostream>

template<typename T> struct A{
protected:
    T v;
public:
    const T get() const{
        return v;
    }

    A(T v_)
    :v(v_){
    }
};

class B: public A<const int*>{
public:
    void doSomething() const{
        const int* tmp = get();
        std::cout << *tmp << std::endl;
    }

    B(const int* p)
    :A<const int*>(p){
    }
};

int main(int argc, char** argv){
    int a = 134;
    B b(&a);
    const B& c = b;
    b.doSomething();
    c.doSomething();
    return 0;
}

However, as I understand it using A<const int*> should result in const const int* A::get() const;. And I’m pretty sure I haven’t seen anything like that in real code. Is using template in such way “legal”?

If not, what are alternatives? In my code I need a template class that provides two “getter” methods (const/non-const), and can take a “const int*” as a type. Something like this:

template<typename T> struct A{
protected:
    T v;
public:
    const T get() const{
        return v;
    }

    T get(){
        return v;
    }

    A(T v_)
    :v(v_){
    }
};

Any ideas?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T05:58:54+00:00Added an answer on May 26, 2026 at 5:58 am

    If T = const int *, then const T is const int * const.

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