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Home/ Questions/Q 7793317
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Editorial Team
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Editorial Team
Asked: June 1, 20262026-06-01T22:26:57+00:00 2026-06-01T22:26:57+00:00

To me it looks perfectly safe to cast a void(Derived::*)() to a void(Base::*)() ,

  • 0

To me it looks perfectly safe to cast a void(Derived::*)() to a void(Base::*)(), like in this code:

#include <iostream>
#include <typeinfo>
using namespace std;
struct Base{
    void(Base::*any_method)();
    void call_it(){
        (this->*any_method)();
    }
};
struct Derived: public Base{
    void a_method(){
        cout<<"method!"<<endl;
    }
};
int main(){
    Base& a=*new Derived;
    a.any_method=&Derived::a_method;
    a.call_it();
}

But the compiler complains about the cast at a.any_method=&Derived::a_method;. Is this a roadblock to prevent subtle programming errors, or just something to make life easier for compiler writers? Are there workarounds to let the Base class have a pointer to member functions of Derived without type knoweledge (that is, I cannot make Base a template with template argument Derived).

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  1. Editorial Team
    Editorial Team
    2026-06-01T22:26:58+00:00Added an answer on June 1, 2026 at 10:26 pm

    What happens if your Derived::a_method() attempts to use a data member only present in Derived, not in Base, and you call it on a Base object (or an object derived from Base but not related to Derived)?

    The conversion the other way around makes sense, this one doesn’t.

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