2

What's going on here -- why does this not compile?

#include <iostream>
class Base {
    void print(double d) {
        std::cout << "Base: " << d << std::endl;
    }
};

class Derived : public Base {
void print(std::string const & str) {
        std::cout << "Derived: " << str << std::endl;
    }
};

int main(int argc, char* argv[]) {
    Derived d;
    d.print(2.);
    d.print("junk");
}

(Errors in both MinGW and VC11 with something equivalent to No conversion from double to std::string.)

If I change the name of the print function in Derived, it compiles successfully, so clearly Derived::print(string const &) is masking Base::print(double) somehow. But I was under the impression that the function signature included the argument type(s) and so this masking should occur here. Is that not correct in the case of base class methods?

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1 Answers1

3

No, it is not correct : name hiding occurs when a function with the same name is declared in the same scope : here print in Derived hides the print of Base.

Arguments names/types are ignored in this (first) step of functions lookup.

You can bring the declaration of the Base function within the Derived one with a using declaration:

using Base::print

If you do so, regular overload resolution will take place.

For details about why hiding occurs in this Derived/Base situation, I suggest this other SO post that perfectly answers it.

As for the standard, this particular rule is defined in section 3.3.1 :

Name hiding [basic.scope.hiding]

A name can be hidden by an explicit declaration of that same name in a nested declarative region or derived class (10.2).

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quantdev
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  • Also in the other SO post (further down) http://stackoverflow.com/a/1628786/1821972 shows how you could put class Derived : public Base { using Base::print; ... } to get what you originally expected. Excellent tutorial p.901-904 in http://www.amazon.com/Primer-3rd-Edition-Stanley-Lippman/dp/0201824701 – Joe Kul Jun 09 '14 at 21:50