The overload that hid the other one

#include <iostream>

struct Base {
    void f(double x) { std::cout << "base double: " << x << '\n'; }
};
struct Derived : Base {
    void f(int x) { std::cout << "derived int: " << x << '\n'; }
};

int main() {
    Derived d;
    d.f(3.5);
}

A derived-class member hides every base-class member of the same name, regardless of signature — overload resolution never sees Base::f. So 3.5 converts to int and the derived version runs.

Observed output: derived int: 3.

Fix: using Base::f; inside Derived to bring the base overloads back into scope.

Lesson: hiding is by name, not by signature.

Facts verified 2026-11-04. The code on this page was compiled and run on the box with GCC 16.2.0 at -std=c++23 -Wall -Wextra; the outputs quoted are the observed outputs. MSVC was not tested.

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