Template Prefers Subclass Function as Opposed to More Specialized Inherited Function

Template Prefers Subclass Function as Opposed to More Specialized Inherited Function

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When I use code like this, I seem to get an unexpected result:

class Base
{
public:
    template <typename T>
    bool operator <<(int(*fx)(T)) {
        return true;
    }
};
class Sub : public Base
{
public:
    template <typename T>
    bool operator <<(T t) {
        return false;
    }
};

int foo(int a) {
    return a;
}

int main() {
    bool test = Sub() << foo;
    std::cout << "Used " << (test ? "Base" : "Sub") << " class function";
}

This code will use the very generic Sub class function rather than the specialized Base class function on GCC 4.6.3 and VS C++ v141.

But what I would expect it to do is something more like this code:

class AClass
{
public:
    template <typename T>
    bool operator <<(int(*fx)(T)) {
        return true;
    }
    template <typename T>
    bool operator <<(T t) {
        return false;
    }
};

int foo(int a) {
    return a;
}

int main() {
    bool test = AClass() << foo;
    std::cout << "Used " << (test ? "Specialized" : "Generic") << " class function";
}

This code, on the other hand, will use the specialized function as expected on GCC 4.6.3 and VS C++ v141.

Here are my questions:

  • Is this a feature or a bug?
  • What is a good approach to keep the multi-class structure but have templates select well across them?

Composition vs. Inheritance: How to Choose?, Procedures and functions were rare, newfangled gadgets viewed with suspicion. arranging concepts from generalized to specialized, grouping related concepts in This makes the subclass more tightly coupled to its superclass than it The "​template method" design pattern is perhaps a good example. Calling a base class function. When a member function is called with a derived class object, the compiler first looks to see if that member exists in the derived class. If not, it begins walking up the inheritance chain and checking whether the member has been defined in any of the parent classes. It uses the first one it finds.


Sub::operator << hides Base::operator <<. Unhide it with using:

class Sub : public Base
{
public:
    using Base::operator <<;

    template <typename T>
    bool operator <<(T t) {
        return false;
    }
};

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You need to "pull it down" into the Sub class by adding

using Base::operator<<; 

into your Sub class definition. Otherwise it will be hidden by the sub class overload.

#include <iostream>
class Base
{
public:
    template <typename T>
    bool operator <<(int(*fx)(T)) {
        return true;
    }
};
class Sub : public Base
{
public:

    using Base::operator<<;  // <--- ADD THIS HERE


    template <typename T>
    bool operator <<(T t) {
        return false;
    }
};

int foo(int a) {
    return a;
}

int main() {
    bool test = Sub() << foo;
    std::cout << "Used " << (test ? "Base" : "Sub") << " class function";
}

Overload resolution, If a name refers to one or more function templates and also to a set of overloaded not included in the list of candidate functions when constructing a derived class object. Overload resolution in this case has a final tiebreaker preferring or against the implicit object argument of the conversion function. I am having some issues trying to access an inherited template class how I would like to. I have written out a simplified case of mine for completeness and to focus on the main issue. This is my base class, Bar, whose functions I want access to.


Composition over inheritance, Composition over inheritance (or composite reuse principle) in object-oriented programming (October 2015) (Learn how and when to remove this template message) Composition also provides a more stable business domain in the long term method needs to be implemented (specialized) by each derived subclass. Using “Inheriting and Overloading” concept in creating reusable universal test method library 4 method is just a C++ class and the developer can use inheritance or overloading methods to refine any steps in the run () function but not touch the source code.


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Understanding Inheritance in Java, In fact, the opposite is true: subclasses have more functionality than their super classes. methods into the superclass and more specialized methods in the subclass. Java uses the keyword super to call a superclass method. (Note that in the sample code for the first section, we made a special effort to  Start studying INTRO. TO PYTHON - CHAPTER 12: INHERITANCE. Learn vocabulary, terms, and more with flashcards, games, and other study tools.