C++ - using base class as template parameter

Question

I'd like my template function to accept as an argument only a class that inherits from a base class. I think a code snippet explains it better.

class Base
{
    // Some magic
}

class Derived : public Base
{
    // Even more magic
}

class Foo
{}

// Is it possible to tell template to accept only classes derived from Base?
template<class T>
do_something(T obj)
{
    // Perform some dark magic incantations on obj
}

int main()
{
    Foo foo;
    Derived derived;
    do_something<Derived>(derived); // Normal
    do_something<Foo>(foo); // Compilation error if I understand templates correctly
}

@AndyG C++ 17 only I suppose – Sherlock Holmes Jan 04 '21 at 21:28 — Sherlock Holmes, Jan 04 '21 at 21:28
Should `do_something` allow instances of `Base` itself? – AndyG Jan 04 '21 at 21:38 — AndyG, Jan 04 '21 at 21:38

AndyG · Accepted Answer · 2021-01-04T21:42:22.487

Pre-C++20 you can use enable_if coupled with checking is_base_of:

template<class T, std::enable_if_t<std::is_base_of_v<Base, T> && !std::is_same_v<Base, T>, int> = 0>
void do_something(T obj)
{
    // Perform some dark magic incantations on obj
}

Note that I've explicitly disallowed the type to be an instance of Base (because is_base_of considers a type to be a base of itself). If you want to allow instances of Base, then remove && !std::is_same_v<Base, T>

Live Demo

In C++20 we can almost directly translate our enable_if into a requires expression:

template<class T>
requires (std::is_base_of_v<Base, T> && !std::is_same_v<Base, T>)
void do_something(T obj)
{
   // ...
}

Concepts Demo

Or, if you want to allow instances of Base, you can use the built-in derived_from Concept:

template<class T>
requires std::derived_from<T, Base>
void do_something(T obj)
{
   // ...
}

C++ - using base class as template parameter

1 Answers1

Live Demo

Concepts Demo

Concepts Demo 2