Checking function with given signature then compile differently - c++

Question

In a templated class, can we check a member function's signature and define different compilation behaviors for different subclasses? To be more specific, consider the following simple example:
template <typename T>
class Foo {
// ingore all other functions
virtual std::shared_ptr<T> do_something(int a) {
return std::make_shared<T>(a);
}
};
This should just work fine with the T1 class:
class T1 {
T1(int a) {
// implememntation
}
// ingore all other functions
};
// Foo<T1> would just work
However, it will fail the compilation with T2 because Foo's do_something function is clearly implemented to call T's constructor with one argument only:
class T2 {
T2(int a, int b) {
// implememntation
}
// ingore all other functions
};
// Foo<T2> would fail to compile
So the question is, can we rework on Foo to let it work for both T1 and T2, in the way that, for T1 and classes whose constructor takes one int argument, it will compile with a default implementation, whereas for T2 and classes whose constructors are different, it will compile with a virtual function and enforce subclass to implement it with override. Somewhat like below:
Template <typename T>
class Foo {
// ingore all other functions
/* if T's signature takes one int input only
* compile like the below
*/
virtual std::shared_ptr<T> do_something(int x) {
return std::make_shared<T>(x);
}
/* if T's signature is different
* compile like the below and let the subclass implement it
*/
virtual std::shared_ptr<T> do_something(int x) = 0;
}
Is this possible without using any 3rd-party library? It's acceptable if we have to use macros/preprocessors.
It is also acceptable if there must be one function do_something there but for the case of mismatched signature, it just raises runtime exception, something like:
Template <typename T>
class Foo {
// ingore all other functions
virtual std::shared_ptr<T> do_something(int x) {
/* if T's signature takes one int input only
* compile like the below
*/
// return std::make_shared<T>(x);
/* if T's signature is different
* compile like the below and let the subclass implement it
*/
// std::throws...
}
}

Answer 1

As far as I can tell, we need class template specialization here. Not even C++20 requires-clauses can be applied to virtual functions, so the only thing we can do is have the whole class change.
template<typename T> // using C++20 right now to avoid SFINAE madness
struct Foo {
virtual ~Foo() = default;
virtual std::shared_ptr<T> do_something(int a) = 0;
};
template<std::constructible_from<int> T>
struct Foo<T> {
virtual ~Foo() = default; // to demonstrate the issue of having to duplicate the rest of the class
virtual std::shared_ptr<T> do_something(int a) {
return std::make_shared<T>(a);
}
};
If there is a lot of stuff in Foo, you can avoid duplicating it with a heap of upfront cost by moving do_something to its own class.
namespace detail { // this class should not be touched by users
template<typename T>
struct FooDoSomething {
virtual ~FooDoSomething() = default;
virtual std::shared_ptr<T> do_something(int a) = 0;
};
template<std::constructible_from<int> T>
struct FooDoSomething<T> {
virtual ~FooDoSomething() = default;
virtual std::shared_ptr<T> do_something(int a);
};
}
template<typename T>
struct Foo : detail::FooDoSomething<T> {
// other stuff, not duplicated
// just an example
virtual int foo(int a) = 0;
};
namespace detail {
template<std::constructible_from<int> T>
std::shared_ptr<T> FooDoSomething<T>::do_something(int a) {
Foo<T> &thiz = *static_cast<Foo<T>*>(this); // if you need "Foo things" in this default implementation, then FooDoSomething is *definitely* unsafe to expose to users!
return std::make_shared<T>(thiz.foo(a));
}
}
Godbolt
To convert this down from C++20, replace the concept-based specialization with old-type branching:
// e.g. for the simple solution
template<typename T, bool = std::is_constructible_v<T, int>>
struct Foo { // false case
// etc.
};
template<typename T>
struct Foo<T, true> { // true case
// etc.
};
// or do this to FooDoSomething if you choose to use that
A runtime error is much easier, at least in C++17 and up, since you can just use a if constexpr to avoid compiling the problematic code
template<typename T>
struct Foo {
virtual ~Foo() = default;
virtual std::shared_ptr<T> do_something(int a) {
if constexpr(std::is_constructible_v<T, int>) return std::make_shared<T>(a);
else throw std::logic_error("unimplemented Foo<T>::do_something");
}
};