check if variable is of a class that has specific base class

Question

There is is_base_of to check if a class A is a base class of B. But what if I don't have the type of a variable?

Let's say I have five classes A, B, C, D and E, and D and E are derived from both A and B respectively. C is only derived from A. Now I have a pointer to A and I want to know if I can cast the pointer to B. In this case I want to know if my A pointer points to a D or E object, so that I can cast the pointer to a B object.

I tried the following which did not work.

void foo(const std::shared_ptr<A> & ptr) {
    if (std::is_base_of<B, decltype(*ptr)>::value) {
        doSomething(std::static_pointer_cast<B>(ptr));
    }
}

Edit: foo is not a template function, A and B are just placeholders for my existing classes.

to narrow, use `dynamic_cast`. It checks for you (will throw `std::bad_cast `) — Red Alert, Mar 26 '15 at 19:53
@RedAlert Exactly ! Provided there's at least one virtual function. — Christophe, Mar 26 '15 at 19:54
Well, `std::dynamic_pointer_cast`, and it'll be a null pointer returned, not `std::bad_cast` thrown in case it doesn't work. But yes, this requires a runtime check. — Wintermute, Mar 26 '15 at 19:56
`dynamic_cast` also works with pointers, and returns null on failure. It is only when casting references that `std::bad_cast` is thrown, since a reference cannot be null. — Remy Lebeau, Mar 26 '15 at 19:57
possible duplicate of [Check if class is derived from a specific class (compile, runtime both answers available)](http://stackoverflow.com/questions/18099241/check-if-class-is-derived-from-a-specific-class-compile-runtime-both-answers-a) — Julian, Mar 26 '15 at 20:07

score -1 · Answer 1 · answered Mar 26 '15 at 19:55

If foo is a template function, you must define B before A in the template parameter list. You will also have to specifically type B when calling foo.

If your code is provided as is, then you must make foo a template with A being an argument for the template.

Example:

#include <memory>

template <typename B>
void doSomething(std::shared_ptr<B> &ptr)
{

}

template <typename B, typename A>
void foo(const std::shared_ptr<A> & ptr) {
    if (std::is_base_of<B, decltype(*ptr)>::value) {
        doSomething(std::static_pointer_cast<B>(ptr));
    }
}

class A { };

class B : public A { };

int _tmain(int argc, _TCHAR* argv[])
{
    std::shared_ptr<A> a(new B);

    foo<B>(a);

    return 0;
}

foo is not a template function. – gartenriese Mar 26 '15 at 19:57 — gartenriese, Mar 26 '15 at 19:57

check if variable is of a class that has specific base class

1 Answers1