如果类型是多态的，如何在编译时检查

Question

I have template function. In the template function I am using dynamic_cast on the template argument. But since you can't use dynamic_cast on non polymorphic type, I want to check if type is polymorphic ( has at least one virtual function ) at compile time, and if type is not polymorphic I will skip using dynamic_cast. Is this possible ?

Answer 1

You can use std::is_polymorphic :

struct Foo {};

std::cout << std::is_polymorphic<Foo>::value << std::endl;

You can use this in combination with std::enable_if to use different code depending on its value.

Answer 2

Another way compared to @juanchopanza

template<class T>
struct IsPolymorphic
{
    struct Derived : T {
        virtual ~Derived();
    };
    enum  { value = sizeof(Derived)==sizeof(T) };
};

class PolyBase {
public:   
    virtual ~PolyBase(){}
};

class NPolyBase {
public:
    ~NPolyBase(){}
};

void ff()
{
    std::cout << IsPolymorphic<PolyBase >::value << std::endl;
    std::cout << IsPolymorphic<NPolyBase>::value << std::endl;
}