C++ gcc 9.3.0 typeid of derived pointer always return typeid(BaseClass*) but dynamic_cast<derived_ptr> works fine

Question

There are a few similar posts, here is my minimal case code:

bool useDerived=true;
BaseClass* maker;
if (useDerived) {
    maker = new DerivedClass();
}
else {
    maker = new BaseClass();
}
if (typeid(maker) == typeid(DerivedClass*)) {
    cerr << "is derived type\n";
}
else {
    cerr << "not derived type\n";
}
DerivedClass* x = dynamic_cast<DerivedClass*>(maker);
if (x == nullptr) {
    cerr << " cannot cast back to derived\n";
}
else {
    cerr << " cast to derived OK\n";
}

The result is:

not derived type
cast to derived OK

Am I missing something here? I include the <typeinfo> header. Could it be a compiler version-specific bug?

The DerivedClass is derived from the BaseClass, and the DerivedClass implements several virtual functions from the BaseClass. The idea to use typeid is based on the blog section: Typeid as faster dynamic_cast.

Answer 1

typeid is evaluated statically in this case. You should have dereferenced so you would give it a glvalue of polymorphic type, instead of some pointer type.

This line:

if (typeid(maker) == typeid(DerivedClass*)) {

should be:

if (typeid(*maker) == typeid(DerivedClass)) {

Answer 2

There is no bug here.

typeid() is evaluated at compile-time. BaseClass* and DerivedClass* are two completely different pointer types, so comparing their respective std::type_info structs will never compare as equal.

dynamic_cast is evaluated at runtime. Casting a BaseClass* pointer to a DerivedClass* pointer will invoke an RTTI lookup to find and return a pointer to the DerivedClass portion of the object that the BaseClass* pointer is pointing at. If the object does not implement DerivedClass, dynamic_cast will return nullptr.