Using template specialization to compare pointer references

Question

Based on this SO answer, I was experimenting with something similar but with a pointer:

#include <iostream>

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

class Foo: Bar {
public:
    Foo() { std::cout << "Foo::Foo()" << std::endl; }
    ~Foo() override { std::cout << "Foo::~Foo()" << std::endl; }
};

class Faz {
public:
    Faz() { std::cout << "Faz::Faz()" << std::endl; }
    ~Faz() { std::cout << "Faz::~Faz()" << std::endl; }
};

template <typename T>
typename std::enable_if<std::is_base_of<Bar, std::remove_pointer<T>>::value>::type
func(char const* type, T) {
    std::cout << type << " is derived from Bar" << std::endl;
}

template <typename T>
typename std::enable_if<!std::is_base_of<Bar, std::remove_pointer<T>>::value>::type
func(char const* type, T) {
    std::cout << type << " is NOT derived from Bar" << std::endl;
}

int main()
{
    func("std::unique_ptr<Foo>",  std::unique_ptr<Foo>());
    func("std::unique_ptr<Faz>",  std::unique_ptr<Faz>());
}

cout is :

std::unique_ptr<Foo> is NOT derived from Bar
std::unique_ptr<Faz> is NOT derived from Bar

Why does !std::is_base_of<Bar, type_identity<std::remove_pointer<T>>>::value always evaluate as true? I assumed (as a beginner):

std::unique_ptr<Foo> is derived from Bar
std::unique_ptr<Faz> is NOT derived from Bar

I'm probably missing something stupid.

score 0 · Answer 1 · answered Oct 04 '19 at 22:01

0

std::remove_pointer<> acts on raw pointer types.
The smart pointer's type is not related to the class hierarchy of referent type

answered Oct 04 '19 at 22:01

WilliamClements

350
3
8

So I tried it with `new Foo()` and `new Faz()` and it printed the same output. Shouldn't it have worked if they're raw pointers? – Sean McVeigh Oct 04 '19 at 22:31
@SeanMcVeigh - `std::remove_pointer<>::type`, not `std::remove_pointer<>`. You need to actually access the transformed type. – StoryTeller - Unslander Monica Oct 04 '19 at 22:35
For example, `std::enable_if::type>::value>::type`? – Sean McVeigh Oct 04 '19 at 22:47
I get an error compiling `error: template argument for template type parameter must be a type; did you forget 'typename'?` on `std::remove_pointer` – Sean McVeigh Oct 04 '19 at 22:49
@SeanMcVeigh - As the error says `typename std::remove_pointer<>::type` – StoryTeller - Unslander Monica Oct 04 '19 at 23:11
Okay, great. I understand the purpose of helper types now lol. – Sean McVeigh Oct 05 '19 at 00:01
So the reason this only works for raw pointers is that `remove_pointer` is looking for `type` i.e. `template struct _LIBCPP_TEMPLATE_VIS remove_pointer {typedef _Tp type;};` – Sean McVeigh Oct 05 '19 at 00:03
So to get this to work for smart pointers there would need to be `remove_smart_pointer {typedef _Tp element_type;}` ? – Sean McVeigh Oct 05 '19 at 00:04

Using template specialization to compare pointer references

1 Answers1