Check if a class has a pointer data member

Question

Is there a way to test if a class has a pointer data member?

class Test
{
  int* p;
}

template< typename T >
foo( T bla )
{
}

This should not compile. because Test has a pointer data member.

Test test;
foo( test )

Maybe I can use a trait to disable the template? Or is my only option macros? Maybe someone knows if boost can do it?

What do you really want to do in your template function? How it associated with pointer in class? — αλεχολυτ, Feb 01 '14 at 13:18
"Check if a class has a pointer data member" - why? I don't see why a pointer would make a difference. — Karoly Horvath, Feb 01 '14 at 13:34
@Ahmed because of the missing semicolon :) Seriously, he's trying to say "I want to static assert that the class has no pointer member" — sehe, Feb 01 '14 at 19:03

Daniel Frey · Accepted Answer · 2014-02-01T16:09:26.197

2

The following can work as a protection, but the member variable has to be accessible (public), otherwise it won't work:

#include <type_traits>

class Test
{
public:
  int* p;
};

template< typename T >
typename std::enable_if< std::is_pointer< decltype( T::p ) >::value >::type
foo( T bla ) { static_assert( sizeof( T ) == 0, "T::p is a pointer" ); }

template< typename T >
void foo( T bla )
{
}

int main()
{
    Test test;
    foo( test );
}

Live example

Of course you need to know the name of the member variable to check, as there is no general reflection mechanism built into C++.

Another way which avoid the ambiguity is to create a has_pointer helper:

template< typename, typename = void >
struct has_pointer : std::false_type {};

template< typename T >
struct has_pointer< T, typename std::enable_if<
                         std::is_pointer< decltype( T::p ) >::value
                       >::type > : std::true_type {};

template< typename T >
void foo( T bla )
{
    static_assert( !has_pointer< T >::value, "T::p is a pointer" );
    // ...
}

Live example

Note that I simply added a static_assert as the first line to the function to get a nice, readable error message. You could, of course, also disable the function itself with something like this:

template< typename T >
typename std::enable_if< !has_pointer< T >::value >::type
foo( T bla )
{
    // ...
}

edited Feb 01 '14 at 16:09

answered Feb 01 '14 at 13:23

Daniel Frey

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Maybe you should remove the ambiguity from the call to `foo`. – dyp Feb 01 '14 at 15:49
@dyp Updated the answer, showing two ways to improve the error/overlead situation. – Daniel Frey Feb 01 '14 at 16:10
Of course, this assumes there'll be one member only, and it's name shall be p. – sehe Feb 01 '14 at 19:04
Thank you for your answer. I would like to know if there is a way to check for a pointer data member of which we do not know the name. I gave p as an example but the idea is that I do not know what the name will be or if there is only 1 pointer data member. (Same as what @sehe mentioned) – user985611 Feb 01 '14 at 19:05
@user985611 As I already wrote in the answer, you do have to know the name as C++ has no general reflection mechanism. Sorry. – Daniel Frey Feb 01 '14 at 19:26

score 1 · Answer 2 · answered Feb 01 '14 at 19:12

I would like to know if there is a way to check for a pointer data member of which we do not know the name. I gave p as an example but the idea is that I do not know what the name will be or if there is only 1 pointer data member

I don't think you can.

You could use gcc's -Weffc++ which will warn about classes with pointer members (that don't have the minimum special members defined).

What I really think you're after, though, is "Does this class have value semantics" ("Do I need to deep clone this"). In C++ you have to assume this unless copying/assignment are prohibited.

Check if a class has a pointer data member

2 Answers2

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