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Should std::unique_ptr<void> be permitted

This is a very simple question. Consider the following code:

#include <iostream>
#include <memory>

typedef std::unique_ptr<void> UniqueVoidPtr;

int main() {
    UniqueVoidPtr p(new int);
    return 0;
}

Compiling with cygwin (g++ 4.5.3) with the following command g++ -std=c++0x -o prog file.cpp works just fine. However, compiling with the microsoft compiler (either VS 2010 or 2013) I get this error:

C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\INCLUDE\memory(2067) : error C2070: 'void': illegal sizeof operand
        C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\INCLUDE\memory(2066) : while compiling class template member function 'void std::default_delete<_Ty>::operator ()(_Ty *) const'
        with
        [
            _Ty=void
        ]
        C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\INCLUDE\type_traits(650) : see reference to class template instantiation 'std::default_delete<_Ty>' being compiled
        with
        [
            _Ty=void
        ]
        C:\Program Files (x86)\Microsoft Visual Studio 10.0\VC\INCLUDE\memory(2193) : see reference to class template instantiation 'std::tr1::is_empty<_Ty>' being compiled
        with
        [
            _Ty=std::default_delete<void>
        ]
        foo1.cpp(7) : see reference to class template instantiation 'std::unique_ptr<_Ty>' being compiled
        with
        [
            _Ty=void
        ]

Is this expected? I'm writing a class where I wanted to have a unique pointer in the in the class. While trying to work out the semantics of a move constructor for the class, I ran into this (I assume because I finally got my move constructor coded correctly: i.e. the other errors were fixed).

like image 708
Andrew Falanga Avatar asked Nov 07 '13 16:11

Andrew Falanga


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When should we use unique_ptr?

Use unique_ptr when you want to have single ownership(Exclusive) of the resource. Only one unique_ptr can point to one resource. Since there can be one unique_ptr for single resource its not possible to copy one unique_ptr to another. A shared_ptr is a container for raw pointers.

What happens when unique_ptr goes out of scope?

std::unique_ptr is a smart pointer that owns and manages another object through a pointer and disposes of that object when the unique_ptr goes out of scope. The object is disposed of, using the associated deleter when either of the following happens: the managing unique_ptr object is destroyed.

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Does unique_ptr delete itself?

unique_ptr objects automatically delete the object they manage (using a deleter) as soon as they themselves are destroyed, or as soon as their value changes either by an assignment operation or by an explicit call to unique_ptr::reset.


2 Answers

MSVC is right while GCC is wrong:

Standard(3.9/5):

Incompletely-defined object types and the void types are incomplete types

Standard(20.7.1.1.2/4):

If T is an incomplete type, the program is ill-formed

like image 103
ixSci Avatar answered Sep 18 '22 19:09

ixSci


GCC actually has code to prevent it, but it didn't work until recently.

GCC's unique_ptr has a static assertion in default_deleter::operator() that should reject incomplete types:

    static_assert(sizeof(_Tp)>0,
                  "can't delete pointer to incomplete type");

However, as an extension GCC supports sizeof(void), so the assertion doesn't fail, and because it appears in a system header doesn't even give a warning (unless you use -Wsystem-headers).

I discovered this problem myself recently so to fix it I added this 10 days ago:

    static_assert(!is_void<_Tp>::value,
                  "can't delete pointer to incomplete type");

So using the latest code on trunk your example fails to compile, as required by the standard.

like image 25
Jonathan Wakely Avatar answered Sep 21 '22 19:09

Jonathan Wakely