c ++ unique_ptr移动构造函数

Question

In another post was mentioned this doesn't work due to the different deleter types.

std::unique_ptr<char[]> ptr(nullptr);
std::unique_ptr<const char[]> ptr_2(std::move(ptr));

But there was no solution to achieve this behavior in VSS2013. Does somebody know a short and clean solution?

Edit::

std::unique_ptr<const char[]> ptr_2(ptr.release());

doesn't work and compiles with the error:

error C2280: 'std::unique_ptr<const char [],std::default_delete<_Ty>>::unique_ptr<char*>(_Ptr2)' : attempting to reference a deleted function
1>          with
1>          [
1>              _Ty=const char []
1>  ,            _Ptr2=char *
1>          ]
1>          C:\Program Files (x86)\Microsoft Visual Studio 12.0\VC\include\memory(1612) : see declaration of 'std::unique_ptr<const char [],std::default_delete<_Ty>>::unique_ptr'
1>          with
1>          [
1>              _Ty=const char []
1>          ]

Answer 1

The wording in the published C++ standard says that conversions involving changes in const-qualification are not allowed for unique_ptr<T[]> . This is a defect, DR 2118 , which was resolved by the changes in N4089 .

Older compilers may not implement the new rules (I implemented an earlier version of the fix for GCC 4.8, but GCC didn't support it before then either).

Answer 2

The straightforward release answer does not compile since Visual Studio defines constructor from another pointer type template<class _Ptr2> explicit unique_ptr(_Ptr2) as private so you must do it like this:

std::unique_ptr<const char[]> ptr_2(const_cast<const char*>(ptr.release()));

to make the pointer types match.

Answer 3

Sure, simply avoid the move and do it manually:

std::unique_ptr<char[]> ptr(nullptr);
std::unique_ptr<const char[]> ptr_2(ptr.release());

Answer 4

You can always release:

std::unique_ptr<const char[]> ptr_2(ptr.release());

But be sure to add a comment pointing out why a simple move doesn't work.