pointers inside derived types during assignment

[forall]

what happens to pointers during assignment of derived types? eg, given definition
!----
type gtype
integer::i=0
real,pointer::p(:)=>null()
endtype gtype
type(gtype)::v1,v2
!-----
will the following
case A:
v1%i=5;allocate(v1%p(10))
v2=v1
result in v2%p being dereferenced to the same unnamed storage area as v1%p? (as well as the value of "i" being copied across)?
I guess this kind of code is a great way to create a memory leak...

[Steven_L_Intel1]

Yes, indeedy, it will. This is why TR15581 came into being, allowing ALLOCATABLE components and making assignment of derived types containing such components "do the right thing". Make p an allocatable instead of a pointer, and you'll get the behavior you want, with v2%p being an independent copy of whatever was in v1%p.

[forall]

thanks Steve - at this point I may actually neeed both behaviours (for different variables, of course).

Also, just to make sure I understand whats going on:

caseB:
v1%i=5 ! (ie, with unassociated v1%p)
v2=v1
should leave v2%p unassociated

caseC:
v1%i=5 ! (ie, with unassociated v1%p)
v2%p=>sometarget
v2=v1
should copy i but nullify v2%p

[Steven_L_Intel1]

Is p pointer or allocatable? If allocatable, case C is illegal. I believe you are correct on case B.

The semantics are: deallocate the left-hand-side,then allocate the left-hand-side to match the right and copy the data. If right side is not allocated, then the allocate/copy doesn't happen.