Re: Allocatable arrays in derived types

On 29 apr, 09:57, Gib Bogle <bo...@xxxxxxxxxxxxxxxxxxxxxxxx> wrote:
Consider a derived type used thus:

type mytype
     integer :: a,b,c
     real :: x,y,z
     real, allocatable :: mydata(:)
end type

type(mytype), allocatable :: cell(:)
type(mytype) :: acell

allocate(cell(1000))
do i = 1,1000
     if (i <= 500) then
         allocate(cell(i)%mydata(10))
     else
         allocate(cell(i)%mydata(100))
     endif
enddo

acell = cell(100)
...
acell = cell(600)

Something like this seems to work fine, so I guess it's OK.  I'm
uncertain about how/where mydata is stored.  Perhaps I shouldn't even be
thinking about this, but old habits die hard.  When cell is allocated
how much space is reserved for mydata?  I suppose this is
compiler-dependent, but what would be a typical number?  When mydata is
allocated, where is it stored?  Finally, when acell is equated to an
element of the cell array, where is its mydata?

Thanks in advance for a free Fortran lesson :-)

An allocatable array (just like a pointer to an array) is likely to
have a bit of overhead - otherwise the size()
function and friends would not work.

If you simply allocate an array of cells but not the
mydata field, then the mydata field will be unallocated.
So I guess the overhead is in the array descriptor
or whatever structure is used.

Using this program:

program extra_mem

type cell_data
integer, allocatable, dimension(:) :: mydata
integer :: some_int
end type cell_data

type(cell_data) :: cell

integer, dimension(1) :: dummy

write(*,*) size(transfer(cell,dummy))
end program

with g95 as the compiler I get "8" as the answer to the
size. So, I deduce that in this case each allocatable
component involves 7 integers of extra memory.

Note: if you allocate cell%mydata the reported size of cell
is still 8 integers. The allocated memory is _not_
transferred.

Regards,

Arjen
.

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