Error: The part-name to the right of a part-ref with nonzero rank has the ALLOCATABLE attribute

Question

I would like to access the the elements of an array in a an arrayed derived type using the subroutine sum_real. That is: sum over first entry in the weight for all people.

type my_type
   real, dimension(:), allocatable :: weight
   real :: total_weight
end type my_type

type (my_type), dimension (:), allocatable :: people
type (my_type) :: answer

allocate (people (2))
allocate (people (1)%weight(2))
allocate (people (2)%weight(2))

people (1) % weight(1) = 1
people (2) % weight(1) = 1
people (1) % weight(2) = 3
people (2) % weight(2) = 3

call sum_real ( people (:) % weight(1), answer % total_weight )

What I want to do is similar to the answer found in Array of derived type: select entry, except for the fact that I have an allocated array in an arrayed derived type instead of an single element.

But, I get a compiler error:

error #7828: The part-name to the right of a part-ref with nonzero rank has the ALLOCATABLE attribute (6.1.2).   [WEIGHT]

Answer 1

What you try is not possible if your component is allocatable. The reference (6.1.2) is actually a reference to the official standard documents, which prohibits this.

The reason is simple, the allocatable components (scalar or arrays) are stored in a different part of memory than the derived type itself. Therefore if you write

sum(people%total_weight)

or

people%total_weight = 0

it is no problem, total_weight is not allocatable, it is stored within the derived type and the compiler just goes in a simple loop and sets one field after another to zero. You can know the address of each %totalweight beforehand.

On the other hand

sum(people%weight)

or

people%weight = 0

each %weight is stored elsewhere and you don't have any simple formula to compute where is each %weight(i).

The solution is either, to fix the size of the array, if possible

real, dimension(2) :: weight

or use a do loop

s = 0
do i = 1, size(people)
  S = S + sum(people(i)%weight)
end do