arrays - Preserve bounds in allocation in intrinsic assignment

Question

I am using automatic allocation on assignment to calculate the difference of two arrays, with bounds starting at 0:

program main
   implicit none
   integer, allocatable :: a(:), b(:), c(:)

   allocate(a(0:10))
   allocate(b(0:10))

   a = 1
   b = 2

   write (*,*) lbound(a)
   write (*,*) lbound(b)

   c = b - a

   write (*,*) lbound(c)
end program main

Both, gfortran and ifort give the output:

       0
       0
       1

Why doesn't c have the same bounds as a and b? Is there a short and concise (without an explicit allocate) way of making sure c has the same bounds?

Answer 1

Consider an intrinsic assignment statement where the left-hand side is not allocated and is an array:

variable = expr

In such a case, the array variable is allocated to the shape of the expression expr, with lower bound equal to the value of LBOUND(expr) (see Fortran 2018 10.2.1.3).

For the example of the question

c = b - a

the right-hand side is the expression b-a. For this expression LBOUND(b-a) is equal to 1: b-a is not a whole array (F2018 16.9.109). The lower bound of c on allocation is thus 1.

The only way the variable assigned to in intrinsic assignment with allocation to have lower bound not 1 is for the right-hand side to have lower bound not 1. In the assignment (c not allocated)

c = b

then c has lower bound that of b.

You could avoid an explicit allocate statement with

c = b
c = c - a

but that is rather unclear and (to repeat a point in Vladimir F's answer) would you want the lower bound to be that of b or a if these differ.

For completeness:

allocate(c(7:17), source=b-a)