Reputation: 143765
I have the following program
module test
contains
subroutine foo()
integer, allocatable :: a(:)
allocate(a(-5:5))
call bar(a)
print *, a
end subroutine
subroutine bar(a)
integer, intent(out) :: a(:)
a = 0
a(-4) = 3 ! here
a(2) = 3
end subroutine
end module
program x
use test
call foo()
end program
In the line marked with "here" I am doing something wrong. The fact is that when I receive the array a
(in the caller allocated from -5 to +5), the callee uses conventional numbering (1 to n), meaning that assigning -4 I am doing an out of boundary assignment. How can I instruct the compiler that, within the bar
routine, the numbering of the a
array must be the same as in the caller ?
Upvotes: 5
Views: 1636
Reputation: 29381
The type of dummy argument that you are are using in the subroutine, with the dimension specified with a colon, is called "assumed shape". This name is the clue -- Fortran passes only the shape and not the lower and upper bounds. The lower bound is assumed to be one unless you override it as shown in the answer by kemiisto. If the lower bound is not fixed, you can pass an argument to use as the lower bound.
Later addition: a code example if the lower dimension isn't known at compile time:
subroutine example (low, array)
integer, intent (in) :: low
real, dimension (low:), intent (out) :: array
Upvotes: 5
Reputation: 31
There are two common options:
Upvotes: 3
Reputation: 8870
How can I instruct the compiler that, within the bar routine, the numbering of the a array must be the same as in the caller ?
Not sure but according to the standard you can specify the lower bound for an assumed-shape array.
subroutine bar(a)
integer, intent(out) :: a(-5:)
Upvotes: 2