I currently write a small Fortran code (90/03). I have a serious doubt with a simple syntax:
INTEGER :: BB
DOUBLE PRECISION,DIMENSION(10) :: CC
INTEGER,DIMENSION(:),ALLOCATABLE :: AA
BB=10
... (many things to declare double precision values in CC)
AA=pack([( i,i=1,BB )],mask=CC.GT.0.0) ! size(CC,1)=BB
Could I use this syntax? Do I need to allocate AA before call pack? Is there any risk of using this syntax?
The syntax is correct (except for the part that is left out where we cannot anything about it).
Could I use this syntax?
Yes.
Do I need to allocate
AAbefore callpack?
No. Allocatable arrays will be automatically allocated and deallocated when out-of-scope. Regardless of whether the object on the right-hand is the result of any function or something else, e.g. an array of literals.
Is there any risk of using this syntax?
I cannot think of anything risky here.
I guess you are confused about this line
AA=pack([( i,i=1,BB )],mask=CC.GT.0.0)
I will try to explain it:
[( i,i=1,BB )] Builds an array of integers from 1 through 10 by an implied-do loop. Lets call this array B=B(i)mask=CC.GT.0.0 Only the elements B(i) of the previous array are choosen where CC(i)>0 holds.pack(...) returns a 1-dimensional array of the previous result.
pack line was clear for me but I had a serious doubt concering the auto-allocation.