Fortran: Counter inside DO-loop

Question

I have an array, from which I want to have some information.

I wrote a small DO-loop, but I don't know why it always returns

integer :: inn=0
parameter :: m=115200
real*8 :: da1(m)

DO i=1, 115200
    IF( i<=19200 .and. da1(i)>1 .and. da1(i)<999.9999 .and. da1(i)<-1 )then
        inn=inn+1
    END IF
END DO
write(*,*) 'inn=',inn

why it always prints 0, whereas, I checked in the file and this array indeed has many values in the defined range
If the fault is in logic, could someone please give me some pointers on not making such mistakes in the future?

score 2 · Accepted Answer · answered Feb 20 '15 at 09:39

2

The problem is your conditional:

da1(i).gt.1.00 .and. da1(i).lt.999.9999 .and. da1(i).lt.-1.000

How can a number (here: da1(i)) be > 1 and < -1 at the same time? That conditional is always false, and inn is never incremented.

answered Feb 20 '15 at 09:39

Alexander Vogt

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I would strongly suggest the OP to use a more readable syntax to see it. I.e., use spaces and symbols `if ( i<19200 .and. da1(i)>1 .and. da1(i)<999.9999 .and. da1(i)<-1 ) then` – Vladimir F Героям слава Feb 20 '15 at 09:42
@VladimirF Sure, I'll be careful next time. Thanks for the suggestion. – PT2009 Feb 20 '15 at 09:43
2

@PT2009 No need to change the question. It is important to use it in your code to avoid future errors. – Vladimir F Героям слава Feb 20 '15 at 09:44

score 2 · Answer 2 · answered Feb 20 '15 at 12:22

Extended comment rather than answer. The loop could be entirely replaced, probably by

count( da1(1:19200)>1 .and. da1(1:19200)<999.999 )

I write probably because I haven't tested this.

I think looping over a range i = 1, 115200 and then, inside the loop, testing that i<=19200 is just wrong. If you want to loop, write i = 1, 19200.

Fortran: Counter inside DO-loop

2 Answers2