Fortran calculates 0 for a fraction

Asked Sep 23 '16 at 01:19

Active Sep 23 '16 at 01:19

Viewed 44 times

PROGRAM TRYY

implicit none

REAL:: x_value=2
REAL:: y_value=3
REAL:: answer

IF ((x_value >= 0 .AND. x_value < 10) .AND. (y_value >= 0 .AND. y_value < 10)) THEN
  answer = (1/3)*(x_value) + (1/2)*(y_value)
  WRITE(*,*) answer
ELSE
  WRITE(*,*)
END IF

STOP
END PROGRAM TRYY

This should read 2.16667, but instead it reads 0 because there are fractions in there. The fractions are (1/3) and (1/2). When I change them to 0.3 and 0.5 respectively, it gives me 2.1

I don't want that. I want (1/3) and (1/2) to work. What is happening?

asked Sep 23 '16 at 01:19

Master

1

The compiler is performing integer division for 1/3 and 1/2. Use 1.0/3.0 and so on. – Mick Sep 23 '16 at 01:24
I got it. I just changed them to (1.0/3.0) and (1.0/2.0) – Master Sep 23 '16 at 01:24
@MickSharpe thanks! – Master Sep 23 '16 at 01:25
That's the problem with computers. They don't always do what you want them to do, only what you tell them to do. ;-) – Mick Sep 23 '16 at 01:26

Fortran calculates 0 for a fraction

0 Answers0