The code is below:
program error
implicit none
real :: l,h
integer :: r
l=1.0
h=0.2
r=l/h
print*,r
end program error
The problem is that the answer is 4. When I explicitly divide 1.0 by 0.2, the answer is 5, but when I use symbols, it is 4. In addition, for instance, 1.0/h and l/0.2 is also 4. I am really confused.
When I define r as real number, the problem is gone; but I need it to be an integer since I am going to use those variables to define a two dimensional array.
You are hitting a problem with rounding of real numbers.
0.2 is not a number that can be represented exactly in binary, so 1/0.2 is also not exactly 5, but can be slightly different. So, it is possible that it is evaluated to 4.999997 (or something like that).
Second, assignment of a real to an integer truncates, so in your case, it probably happened to be 4 instead of 5. If you want to assign to the nearest integer, use r = nint(l/h) instead.
Also, read What Every Computer Scientist Should Know About Floating Point Arithmetic.
This has nothing to do with Fortran specifically, it is a general property of floating point arithmetic. The same thing could happen, for example, in C.
Use the NINT (nearest integer) function to get the nearest integer. The problem with your code is that setting an integer variable to a real number even slightly less than 5 will set it to 4, as demonstrated by the program below. The expression 1.0/0.2 may be computed as slightly less than 5 due to the finite precision of floating point arithmetic.
program error
implicit none
real :: l,h
integer :: r
l=1.0
h=0.2
r=nint(l/h)
print*,r ! gives 5
r = 4.99999
print*,r ! gives 4
end program error
