Using MINLOC with Fortran: Incompatible ranks 0 and 1 in assignment

Question

Version that gives an error message

program hello
   integer a(9)
   integer index; ! note no dimension here
   a=(/1, 3, 4, 5, 6, 7, 8, 9, 10/)
   index =         


        

Answer 1

You can fix the first version, obtaining a scalar return from minloc, by using the DIM argument:

index = MINLOC(a, DIM=1, MASK=(a > 5))

P.S. No need for semicolons to end statements unless you place multiple statements per line. Fortran isn't C.

Answer 2

This discussion brings up an important point: MINLOC returns an array, even if it's only one number, it's still an array.

It may be possible to declare index as an array, like mentioned above, or to use a temporary array.

integer :: temp(1)
...
temp=minloc(dist)
index=temp(1)

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It is also possible to use DIM argument to avoid manual fiddling with data types, like mentioned in M.S.B's answer.