Declaring an integer FORTRAN

Question

I'm new to Fortran, and sorry for this noobish question, I didn't find an answer for it. In the code:

  integer ( kind = 4 ) k
  integer ( kind = 4 ) v(k)
  integer ( kind = 4 ) list(*)

What does (k) and (*) do in the second, third line?

Thanks

casey · Accepted Answer · 2014-08-05T14:54:06.877

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The first integer, k is a scalar. The second integer v(k) is an array v with k elements. The last integer list(*) an assumed size array that is a dummy argument to a procedure. Its length (number of elements) will be determined by the actual argument passed to the procedure.

Note that kind = 4 is not portable and you should instead use the intrinsics kind() or selected_int_kind() to specify the size of your integers.

edited Aug 05 '14 at 14:54

answered Aug 05 '14 at 14:05

casey

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`list(*)` is an assumed size array and the user must now the size from some other variable. – Vladimir F Героям слава Aug 05 '14 at 14:13
As stated, `kind=4` is not portable. See http://stackoverflow.com/questions/3170239/fortran-integer4-vs-integer4-vs-integerkind-4 – M. S. B. Aug 05 '14 at 14:49
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As Vladimir suggests, you cannot invoke the SIZE intrinsic on an assumed size array. – IanH Aug 05 '14 at 14:52

Stefan · Answer 2 · 2014-08-08T08:16:46.557

3

Complementing the answer of @casey:

The definition of

INTEGER(KIND=4) list(*)

is only valid as definition of a dummy argument. Though, you can define this list with the help of a constant as a named constant (specified by the PARAMETER keyword):

INTEGER(KIND=4), PARAMETER :: list(*) = [1,2,3,4,5]

In this case, this is called an implied-shape array (5.3.8.6) which gets its length implicitely from the constant array.

edited Aug 08 '14 at 08:16

answered Aug 05 '14 at 15:13

Stefan

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Although an implied-shape array must be a named constant. – francescalus Aug 06 '14 at 18:31

Declaring an integer FORTRAN

2 Answers2