Array puzzle

I'm a relatively inexperienced Fortran programmer, so please go gently.

While looking through the IVF documentation, I learned that it's inefficient to pass an assumed shape or allocatable array to a procedure, inside of which it's referenced as a non-deferred shape array. Further reading discourages using assumed size arrays. So I looked through the very old code I'd been using for some time, and found this:

The original code had basically explicit shape arrays, passed into procedures as automatic or assumed-size arrays. Example:

Program Old
REAL A
INTEGER N
DIMENSION A(30)

N=6
.. . .

CALL P1(A)
CALL P2(A,N)
END

SUBROUTINE P1(B)
REAL B
DIMENSION B(*)

B(23) = . . .
END

SUBROUTINE P2(B,N)
REAL B
INTEGER N
DIMENSION B(N,*)

B(3,4) = . . .
END

Some time back I made array A in the main program allocatable with rank 1. With no other changes, everything works ok. But recent attempts to conform to the IVF guidelines have been a failure. First off, some of these arrays are large, so it's undesirable to make unnecessary copies. The IVF documentation says that if an assumed shape or allocatable array is passed to a procedure where it's referenced as an explicit shape or assumed size array, a temporary array is created. So I decided to try sending the array in as an assumed shape array and declaring it that way inside the procedure also.

It works fine for procedures like P1, where the rank of the array inside the procedure is the same as the array in the calling program. The problem comes with subroutine P2 in the example above, where the rank of the array inside the procedure is different from the rank of the array passed in by the calling program. I made INTERFACE blocks, but found that the rank in the interface block has to match both the rank inside the procedure and the rank of the array in the calling program or procedure. If I simply make all the ranks one, an error occurs at the B(3,4) statement.

Is there a way to do this, that is, to pass an assumed shape array into a procedure where the array inside is a different rank than in the calling procedure? Even if the effort of changing things isn't worth the trouble, I'd like to learn more about this.

Thanks!

Roy Lewallen


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Relevant Pages

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