Problem with large integer constant

iev · ‎10-08-2003

Hello,

First of all, I would like to mention that I don't know much about Fortran, but I have to compile a program written in it with the Intel compiler. In the source code there is a definiton:

INTEGER *8 mask

and later in the code the assignment:

mask = 15815322240

When I try to compile the code with the command line:

ifc -c -O2 code.f

I get the error:

mask = 15815322240
^
Error 18 at (:code.f> : invalid integer constant

I replaced the assignment with:

mask = Z'3AEAAAA80'

and now the program compiles correctly (The hexadecimal equivalent of 15815322240 is 3AEAAAA80).My questions are:

1) Is the change I made correct?
2) If it is correct, why does it compile in one case and doesn't in the other?
3) I noticed that the above value needs 5 bytes to be saved. How many bytes are allocated for an INTEGER *8 on an IA-32 system?

Executing an ifc -V gives:
Intel Fortran Compiler for 32-bit applications, Version 7.1 Build 20030922Z
Copyright (C) 1985-2003 Intel Corporation. All rights reserved.
FOR NON-COMMERCIAL USE ONLY

GNU ld version 2.14.90.0.4 20030523
Supported emulations:
elf_i386
i386linux

Thank you in advance for your help.

TimP · ‎10-08-2003

Your untyped integer constant is a default integer, thus it must fit in 4 bytes with the Intel compilers. INTEGER*8 is a non-standard but well entrenched notation for an integer stored in 8 bytes.
Ideally, you would use the standard (since f90) notation such as
INTEGER,PARAMETER :: LONG = SELECTED_INT_KIND(11)
! integer kind big enough for 11 decimal digits
INTEGER(LONG) :: mask
mask = 15815322240_LONG

With the Intel compilers, the value of LONG happens to be 8, but assuming that makes your code non-portable.