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Q111234: FIX: ISHFT and ISHL Always Assume 4-Byte Integers

Article: Q111234
Product(s): Microsoft Fortran Compiler
Version(s): 1.0,1.0a
Operating System(s): 
Keyword(s): kbFortranPS kbLangFortrankbbuglist
Last Modified: 24-MAR-2000

-------------------------------------------------------------------------------
The information in this article applies to:

- Microsoft FORTRAN PowerStation for MS-DOS, versions 1.0, 1.0a 
- Microsoft Fortran Powerstation 32 for Windows NT, version 1.0 
-------------------------------------------------------------------------------

SYMPTOMS
========

The ISHL and ISHFT intrinsic functions may return incorrect results when
shifting INTEGER*1 or INTEGER*2 variables. The incorrect results occur only when
right-shifting negative values.

CAUSE
=====

The compiler converts the value to be shifted to a 4-byte value. For 1- and
2-byte negative integers, the higher order bits are filled with 1s, which can be
incorrectly shifted into the low order bytes.

RESOLUTION
==========

Convert the bit pattern of the negative INTEGER*1 or INTEGER*2 value to an
unsigned INTEGER*4 value. This is done by adding 256 to a negative INTEGER*1
value or by adding 65536 to a negative INTEGER*2 value. Then shift the resulting
INTEGER*4 value.

STATUS
======

Microsoft has confirmed this to be a problem in FORTRAN PowerStation 32 for
Windows NT version 1.0 and MS-DOS version 1.0 and 1.0a. This problem was fixed
in FORTRAN PowerStation 32, version 4.0.

MORE INFORMATION
================

The following sample illustrates the problem:

Sample Code #1
--------------

  c Compile options needed: none
  c
        integer*2 i2
        integer*1 i1
        i1 = -123
        i2 = i1
        write(*,'('' Before shift '',Z2,2x,z4)') i1, i2
        write(*,'('' After shift  '',Z2,2x,z4)') ishl(i1,-2), ishl(i2,-2)
        end

Output
------

  Before shift 85 FF85
  After shift E1 FFE1

The following code illustrates the article's resolution:

Sample Code #2
--------------

  c Compile options needed: none
  c
        integer*4 j,k
        integer*2 i2
        integer*1 i1
        i1 = -123
        i2 = i1
        j = 256 + i1
        k = 65536 + i2
        write(*,'('' Before shift '',Z2,2x,z4)') i1, i2
        write(*,'('' After shift  '',Z2,2x,z4)') ishl(j,-2), ishl(k,-2)
        end

Output
------

  Before shift 85 FF85
  After shift 21 3FE1

Additional query words: 1.00

======================================================================
Keywords          : kbFortranPS kbLangFortran kbbuglist
Technology        : kbAudDeveloper kbFortranSearch kbZNotKeyword2 kbZNotKeyword3 kbFORTRANPower32100NT kbFORTRANPower100DOS kbFORTRANPower100aDOS
Version           : :1.0,1.0a
Issue type        : kbbug
Solution Type     : kbfix

=============================================================================

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