Q58499: "Overflow" with Integer Division and MOD Operator; Workaround

Article: Q58499
Product(s): See article
Version(s): 4.00 4.00b 4.50
Operating System(s): MS-DOS
Keyword(s): ENDUSER | SR# S900113-13 B_BasicCom | mspl13_basic
Last Modified: 7-FEB-1990

The integer division operator (\) and the modulo arithmetic operator
(MOD) correctly produce an "Overflow" error if an operand is a
negative number less than -2,147,483,648 or a positive number greater
than +2,147,483,647 (outside the limits for long integers).

This information applies to Microsoft QuickBASIC Versions 4.00, 4.00b,
and 4.50 for MS-DOS, to Microsoft BASIC Compiler Versions 6.00 and
6.00b for MS-DOS and MS OS/2, and to Microsoft BASIC Professional
Development System (PDS)  Version 7.00 for MS-DOS and MS OS/2.

The following program shows how to do integer division and modulo
arithmetic when the size of an operand causes overflow:

   x# = 2147483648                ' numerator
   y# = 123                       ' denominator
   x# = INT(x# + .5)              ' round off the numerator
   y# = INT(y# + .5)              ' round off the denominator
   PRINT FIX(x# / y#)             ' Emulate integer division
   PRINT x# - ( y# * FIX(x# / y#) )  ' Emulate modulo arithmetic

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