Q68599: Integer Divide by 0 Only on 8088/86 Machine

Article: Q68599
Product(s): See article
Version(s): 6.00 6.00a
Operating System(s): MS-DOS
Keyword(s): ENDUSER | s_quickc | mspl13_c
Last Modified: 24-JAN-1991

The code below produces an error on computers with Intel 8088/86
processors, but runs correctly on later versions of the Intel
processors, such as the 80286. When the program is run on an 8088/8086
machine, the following error will occur:

   run-time error R6003
   - integer divide by 0

The problem is caused by the difference between the IDIV instruction
on the processors. The Intel programmer's reference manual for the
8088/86 processor states the following:

   For word integer division, the maximum positive quotient is +32767
   and the minimum negative quotient is -32767. If the quotient is
   positive and exceeds the maximum, or is negative and is less than
   the minimum, the quotient and remainder are undefined, and a type 0
   interrupt is generated.

A type 0 (zero) interrupt is an "Integer Divide By 0" error, which
means that the lowest negative number that can be used on an 8088/86
CPU machine for integer division is -32767. The developer should not
allow an integer to become -32768. This limitation has changed with
later versions of the Intel processors, which allow integer division
with a quotient of -32768.

Note: This is not a problem with the code generated by the compiler.
Because the problem is exhibited at execution time, it is up to the
programmer to ensure that an integer will not take on a value of
-32768 before it is used as a quotient.

Sample Code
-----------

#include <stdio.h>
void main(void)
{
  int numerator=-32768;
  int denominator=1;
  int result;
  result=numerator/denominator;
}

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