• C

Algorithm in Detecting Integer Overflows in C (on Unix)

How do I detect integer overflows in C, on Unix.  The following is a sample code.

#include <stdio.h>

void main(void)
{
   unsigned short int op1, op2, answer;

   op1 = 65500;
   op2 = 10;
   answer = op1 * op2;
}

Is there a way that I can detect and prompt the user that the data being used to compute the multiplication in line #9 WILL CAUSE an integer overflow if computed.  Is there any algorithm or method in detecting this ?
The algorithm I use must be able to handle unsigned long int as operand types instead of unsigned short int too !!!
Please help !! thanks
rdosaniAsked:
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imladrisCommented:
There are no direct methods provided by C for detecting overflow, as I'm guessing you're aware, since you're asking for an algorithm. So, once the primitive operation op1*op2 has occurred the fat is already in the fire. There are two alternatives I can think of:

1. If portability is not a concern, you could probably use assembler to examine the overflow bit directly (probably since it depends on the tools you have available), and do something relevant if you determine that it has been set.

2. Implement the multiplication by hand, i.e. as repetitive addition. E.g. take the largest number, add it to an accumulator the number of times indicated by the smallest number. Before each addition check that the difference between MAXINT (65536 in the case of unsigned short) and the number being added is less than the number being added. If not you will get an overflow, i.e.

if(MAXINT-largeop<largeop)dooverflowwarning();


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rdosaniAuthor Commented:
I don't specifically have to use mutiplication.  Any operator should work for my algorithm.  It should be operator independent.  It should be able to handle (+ - / *) anything, and still detect an overflow. Example, What if I use addition instead of multiplication. For example
op1 = 65500;
op2 = 10000;
** Note **
At this point in the code.  I should realize that adding these two number will cause an overflow, and hence prompt the user that an overflow has occured.  Question is how do I realize this ?
ans = op1 + op2;
What then ? Your solution will not work for addition ?
Yes portability is a concern.  I can't use assembler only C
I realize I need some sort of algorithm in detecting this ?  I know there is not simple way around it.
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imladrisCommented:
Carrying on along these lines the solution would be to code seperate checks for each operation (there's only 4 of them).
For addition:
     if(MAXINT - op1<op2) dooverflow();

For subtraction:
    if(op2>op1)dounderflow()

For multiplication as specified.

For division:
    if(op2==0)dodividebyzero();

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jceaCommented:
Unsigned integer arithmetic:

if(op1+op2<op1) add_overflow();

if(op1-op2>op1) sub_overflow();

if((!op2)&&(MAXINT/op2>op1) mul_overflow(); /* Try op1*op2=MAXINT */

Integer division has no overflow (unless divide by zero).
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imladrisCommented:
You are right. I wrote mi answer without atention. Try (second round :-):

Unsigned integer arithmetic:

if(op1+op2<op1) add_overflow();
if(op1<op2) sub_overflow();
if((op2)&&(MAXINT/op2<op1-1) mul_overflow(); /* Try op1*op2=MAXINT */

Integer division has no overflow (unless divide by zero).
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jceaCommented:
Sorry, I see what jcea is doing now. He had the comparison reversed too. If it's flipped (or the clauses):

if((op2) && op1>(MAXINT/op2)) dooverflow();

it works admirably.

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imladrisCommented:
You can also test to see if the opposite function yields the same results.

result = X + Y;
if( result/X != Y && result/Y != X )  /* overflow */

This assumes that you just want to know about the overflow and are not looking to head it off at the pass.
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emmonsCommented:
Uhm, of course that should have been
result = X * Y
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