how to handle integer overflow in C

How to handle integer overflow in C

Like in the following multiplication program if u give the numbers as 50000 100000 the result is 7502634 which is not correct how do we handle it

#include<stdio.h>

int main()
{
int a,b;
long int result;
printf("enter the numbers to multiply\n");
scanf("%d\n %d", &a, &b);
result = a * b;
if(result > 0){
printf("result is %d\n", result);
return 0;}
else{
printf("result not in range\n");
return -1;
}
}

It displays 500000000 on my compile. It could be the printf on your compile!
Try %ld

Code:

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printf("result is %ld\n", result);

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Quote:

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Like in the following multiplication program if u give the numbers as 50000 100000 the result is 7502634 which is not correct how do we handle it

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If we are assuming that a and b are positive, we know the multiplication would be incorrect if a * b > INT_MAX, so would checking to see if b > INT_MAX / a do it? Something like this.

Code:

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#include<stdio.h>
#include<limits.h>

void foo(int a, int b)
{
    printf("a = %d, b = %d, ", a, b);
    if ( b > INT_MAX / a )
    {
        puts("overflow warning");
        return;
    }
    printf("a * b = %d\n", a * b);
}

int main(void)
{
    int a, b;
    for(a = 10, b = 5; b < a && a < INT_MAX / 10; a *= 10, b *= 10)
    {
        foo(a,b);
    }
    printf("INT_MAX = %d\n", INT_MAX);
    return 0;
}

/* my output
a = 10, b = 5, a * b = 50
a = 100, b = 50, a * b = 5000
a = 1000, b = 500, a * b = 500000
a = 10000, b = 5000, a * b = 50000000
a = 100000, b = 50000, overflow warning
a = 1000000, b = 500000, overflow warning
a = 10000000, b = 5000000, overflow warning
a = 100000000, b = 50000000, overflow warning
INT_MAX = 2147483647
*/

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Thanks a lot guys. I appreciate ur help.

But the code which u wrote doesnt work for negative values

How to handle integer overflow for both positive and negative numbers can anyone help me pleaseeeeeeeee.

You've already got 1/4 of it. Here are some hints for the missing pieces.

• A negative number is less than zero
• positive * positive = positive and positive / positive = positive (the part you've got)
• positive * negative = negative and positive / negative = negative
• negative * positive = negative and negative / positive = negative
• negative * negative = positive and negative / negative = positive
• INT_MAX is the most positive int
• INT_MIN is the most negative int

What code have you tried that didn't work?

[edit]
Aw, heck. The prodding doesn't seem to be working. The hinting to kate1234 to post some code, any code, that makes a vague attempt at expanding on my previous post seems to have been in vain.

I had written the following code shortly after I posted the first, anticipating a followup question for handling negative numbers (this is why I spelled out this intention by stating, "assuming that a and b are positive").

With the subsequent posts, it seemed that a solution was wanted with no effort. Ten days or so have not seemed to produce any code on that end. Nor even a question as to why things that were tried did not seem to work.

Well, maybe I'm in a good mood, or maybe I just want to justify some of the smart-ass comments above. Or maybe I want to see if anyone can point out things that I've overlooked. But here is what I'd come up with. (I added the comments after this original post.)

Code:

---

#include<stdio.h>
#include<limits.h>

int foo(int a, int b, int *result)
{
    if(a > 0 && b > 0)
    {
        /*
        * 'a' is positive, 'b' is positive, product will be positive.
        * See if one positive value 'b' is more than the most positive int
        * divided by the other positive value 'a' (positive quotient).
        */
        if ( b > INT_MAX / a )
        {
            return -1;
        }
    }
    else if (a > 0 && b < 0)
    {
        /*
        * 'a' is positive, 'b' is negative, product will be negative.
        * See if the negative value 'b' is less than the most negative int
        * divided by the positive value 'a' (negative quotient).
        */
        if ( b < INT_MIN / a )
        {
            return -1;
        }
    }
    else if (a < 0 && b > 0)
    {
        /*
        * 'a' is positive, 'b' is negative, product will be negative.
        * See if the negative value 'a' is less than the most negative int
        * divided by the positive value 'b' (negative quotient).
        */
        if ( a < INT_MIN / b )
        {
            return -1;
        }
    }
    else /* if (a < 0 && b < 0) */
    {
        /*
        * 'a' is negative, 'b' is negative, product will be positive.
        * See if one negative value 'b' is less than the most positive int
        * divided by the other negative value 'a' (negative quotient).
        */
        if ( b < INT_MAX / a )
        {
            return -1;
        }
    }
    *result = a * b;
    return 0;
}

void bar(int a, int b)
{
    int product;
    printf("a = %d, b = %d, ", a, b);
    if(foo(a, b, &product) < 0)
    {
        puts("overflow warning");
    }
    else
    {
        printf("a * b = %d\n", a * b);
    }
}

int main(void)
{
    int a, b;
    printf("INT_MAX = %d\n", INT_MAX);
    printf("INT_MIN = %d\n", INT_MIN);
    for(a =  10, b =  5; a < INT_MAX / 10; a *= 10, b *= 10)
    {
        bar(a,b);
    }
    for(a =  10, b = -5; a < INT_MAX / 10; a *= 10, b *= 10)
    {
        bar(a,b);
    }
    for(a = -10, b =  5; a > INT_MIN / 10; a *= 10, b *= 10)
    {
        bar(a,b);
    }
    for(a = -10, b = -5; a > INT_MIN / 10; a *= 10, b *= 10)
    {
        bar(a,b);
    }
    return 0;
}

/* my output
INT_MAX = 2147483647
INT_MIN = -2147483648
a = 10, b = 5, a * b = 50
a = 100, b = 50, a * b = 5000
a = 1000, b = 500, a * b = 500000
a = 10000, b = 5000, a * b = 50000000
a = 100000, b = 50000, overflow warning
a = 1000000, b = 500000, overflow warning
a = 10000000, b = 5000000, overflow warning
a = 100000000, b = 50000000, overflow warning
a = 10, b = -5, a * b = -50
a = 100, b = -50, a * b = -5000
a = 1000, b = -500, a * b = -500000
a = 10000, b = -5000, a * b = -50000000
a = 100000, b = -50000, overflow warning
a = 1000000, b = -500000, overflow warning
a = 10000000, b = -5000000, overflow warning
a = 100000000, b = -50000000, overflow warning
a = -10, b = 5, a * b = -50
a = -100, b = 50, a * b = -5000
a = -1000, b = 500, a * b = -500000
a = -10000, b = 5000, a * b = -50000000
a = -100000, b = 50000, overflow warning
a = -1000000, b = 500000, overflow warning
a = -10000000, b = 5000000, overflow warning
a = -100000000, b = 50000000, overflow warning
a = -10, b = -5, a * b = 50
a = -100, b = -50, a * b = 5000
a = -1000, b = -500, a * b = 500000
a = -10000, b = -5000, a * b = 50000000
a = -100000, b = -50000, overflow warning
a = -1000000, b = -500000, overflow warning
a = -10000000, b = -5000000, overflow warning
a = -100000000, b = -50000000, overflow warning
*/

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Posting this as an edit, rather than a new reply, will leave a minimal effort to kate1234. A solution is now here if you bother to look.
[/edit]

Type casting. Promote integers to long.

Quote:

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Originally posted by Dev
Type casting. Promote integers to long.

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But a long isn't guaranteed to be any bigger than an int.