Overflow not beeing recognized, p is not printing warning

Question

I'm having a problem with my little c program.

I'm trying to print out a warning if an overflow occurs, I'm using the limits.h library to recognise an overflow with INT_MAX. I guess the problem is in my if loop, but I can't really find the problem...

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

int main()
{
    int x = 1627964;
    int y = 9;

    for(int i=1; i<y; ++i){
        x*= i; // x= x * i
        printf("%d * %d \n",x , i+1);
        if(x >= INT_MAX){
            printf("An Overflow has occured!\n");
            return 0;
            break;
        }
    }
}

I think the overflow occurs after multiplying with 7:

1627964 * 2
3255928 * 3
9767784 * 4
39071136 * 5
195355680 * 6
1172134080 * 7
-384996032 * 8
1214999040 * 9

I can't really tell why the warning is not printed out...

Some help would be really appreciated, thanks :)

Because x doesn't equal INT_MAX. It's like if you write `if(x == 5)` and x isn't 5. What were you expecting? — user253751, Nov 22 '16 at 19:45
also note , by definition x can never be > INT_MAX. Thats why is call INT_MAX — pm100, Nov 22 '16 at 19:49
Think about it. The max value for an `int` is `INT_MAX`. So will checking whether it is greater than that value ever work? — kaylum, Nov 22 '16 at 19:50
one clue to notice is that when you overflow you end up with a negative value. THat lets you know something odd happened (overflow) — pm100, Nov 22 '16 at 19:51

score 0 · Answer 1 · answered Nov 22 '16 at 19:52

0

Only unsigned int has defined overflow behavior, signed int does not, so compiler could throw away half of your code if it detect it. Anyway in your code int never will be bigger than INT_MAX because on overflow it should be negative value.

answered Nov 22 '16 at 19:52

Mateusz Drost

1,171
10
23

score 0 · Answer 2 · answered Nov 22 '16 at 19:53

0

If you try int max / i < x for the overflow ?

answered Nov 22 '16 at 19:53

Aodren BARY

121
9

Govind Parmar · Accepted Answer · 2016-11-22T19:52:23.870

-3

x == INT_MAX only tests if x is exactly equal to INT_MAX. Not less, not more, exactly equal.

What you should be testing for is signed integer overflow, which happens when you multiply by 1627964 by 8: if (x==INT_MAX || x < 0) - the sign bit gets incremented which makes your value negative.

edited Nov 22 '16 at 19:52

answered Nov 22 '16 at 19:46

Govind Parmar

20,656
7
53
85

Given that `x` is an `int`, how can it *ever* be greater than `INT_MAX`? – Andrew Henle Nov 22 '16 at 19:48
you cant test for > INT_MAX by definition – pm100 Nov 22 '16 at 19:48
Sorry guys, updated my answer – Govind Parmar Nov 22 '16 at 19:52
1

signed integer overflow is Undefined Behaviour. The compiler is not obliged to produce a negative number. Suggest you edit or delete this answer as it sounds correct but is technically not. See the duplicate question. – kaylum Nov 22 '16 at 19:54
Thanks! I understand now and its working – werther Nov 22 '16 at 19:54

Overflow not beeing recognized, p is not printing warning

3 Answers3