Why Floating Point Exception(Linux)

Question

I am getting Floating Point Exception when i run this program on Linux.

#include<stdio.h>
#include<math.h>
int main()
{
    int num, num1, num2, n, i, j = 0, t, count = 0;
    int root;

    printf("Enter number of test cases\n");
    scanf("%d", &t);
    while (j < t)
    {
        scanf("%d", &num);
        root = (int)sqrt(num);
        for (i = 1; i < root; i++)
        {
            printf("Inside for");
            if (num % i == 0)
                num1 = i;
            while (num1 > 0)
            {
                n = num1 % 10;
                num1 = num1 / 10;
                if (n == 3 || n == 5 || n == 6)
                    count++;
            }
            if (num % num1 == 0)
            {
                num2 = (int)num / num1;
                while (num2 > 0)
                {
                    n = num2 % 10;
                    num2 = num2 / 10;
                    if (n == 3 || n == 5 || n == 6)
                        count++;
                }
            }
        }
        j++;
        count = 0;
        printf("%d", count);
    }
    return 0;
}

Can anyone please tell me how to correct it

You might want to indent your code, it will be much more easier to read then. — Some programmer dude
– Some programmer dude, Commented Oct 15, 2013 at 10:59
It would help if you could properly indent your code and maybe determine at which line the program crashes. — Tarik
– Tarik, Commented Oct 15, 2013 at 11:01
The only place you use floating point is the sqrt call. When the exception happen, what is the value of num? — Some programmer dude
– Some programmer dude, Commented Oct 15, 2013 at 11:01
@mbratch20 int will automatically be converted by the compiler. — Tarik
– Tarik, Commented Oct 15, 2013 at 11:02
In num % num1, if num1 is 0 (you exit the loop when it becomes zero, so that's more or less an expected case), you divide by zero. Linux is silly and calls that a floating point exception. — user555045
– user555045, Commented Oct 15, 2013 at 11:05

keltar · Accepted Answer · 2013-10-15 11:05:00Z

1

    while (num1 > 0)
    {
        n = num1 % 10;
        num1 = num1 / 10;
        if (n == 3 || n == 5 || n == 6)
            count++;
    }
    if (num % num1 == 0)

You rolling loop unless num1 becomes 0, then dividing by zero. Division by zero is guaranteed SIGFPE (at least on x86 and amd64). Despite name, it have nothing to do with floating point numbers.

answered Oct 15, 2013 at 11:05

keltar

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2 Comments

Eric Postpischil Over a year ago

SIGFPE is a Unix/Posix signal, not an x86 or ARM requirement.

keltar Over a year ago

@EricPostpischil sure thing. Not sure if it is required to trigger when dividing by zero on other CPUs, tho - that's why remark. And there are many other things where SIGFPE could happen, but it results just a quiet NaN.

David Ranieri · Accepted Answer · 2013-10-15 11:06:52Z

0

Using a debugger:

Starting program: /home/david/demo 
Enter number of test cases
2
10

Program received signal SIGFPE, Arithmetic exception.
0x000000000040076d in main () at demo.c:26
26              if (num % num1 == 0)
(gdb) p num1
$1 = 0
(gdb)

As you can see, the value of num1 is 0, division by 0

answered Oct 15, 2013 at 11:06

David Ranieri

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Comments

eran · Accepted Answer · 2013-10-15 11:05:24Z

0

Print the value of num1 before % operation, it gets 0 and then you exception of division by 0

answered Oct 15, 2013 at 11:05

eran

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