 # Why do I get a SIGFPE?

I keep getting SIGFPE error when i try to run my program. Why so?

1 Like

This is the easiest runtime error to debug - it is a floating point error. It is virtually always caused by a division by 0, so check any divisions or modulo operations in your code carefully.

26 Likes

#include<stdio.h>

long int fact(int);
int main()
{
int t,n;
scanf("%d",&t);

while(t–)
{
scanf("%d",&n);
long int r,sum=0,mod=1000000007;
for(r=0;r<=(long int)(n/2.0);r++)
sum+=((fact(n)/(fact®fact(n-r)))%3);
sum
=2;
if(n%2==0)
sum-=((fact(n)/(fact(n/2)*fact(n/2)))%3);

printf("%ld\n",(sum%mod));
}
return 0;
}

long int fact( int n)
{
if(n==0||n==1)
return 1;
long int pro=1;
while(n!=1)
pro*=n–;
return pro;
}

i am getting the same SIGFPE error but cannot debug the program,can anyone please locate the bug

3 Likes

Your code fails for input:

1

68

This is because fact() function overflows resulting in fact(34)=0.

Read about Integer Overflows: http://en.wikipedia.org/wiki/Integer_overflow

2 Likes

SIGFPE may occur due to

1. division by zero
2. modulo operation by zero
3. integer overflow (when the value you are trying to store exceeds the range) - trying using a bigger data type like long.

SIFFPE is very easy to debug.

12 Likes

Why am i getting runtime here??

# include

long long int a,b,c=0;
int calc();
int mod();
int main()

{

``````int t;

scanf("%d",&t);
while(t--)
{
scanf("%lld%lld",&a,&b);

printf("%lld\n",calc());

}
return 0;

``````

}
int mod(){

`````` if(a>b)
{
if(a%b==0)
{
return 1;
}
return 0;

}
else {
if(b%a==0)
{
return 1;
}
return 0;
}

``````

}
int calc()
{
// printf(“a”);
if(mod()==0)
{
while(a!=b)
{
a*=2;
c++;

``````       }
return c;
}
while(mod())
{
if(a%2==0)
{
a/=2;
c++;
}
else{ a=(a-1)/2;
c++;
}
}}

``````

It is a signal representing core dumped means if you do any invalid operation the you will get “SIGFPE”
error. In case floating point error means division by 0 cause this error so please check your code and make sure that you are not operating this type of operation means make sure that you are not dividing by 0

#include<stdio.h>
#include<math.h>

int gcd(int n1, int n2)
{
int min,i, hcf;

``````min=(n1>n2)?n2:n1;

for(i=min; i>=1; --i)
{
if(n1%i==0 && n2%i==0)
{
hcf=i;
break;
}
}
// printf("%d\n",hcf);
return hcf;
``````

}

int fd(int a, int b)
{
int count, i, j, total;
count=total=0;
for(i=a; i<=b; i++)
{
count=0;
for(j=1; j<i; j++)
if(gcd(i,j)==1)
count++;

``````    if(i%count==0)
total++;
}
//printf("%d\n",total);
return total;
``````

}

int main()
{
long long a,j, b,i=0,n, yes;
scanf("%lld", &n);
while(n–)
{
scanf("%lld%lld", &a,&b);

``````    yes=fd(a,b);
printf("%lld\n",yes);
}
return 0;
``````

}
why i m getting sigfpe error in this program…can any1 help??

#include<stdio.h>
main()
{
long int a,b,c,d,e,j,I,n;
scanf("%ld",&a);
for(I=0;I<a;I++)
{
scanf("%ld%ld",&b,&c);
e=bc;
for(j=e;j>1;j–)
{
n=j
j;
if(e%n==0)
{
d=e/n;
printf("%ld",d);
break;
}
}
}
}I am getting sigfpe.Where is the problem in my code .Can any one to it???

#include
using namespace std;
int main()
{
long long int n;
cin>>n;
long long int a[n+1],mul[n+1]={1};
mul=1;
for(long long int i=1;i<=n;i++)
{
cin>>a[i];
mul[i]=1;
}
for(long long int i=1;i<=n;i++)
{
mul[i]=mul[i-1]*a[i];
}

long long int t;
cin>>t;
while(t–)
{
long long int re=1,li,ri,mi;
cin>>li>>ri>>mi;
re=(mul[ri])/(mul[li-1]);
if(re>=mi)
{
re=re%mi;
}
cout<<re<<endl;
}
return 0;
}
iam getting run time error what is the problem

Why am I getting RUNTIME(SIGFPE) error? can anyone help??

Link to my code: https://www.codechef.com/viewsolution/13277608
Thanks in advance!

Note:- In C++ if you fail to typecast the variables properly , you may end up facing a SIGFPE . I faced one as I was storing an int value in a vector of long long int !

3 Likes

use if conditions t ocheck whether anything results in 0 or not

why am i getting RUNTIME SIGFPE error?

link to my code: https://www.codechef.com/viewsolution/18629624

@ashishpm Bro keep dividing the multiplication as its increasing as it is causing Integer Overflow…you are doing whole multiplication at once which is causing To extend the long long range making it negative.

how to reduce time running?

#include
using namespace std;

int main() {
long n,a,i,j,rem,max=0;

`````` cin >> n;

for(i=0;i<n;i++)
{
cin >> a[i];
}
if (n>10)
return 0;
while(i<n)
{
rem=a[i++]%a[j++];
if (rem>max)
max=rem;
if (j==n)
j=0;
}
cout<<"the result of the puzzle"<<max;

return 0;
``````

}

`````````
``````

#include <bits/stdc++.h>

using namespace std;

int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);

``````int t; cin >> t;
while (t--) {
long long l, r, g;
cin >> l >> r >> g;
long long cnt = r / g - (l - 1) / g;
cout << (cnt >= 0 ? cnt : r / g == 1) << "\n";
}
return 0;
``````

}

``I got SIGFPE in above code. When I change "\n" to endl, it become ok. But why ?``
1 Like

#include<bits/stdc++.h>

using namespace std;

#define ll long long

#define I INT_MAX

vector primes;

int total(int i)

{

``````int j=0,done=0,cnt=0;

while(i>1)

{

if(!(i%primes[j]))

{

i/=primes[j];

if(!done) {cnt++;done=1;}

continue;

}

else

{

j++;

done=0;

}

}

return cnt;
``````

}

int main()

{

``````int j;

//finding primes

for(int i=2;i<400;i++)

{

for(j=2;j*j<=i;j++)

{

if(i%j==0) break;

}if(j*j>i) primes.push_back(i);

}

int t,n,a,b,k;

cin>>t;

while(t--)

{

int cnt=0;

cin>>a>>b>>k;

for(int i=a;i<=b;i++)

{

if(total(i)==k) cnt++;

}

cout<<cnt<<endl;

}

return 0;
``````

}

here’s my code. I cannot detect error in my code. error was SIGFPE

I tried this, ain’t getting any error. Can you show me a piece of code where we get error for using %1.