Runtime Error(SIGSEGV)

general
1

MY Program is giving Runtime Error(SIGSEGV).what does it mean???

2

SIGSEGV Error is caused by an invalid memory reference or segmentation fault. The most common causes are accessing an array element out of bounds, or using too much memory. Check this link for more clarity.

1 Like
3

This type of problem generally comes due to array out of index problem.if such problem occurs see your loop consisting array index .happy coding. :slight_smile:

4

Hi
A SIGSEGV is an error(signal) caused by an invalid memory reference or a segmentation fault. You are probably trying to access an array element out of bounds or trying to use too much memory. Some of the other causes of a segmentation fault are : Using uninitialized pointers, dereference of NULL pointers, accessing memory that the program doesn’t own.

source- codechef status code page

5

I am also receiving the same problem. My code works for the CLEANUP task. I have checked on IDEONE.com too and it also compiled it successfully. I couldn’t find any solution .Plz hellp!!
MY CODE:

#include<stdio.h>
using namespace std;
int main(){
    int t;
    scanf("%d", &t);
    if(t>50) return 0;
    for(int j=0;j<t;j++){
        int n,m,a,flag=0;
        scanf("%d%d", &n, &m);
        if(m<0 || m>n || n>1000) return 0;
        int arr[n]={0};
        for(int i=0;i<m;i++){
            scanf("%d",&a);
            arr[a-1]=-1;
        }
        for(int i=0;i<n;i++){
            if(arr[i]==-1) { flag++; }
            else { arr[i] = i-flag; }
        }
        for(int i=0;i<n;i++){
            if((arr[i]>-1) && (arr[i]%2==0)) {printf("%d ",i+1);}
        }
        printf("\n");
        for(int i=0;i<n;i++){
            if((arr[i]>-1) && (arr[i]%2==1)) {printf("%d ",i+1);}
        }
        printf("\n");
    }
    return 0;
}
6

It occurs when you access an element which is out of bound / else using too much memory

7

i am using arr size of n elements specified by the user and if i was accessing any element out of bounds i’d have received an error while it on other platforms too, but i didn’t.

8

Fixed the formatting.

9

int arr[n]={0};

I doubt this statement is allowed…not sure tho.

10

It is allowed, it initiallizes arr with 0 @vijju
@tript change the loop where you scanf to i=0;i<n;i++ it ll probably fix it, if it doesn’t declare a global array for being on the safe side.

11

Not sure. I distinctly remember using that statement once but it didn’t let me. And I had to use a loop to initialise it. I guess it must be that compiler in this case. Thanks for info @neilit1992 :slight_smile:

12

https://www.codechef.com/viewsolution/14108863
can someone pls help me with this one?I am a beginner and stuck in this for a while.I have checked all the possibilities for runtime error but cud nt find any.
Thanks
question : https://www.codechef.com/JUNE17/problems/CLONEME

13

its against rules to discuss problems of live contest…! @akriti17
however runtime error is caused due to
1.using large memory(or declaring large sized arrays) …!! (for dis u have to declare large sized arrays globally like array[1000000]…)
2.accessing elements out of bounds i.e. accessing elements which are not declared !!
happy coding :slight_smile:

14

@tript,

 for(int i=0;i<m;i++){
            scanf("%d",&a);
            arr[a-1]=-1;
        }

This arr[a-1] can cause segmentation fault , check the bounds on a

15

You are accessing any index which is out of the bounds. Try to resolve the problem by carefully looking into the code . Happy Coding

16

#include
#include<bits/stdc++.h>
#include <math.h>
using namespace std;

int main() {
int n,k,sum=0;long a[100005];
// your code goes here
cin>>n>>k;
if(n>=1 && k<pow(2,100000)){
for(int i =0;i<k;i++)
{
a[i]=1;
sum+=a[i];
}
a[k]=k; int t;
for(int p=k+1;p<=n;p++)
{
t=0;
for(int j=1;j<=k;j++)
{
t=t+a[p-j];
}
a[p]=t;//%1000000007;
}
//cout<<a[n]<<endl;

/for(int l=0;l<n;l++)
{
cout<<"\t"<<a[l];
}/
cout<<a[n-1]%1000000007;
}
return 0;

}

for the above code i’m getting SIGSEGV error when i try to submit. Can someone please help ?

17

does declaring an array of size 1000000 give a SIGSEGV because it occupies only 4MB (array type long long int)

18

I am also receiving the same problem. I have checked on IDEONE.com too and it also compiled it successfully. I couldn’t find any solution. It compiled sucessfully everywhere locally as well as on every online platform . But when I submit code it is showing same error as above .Plz hellp!! MY CODE:

#include<bits/stdc++.h>
using namespace std;
int main()

{
long long t;
cin>>t;
while(t–)
{
long long n,p,i,x=0,max=0,j,k,q,a;
vectorarr;
cin>>n>>p;
for(i=1;i<=p;i++)
{
a=n%i;
if(a!=0&&a<n)
arr.push_back(a);
}

   sort(arr.begin(),arr.end(),greater<int>());
    max=arr.front();
    for(i=1;i<=p;i++)
    {
        for(j=1;j<=p;j++)
        {
            for(k=1;k<=p;k++)
            {
                
                if(max==(((n%i)%j)%k)%n)
                x++;
            }
        }
    }
    cout<<x;
}

}

19

This type of error(segmentation fault) means that you are using an undefined memory and accessing it somewhere mostly in loops.

**Most of the time you need to declare an array of a[n+1] instead of a[n] bcz u loop i from 1 to n.