The monitored command dumped core?

Well on a regular basis i get obstructed with this error , the monitored command dumped core.

Which is pretty much alien language to me, hence i cannot not possibly understand what the compiler is saying.

I looked up the internet , for what could be the reason and found out that i could be accessing index which has not been allocated memory, therefore i set on to make a simplest code possible and encounter the same error.

#include<stdio.h> int main() { int n; int a[100000]; scanf("%d",&n); int j=0; for(int i=2;i<=n;i+2) { if (i%2==0) { a[j]=i; j+=1; } } return 0; } 

But i don't understand how i could be accessing non allocated memory.

Also what could be the other reasons for the same error to occur such frequently.

4 Answers

I think the issue could be your for loop, as in the third part you're not updating i. To update, write it as i=i+2 or i+=2.

Your index j gets out of bounds:

Demonstration:

#include<stdio.h> int main() { int n; int a[100000]; scanf("%d", &n); int j = 0; for (int i = 2; i <= n; i + 2) { if (i % 2 == 0) { if (j > 100000) // <<<<<<<<<<<<<< { printf("Bummer\n"); return 1; } a[j] = i; j += 1; } } return 0; } 

Accessing an array with out of bounds results undefined behaviour (google that term).

You are incrementing the value of i. You have written i+2 instead of i+=2.

#include<stdio.h> int main() { int n; int a[100000]; scanf("%d",&n); int j=0; for(int i=2;i<=n;i+=2) { if (i%2==0) { a[j]=i; j+=1; } } return 0; } 

You are writing i+2 ,but you have to write i+=2.