The monitored command dumped core?

user_10118061 0
Viewed 26916

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.

c
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.

user_105472350

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).

user_8983480

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;
}
user_122982050

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

user_139643230
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