why this is happening in for loop

[suryak]

Actually, give an input.. at random .(no zero's)
the printf() which I highlighted is showing out of range array location..
why this is happening ..

# include <stdio.h>

int main()
{
   int n,i,j,flag;
   int arr[3][3];
   
   printf("Enter arr\n");
   for(i=0;i<3;i++)
   {
     for(j=0;j<3;j++)
     {  
       scanf("%d",&arr[i][j]);
     }
   }
      
   for (i=0;i<3;i++)
   {
      for(j=0;j<3;j++)
      {     
          flag = 0;
          if (arr[i][j]==0)
          {  
             printf("[%d,%d]",i,j);
             break;
          }
          else if (arr[i][j]!=0)
          {
             flag = 1; 
             printf("<%d, %d>",i,j); 
          }
      }
   }
         if (flag==1)
         printf("/%d, %d/\n",i,j); 
}

To be specific..
what I am looking for is that.. :
let us assume a number N and a array Arr[][]
now, if the array doesn't have the value N. it should print (say, not present)

if did this way.

for (i=0;i<3;i++)
{   
    for (j=0;j<3;j++)
    {
         flag = 0;
         if ( arr[i][j] == N)
              flag =1;
     }
}
if (flag ==1) {
  printf("not present");
  printf("Last loc (%d %d)",i,j);  //its showing out of range .. ie., showing 3 instead of  
//2 if no number is N
}

where i am going wrong

[Adak]

for (i=0, flag=0;i<3;i++)
{   
    for (j=0;j<3;j++)
    {
         // not here flag = 0;
         if ( arr[i][j] == N) {
              flag =1;
              break;
         }
     }
}
if (flag ==1) {
  printf("N is present in the array");
  printf("Last loc (%d %d)",i,j);  //OK now
} 
else {
   printf("N is not in the array\n");
}

Where i am going wrong?

You were letting the for loops go all the way through the array, instead of breaking out when the number was found. Note that the logic here will only find the first instance of the N number, not all of them.