2D Array Problem...

Hey,

I basically wish to write a function which will accept a pointer (or a pointer to a pointer) and two integers (r and c). The same should display the arr[r][c]th element of the array where the pointer is pointing.

Here is what i have done but the same throws a 'segmentation fault'. Please guide.

#include<stdio.h>
void disp(int **a,int r,int c)
{
printf("%d",a[r][c]);
}
int main()
{
int arr[3][2]={1,2,3,4,5,6};
disp(arr,1,2);
return 0;
}

Thanks
Neville

Hey Nevillemehta!

Check the main! You're declaring a two dimensional array, but initializing a one dimensional array!!
I don't know if that's allowed, but still, you're asking for a value that doesn't exist (memory not reserved and that's why you get a segmentation fault!).



give it a try with this:

int arr[3][2]={{1,2},{3,4},{5,6}};
and ask for arr[1][1] --> I'm a little rusty, but I think that prints out "2"

Good Luck!

Anibal.

I second to what Anibal suggested.He is correct in pointing out the mistake.
A two dimensional array is initialized as { {a,b},
{c,d} }

#include <stdio.h>
#include <stdlib.h>

#define MAXLENGTH 5

void disp(int *a,int r,int c)
{
printf("%d",a[r * MAXLENGTH + c]);
}

int main()
{

int arr[3][MAXLENGTH] ={{1,2,3,4,5},{6,7,8,9,10},{11,12,13,14,15}};
disp((int *)arr, 1, 2);

system("PAUSE");
return 0;
}