I have 2-D array
char arr[2][3]={"sam","ali"}
and pointer to this array
char(*ptr)[3]=arr;
How can I use this pointer to print arr[2][2] which in this case is i.
I've tried * (*(ptr+1)+2) the same way I am dealing with the array but didn't work so can any one help and tell me how to deal with pointer to array in this case to print element [2][2].
You should not print arr[2][2] because you are accessing array out of bounds. Also note that there is no space for '\0' character to be stored in in the given array.
char arr[2][4] = {"sam","ali"}
char(*ptr)[4] = arr;
Using ptr you can access the elements of arr as ptr[1][2].
This:
char(*ptr)[3]=arr;
isn't a pointer to a multi dimensional array, it's a pointer to one-dimensional array of size 3. But that's fine because a pointer can always also point to an array of the type it points to. So you have ptr point to an array of one-dimensional arrays of size 3. So far just for clarifying the terms.
Your immediate problem are just wrong indices. Indices are based on 0, not 1, so with
char arr[2][3]={"sam","ali"}
valid indices for the first dimension of arr are just 0 and 1. The element you're looking for would be at ptr[1][2].
With the pointer arithmetics notation in your question, you actually had the indices right, so I can't see where your problem was in this case. The following prints i as expected:
#include <stdio.h>
int main(void)
{
char arr[2][3]={"sam","ali"};
char(*ptr)[3]=arr;
printf("%c\n", *(*(ptr+1)+2));
}
Note this is completely equivalent to the more readable ptr[1][2].
Side note: if you expected your elements to be strings -- they aren't, see haccks' answer for the explanation (a string must end with a '\0', your string literals do, but your array needs the room to hold it).
'\0'."sam"actually requires 4 bytes, as it includes the string termination character