Reputation:

Why my 2D array is not functioning right?

I am newbie in C programming.I want to print 2 as my first element is 2 in the 2D array.But as i knew that n holds the first address of the array so *n should print the first element that is 2.My code

‪#‎include‬ <stdio.h>
int main()
{
int n[3][3]={2,4,3,6,8,5,3,5,1};
printf("%d\n",*n);
return 0;
}

why it is printing an address.Can anyone explain it to me??

Upvotes: 0

Answers (5)

Anbu.Sankar

Reputation: 1346

I think that you got some warning on %d because you are not tried to print the value - just the address only.

For a 2D array, to get any value, you need to dereference twice. i.e **n. *n is also suitable, but for a 1D array.

Here you can use either **n, *n[0] or n[0][0] instead of *n.

Upvotes: 1

Tom Fenech

Reputation: 74615

You need to declare your array like this:

int n[3][3]={{2,4,3},{6,8,5},{3,5,1}};

Note that the first [3] isn't necessary (but there's nothing wrong with specifying it). By the way, if you enable warnings, e.g. with gcc -Wall, the compiler will warn that there are missing braces in your initialiser.

Then to print the first value you can use:

printf("%d\n",n[0][0]);

Or, if you prefer:

printf("%d\n",*n[0]);

You have an array or arrays, so this takes the zeroth element (which is an array), then dereferences it to get the zeroth value.

Upvotes: 1

Sathish

Reputation: 3870

If you have an 1D array-

int n[9]={2,4,3,6,8,5,3,5,1};
 printf("%d\n",*n);

Because If you dereference the 1D array it will fetch the element. Now It will print 2. But

int n[3][3]={2,4,3,6,8,5,3,5,1};

It is a 2D array so you need to dererence two times. If you dererence one time it will fetch the address of the array only. n, n[0], *n, &n, &n[0] all will represent the starting address of it.

Try -