What is output of this program ?

hare singh nayak · 2012-08-13T01:09:23+00:00

what is output of this program and please explain in detail.. #include<stdio.h> main() { int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} }; int *p,*q; p=&a[2][2][2]; *q=***a; printf("%d----%d",*p,*q); }

What is output of this program ?

hare singh nayak

Member

Updated: Oct 6, 2024

Views: 971

what is output of this program and please explain in detail..

#include<stdio.h>
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}

0

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[Prototype]

Member • Aug 13, 2012

p=&a[2][2][2];

Only if you're trying to take the address of the last element in the array. If you actually want the start address of the array, then it should be only &a.

*q=***a;

How come the "a" became pointer suddenly?

So basically, the output of the program is error because "a" is not a pointer. Not a compilation error but a logical error.

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hare singh nayak

Member • Aug 13, 2012

actually sir there is no error
when i am running this code the output is like that
5641496----10

and i am not able to understand what is happending here so please explain in lit bit detail..

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silverscorpion

Member • Aug 13, 2012

When you declare the array, it's actually stored as follows:

___________________________
| 10 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
-------------------------------

But, since it's declared as a 3-d array, it should be visualized a little differently. You can think of it as an array containing 2 elements, each of which is an array containing 2 elements each of which is a 2-element array again.

So, if you just specify a, it points to the address of the first element of the array. Specifically, 'a' points to the address of the array
_______
|10 | 2 |
| 3 | 4 |
--------

(Note that a only has 2 elements, one is the 2x2 array shown above. The other is the 2x2 array from the other 4 elements 5,6,7 and 8)

Now, *a refers to the value in the 0[sup]th[/sup] position of the array... so, when you put *a, that returns the same 2x2 array shown above.

Now comes **a. As already said, *(some_array) returns the value of the first element in that array. So, **a returns the first element of the 2x2 array, which is a 1x2 array. And that is
_______
|10 | 2 |
--------

So, when you do it 3 times, ie., when you use ***a, it returns the first value of the 1x2 array, which is 10. Voila!!

Hope it's clear! 😀

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Saandeep Sreerambatla

Member • Aug 13, 2012

Hats off, #-Link-Snipped-#! Great explanation!

Correct me if I am wrong, but if "a" is declared as array then it can be used as pointer.

The answer to the question #-Link-Snipped-# asked.

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hare singh nayak

Member • Aug 13, 2012

silverscorpion
When you declare the array, it's actually stored as follows:

___________________________
| 10 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
-------------------------------

But, since it's declared as a 3-d array, it should be visualized a little differently. You can think of it as an array containing 2 elements, each of which is an array containing 2 elements each of which is a 2-element array again.

So, if you just specify a, it points to the address of the first element of the array. Specifically, 'a' points to the address of the array
_______
|10 | 2 |
| 3 | 4 |
--------

(Note that a only has 2 elements, one is the 2x2 array shown above. The other is the 2x2 array from the other 4 elements 5,6,7 and 8)

Now, *a refers to the value in the 0[sup]th[/sup] position of the array... so, when you put *a, that returns the same 2x2 array shown above.

Now comes **a. As already said, *(some_array) returns the value of the first element in that array. So, **a returns the first element of the 2x2 array, which is a 1x2 array. And that is
_______
|10 | 2 |
--------

So, when you do it 3 times, ie., when you use ***a, it returns the first value of the 1x2 array, which is 10. Voila!!

Hope it's clear! 😀

Tank you silverscorpion very nice explanation..

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silverscorpion

Member • Aug 13, 2012

English-Scared
Hats off, #-Link-Snipped-#! Great explanation!

Correct me if I am wrong, but if "a" is declared as array then it can be used as pointer.

The answer to the question #-Link-Snipped-# asked.
We are not actually using a as a pointer. A pointer stores the address of another variable. Here, we are merely trying to get the value stored in a given address, and that happens to be the address where the array is stored..

The * operator gets the value stored in whatever address we use it with. And here, we use it with a. And when a is declared as an array, the name 'a' alone, without any index, 'points' to the first element of the array. So, saying just a and saying &a[0] are same.

We can declare another variable, say b, as an int. And then get the value stored in that variable by using *(&b). In this array, we're doing the same, just that, instead of *(&a[0]), we use *a

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