What is output of this program ?

Replies

• 

  [Prototype]

  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.
• 

  hare singh nayak

  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..
• 

  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! 😀
• 

  Saandeep Sreerambatla

  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.
• 

  hare singh nayak

  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..
• 

  silverscorpion

  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