Programming Question

Question

Find the output printed by this program.

Pls. Explain the Execution of this program in detail.
Thanks 

Comments

Is is printing the sum of all elements?

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arr[0][0[0]+arr[0][0][1]+arr[1][0][0]+arr[1][0][1]=1+2+5+6
Therefore,the answer is 14.

I am not getting how we reached here ?
If you get it. plz explain.

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Yes please see that middle loop is incremented by 2.

Hence first time a[0][0][0] = 1 + a[0][0][1] = 2  then middle loop is incremented by directly 2. hence that won't execute.

Then outer loop will increment by 1 hence we reach a[1][0][0] = 5 + a[1][0][1] = 6

$Total = 1+2+5+6 = 14$

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See starting address of arr (i.e 100 ) is passed to p.
Now, how we proceed further on ?  
What would be the address assigned to array containing 1,2, .......? Would it be contiguous ( If yes how we proceed further?)
 

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here ptr is a pointer to 1-D array and p is pointer to 2-D array...how we can assign p to ptr? wont there be compilation error?

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Here, "arr" points to the base address of the complete 3-d array, "p" points to base address of the first block 0f 2-d arrays which is {{1,2}, {3,4}}. Now, in the first iteration:

(i) Outermost Loop: "ptr" is equal to 'p', thus, 'ptr' points to address of first element of 2-d array which is 1.

(ii) Middle Loop: 'ptr' still points to '1'

(iii)Innermost Loop: '1' is added to the value of sum in first iteration, then value of  'ptr' is incremented by '1' and thus now points to '2' and this value is added, So, at the end of first iteration of complete loop. The value of sum is 3.

(iv)Now, in the second iteration value of 'ptr' is made equal to 'p+1'. Now since 'p' pointed to a complete block of 2-d array 'p+1' will point to the block {{5,6},{7,8}}.

The inner loops will have the same iteration and thus '5' and '6' will be added to the sum.

Hence, sum = 1+2+5+6 = 14

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@debanjan ptr is a pointer which points to base address of p or any incremented address od p.

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just check this out guys...

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@ashwin i didnt get u..can you please elaborate?

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gcc gives warning for it and correctly calculated 14.

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its not warning...its showing compilation error and it is not running...

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@vishal chugh
 

"p" points to base address of the first block 0f 2-d arrays which is {{1,2}, {3,4}}

how we got to  know about it ?
p=arr
so shoudln't  p contain 100 ?
  

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Can anyone explain stepwise execution of this program, in terms of what address is reached when ?
So that i get a clear picture of its execution.

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@yogi_p @debanjan see this. I have printed addresses and statement at each execution hope you will get it! 

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@yogi_p I made the illustration..see if it helps you..

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This video was a great help in making 3d array's concept clear to me. Hope this helps you.

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and also interpret (*p)[2][2] this thing as - "p is a pointer to an array of [2][2]" This will clear the whole thing.

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Thanks Everyone.
There was a problem with my concept ,Got it cleared now.

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@ vishal @MiNiPanda @Ashwin

I am confused. I will be highly grateful if any of u could help..

I am unable to understand few things. Here we know that a[2][2][2] is a 3-D array.

int a[2][2][2] is same as

int *a[2][2] is same as

int **a[2] is same as

int ***a.

So here a is a triple pointer. Here in the above problem and places where I see they give a the address of the 1st element. But if a was containing the address of the 1st element directly than just de-referencing once will give us the element value,right? which is not the case in 3-D array where we have to de-reference thrice to print the value of an element. a shld contain &a[0] and further a[0] contains &a[0][0] and then finally a[0][0] containg &a[0][0][0]. therefore on dereferencing a[0][0] we get the element value.

Then why is arr containing the 1st element address 100 here? Why when we are doing ptr=p then ptr is pointing to 1-D array whereas p is pointing to 2-D array. if ptr =p then both shld contain the same address and point to the same location,right?

What am I missing out? PLease help me..

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arr is holding 100 it is true but this should not be interpreted as arr contains the address of an element(integer variable). See it as, arr is holding the address of the 1st element of the 3D array i.e. the 1st 2D array. Both of them may seem to be same but their type is different as you said.

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@ MiNiPanda

Can u pls pls explain a bit more? If 100 is the 1st addr element and a contains 100 then why de-referencing it once won't give element value? 

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I think it is calculating  sum of 1+2+5+6=14