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UGC-NET | UGC NET CS 2015 Dec – II | Question 16
  • Difficulty Level : Easy
  • Last Updated : 07 Apr, 2021

A three dimensional array in ‘C’ is declared as int A[x][y][z]. Consider that array elements are stored in row major order and indexing begins from 0. Here, the address of an item at the location A[p][q][r] can be computed as follows (where w is the word length of an integer):

(A) &A[0][0][0] + w(y * z * q + z * p + r)
(B) &A[0][0][0] + w(y * z * p + z*q + r)
(C) &A[0][0][0] + w(x * y * p + z * q+ r)
(D) &A[0][0][0] + w(x * y * q + z * p + r)


Answer: (B)

Explanation: According to above question we have to find the address of A[p][q][r]

To reach pth row we must have to cross 0 to p-1 row i.e. p rows and each rows contains y∗z elements

Hence ,

= y∗z∗p



Now to reach qth element in pth row we have to cross q rows and each row contains z(total columns) elements

=z∗q

to reach rth elements we have to cross r elements in (p+1)th row

Total elements to cross =(y∗z∗p+z∗q+r)

Now each element occupies m amount of space in memory

Therefore total space occupied by these elements = m(y∗z∗p+z∗q+r)

Hence , address of A[p][q][r]=base address+ Space Occupied by the Elements Before it.

=&A[0][0][0]+m(y*z*p+z*q+r)

Hence Option (B) is correct.

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