Open In App
Related Articles

UGC-NET | UGC NET CS 2015 Dec – II | Question 16

Improve Article
Save Article
Like Article

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[0][0][0] + w(y * z * q + z * p + r)


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


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


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

Answer: (B)


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.

Quiz of this Question
Please comment below if you find anything wrong in the above post

Last Updated : 07 Apr, 2021
Like Article
Save Article
Similar Reads