In a two-dimensional array, X (9, 7), with each element occupying 4 bytes of memory, with the address of the first element X (1, 1) is 3000, find the address of X (8, 5).

• According to row major:
  LOC(A[J, K])= Base(A)+W[N(J-LB)+(K-LB)]
  where, N is number of columns in the Array, W is number of bytes required to store a single element, and LB is the Lower Bound of the Array.
  So,
  3000+4*[7*(8-1)+(5-1)]
  3000+4*[53]
  3212
  
• 9 years agoHelpfull: Yes(8) No(0)
• 4696
  each element carry 32 bits
  total bits in first row=32*7=224
  total for 7 rows=7*224=1568
  in 8th row=4*32=128
  address of X(8,5)=3000+1568+128
  
• 12 years agoHelpfull: Yes(3) No(13)
• Ans 3248
  [((9-1)*7)+(7-1)]*4 byte
  3000+248
  3248 100% Correct
• 12 years agoHelpfull: Yes(2) No(9)