addressing format

Question

Consider a hypothetical system which has 32 bit instructions and 8 bit addresses. If there are 90 2-address instructions and 200 one address instructions, then how many zero address instructions can be formulated?

A) 2¹⁶ – 90 × 2⁸

  B) ( 2¹⁶ – 90 × 2⁸ – 200) × 2⁸

C) ((2¹⁶ – 90) × 2⁸ – 200) × 2⁸

D) (2¹⁶ – 90) × 2⁸ – 200 × 2⁸

Answer 1

answer(C)
 

opcode 16 bit

address 8 bit

address 8 bit

no of opcodes possible 2^16
no of free opcodes 2^16-90
no of one address instructions (2^16-90)2⁸ 
no of zero address instruction  ((2^16-90)2⁸ -200)*2⁸

Comments

what is the basic concept behind such questions....??