4 votes 4 votes A computer has 32 bit instruction and 12 bit address . If there are 250 two address instructions , the no. of one -address instructions can be ..... Plz formulate a generic solution for this with diagram . CO and Architecture co-and-architecture addressing-modes + – dragonball asked Jan 25, 2018 dragonball 5.0k views answer comment Share Follow See 1 comment See all 1 1 comment reply akash.dinkar12 commented Jul 4, 2018 reply Follow Share answer would be:24576 0 votes 0 votes Please log in or register to add a comment.
5 votes 5 votes ........... abhishekmehta4u answered Jul 4, 2018 abhishekmehta4u comment Share Follow See all 3 Comments See all 3 3 Comments reply the_bob commented Aug 21, 2018 reply Follow Share Could you please give some explanation regarding it ? I was unable to understand the part starting from the "Free Opcode" onwards... 0 votes 0 votes Shanu Gandhi commented Jan 28, 2019 reply Follow Share will the answer be 0 if we take whole 256 instructions of 2 register mode? or will it be 2power 12 0 votes 0 votes Pranavpurkar commented Dec 3, 2022 reply Follow Share @Shanu Gandhiit should be 0 in that case.And, if we are implementing both 2 address and 1 address together then minimum of 1 address should be 1, otherwise only 2 address will be present in it. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes Opcode size = 32 -(12+12)= 8 bits ,given 12 bit address and 32 bit instruction Number of 2 address instructions possible =2^8 =256 (used =250) Therefore available = 256-250 =6 Now, Number of 1 address instructions possible =6*2^12 = 24576 Sanandan answered Sep 5, 2020 Sanandan comment Share Follow See all 0 reply Please log in or register to add a comment.