0 votes 0 votes Assume an instruction set that uses a fixed 31 bit instruction length. Operand specifies are 4 bits in length. If there are m-three operand instructions in total, then how many two instructions are possible at maximum? CO and Architecture co-and-architecture instruction-format + – someshawasthi asked Oct 27, 2022 • retagged Oct 27, 2022 by makhdoom ghaya someshawasthi 525 views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply Kabir5454 commented Oct 28, 2022 reply Follow Share is it D? 1 votes 1 votes someshawasthi commented Oct 28, 2022 reply Follow Share yes 0 votes 0 votes someshawasthi commented Oct 28, 2022 reply Follow Share Can you please share a simple solution? 0 votes 0 votes Kabir5454 commented Oct 28, 2022 reply Follow Share for 3 address instruction we need all 3 address and operand . with 4 bit as operand we can have at most $2^{4}=16$ , $3$ address instruction but it is given only $m$ 3 address instruction is possible . Now we consumed $(2^{4}-m)$ instruction . Two address instruction need 2 operand . so,we can use extra 9 bits to say (add1’s 9 bit) to specify 2-address instruction . so total we can have $(2^4-m)*2^{9}$ , $2$ address instruction . 2 votes 2 votes someshawasthi commented Oct 28, 2022 reply Follow Share In the question, it’s given ”Operand specifies are 4 bits in length” then why we take 4 bits for opcode?? 0 votes 0 votes Kabir5454 commented Oct 28, 2022 reply Follow Share I believe they mean opcode by that . 0 votes 0 votes Abhrajyoti00 commented Oct 28, 2022 reply Follow Share Yes @Kabir5454 Otherwise Opcode comes out to be $19 bits$ which does not match the options 1 votes 1 votes Chandrabhan Vishwa 1 commented Oct 28, 2022 reply Follow Share here operand specifies means opcode 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes Ans:(2^19 -m)*2^4 vishnu777 answered Nov 23, 2022 vishnu777 comment Share Follow See all 0 reply Please log in or register to add a comment.