A computer has 32 bit instructions and 12 bit addresses if there are 250 two address instructions, how many one address instructions can be formulated?
Answers
Answer:
Number of one address instruction
• There are 250 2–address instructions.
• If we have 32–bit instructions allowed, then there are 256 2–address instructions only allowed (two addresses will take 24–bits, leaving 8–bits for the opcode).
• Observing the 8–bit opcode, assume the bit pattern 00000000 (0) to the bit pattern 11111001 (249) are utilized for the 250 2–address instructions.
• Then there are only 6–bit patterns left out for 1–one address instructions.
• Though, each one of these could use the remaining 12–bits attained from possessing only one operand, thus we have 6 * 212 1–address instructions.
Answer:
32 bit instruction and 12 bit two address instruction so total used bit for address are 12+12=24 remaining 32–24=8 bits for opcode . 2^8 combination that means 256 combinations can be made
this you can understand as if we are having 6 bit instruction in which 3 bits are for address so remaining 3 will be for opcode and we can make 2^3 combination as follows
000
001
010
011
100
101
110
110
so accordingly we have total 2^8 combinations according to question that means 256 combination and in that 250 for two address then remaining 256–250=6 are for one address and as 12 bits are for address so total one address instructions are 6*2^12=24,756
Explanation: