Can someone explain how they'd get the answer to this problem step-by-step?
--> for 60 opcodes, 6 bits are needed.
--> 32 registers are there. So, 5 bits are needed for each SR and DR. total 10 bits.
--> remaining bits for IMM is 32 - 6 - 10 = 16
--> So, the range of values are -32768 to 32767 (-215 to 215-1 )
Can someone explain how they'd get the answer to this problem step-by-step? 4.7 Suppose a 32-bit instruction takes the following format: | OPCODE SR DR IMM . If there are 60 opcodes and 32 regist...
OPCODE DR SR IMM Assume a 16-bit instruction with the above format. If there are 32 opcodes and we want to represent values in the range of [-16, 15] in the IMM field, What is the maximum number of registers that this machine can have?
Consider a machine which implements an ISA in which every instruction is 32 bits long and has the following format: Where DR = Destination Register, SR = Source Register, and IMMVAL = Immediate Value. The fields DR, SR are represented using the same number of bits. If there are 7 bits for the opcode and 16 registers, A.) How many unique opcodes can be represented? B.) What is the minimum number of bits required for the source register (SR) field?...