PC memory module contains 512k words. Word has 64 bits. Binary instruction has four segments: indirect bit, operation code, register code part to determine one of the 32 registers and address part. Instruction binary code is stored in word memory.
- How many bits are needed for the opcode, register code and the address segment?
Indirect: 1 bit
Reg: 5 (2^5 = 32)
Address: 19 (2^19 = 512k)
Op-code = 64 - 5 – 19 = 40 bits
I think my answer is correct, but practically an opcode cannot have 40 bits. (That's trillion of instructions). Is 40 correct?