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question on mips and code

hey I have a question on understanding a few things. Im trying to understand
the following structure of a instruction register.
Op code, Address-1, Address-2, are 6 bits, 18 bits, 18 bits respectively

What is the maximum number of distinct operation codes that can be recognized and executed by the processor on this machine?

would this be 64 which is base 2 to the 6th. or am i doing it wrong


What is the maximum memory size on this machine?

and would this be a 32bit machine?


im not sure so if anyone can tell me if im on the rt track
Sep 25 '07 #1
3 2083
Savage
1,764 Expert 1GB
hey I have a question on understanding a few things. Im trying to understand
the following structure of a instruction register.
Op code, Address-1, Address-2, are 6 bits, 18 bits, 18 bits respectively

What is the maximum number of distinct operation codes that can be recognized and executed by the processor on this machine?

would this be 64 which is base 2 to the 6th. or am i doing it wrong


What is the maximum memory size on this machine?

and would this be a 32bit machine?


im not sure so if anyone can tell me if im on the rt track
If the IR's OP code is 6 bytes then there are 64 different OP codes.
It should be 64bit machine,I think.

Savage
Sep 25 '07 #2
RRick
463 Expert 256MB
The address size is only 18 bits, so in theory you'll only need an 18 bit machine. Your available memory is 2 ** 18 "things". "Things" depends on how big your basic chunk of data is.

We typically deal with 8 bit bytes, but your basic chunk could be 6 bits or whatever.
Sep 25 '07 #3
JosAH
11,448 Expert 8TB
would this be 64 which is base 2 to the 6th. or am i doing it wrong
You're correct.

kind regards,

Jos
Sep 25 '07 #4

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