Path: utzoo!utgpu!attcan!uunet!husc6!ukma!rutgers!gatech!gitpyr!tpf
From: tpf@pyr.gatech.EDU (Tom Friedel)
Newsgroups: comp.sys.ibm.pc
Subject: Clock Counts
Message-ID: <6177@pyr.gatech.EDU>
Date: 4 Aug 88 13:17:12 GMT
Organization: Georgia Institute of Technology
Lines: 12


In the text "Programming the 80386" it says `to calculate eleapsed time for 
an instruction, multiply the instruction clock count by the processor clock 
speed'.  

A MOV instrunction from register to register is 2 clock ticks; from register
to memory is 2 and from memory to register is 4.

My question is how is the elapsed time affected if the memory operand is in
(1) a high-speed hardware cache and (2) `slow' memory.

Tom Friedel      - email responses welcome