AnsweredAssumed Answered

Why is moveq better than move.l?

Question asked by Tim Updegrove on Jul 10, 2009
Latest reply on Jul 15, 2009 by Tim Updegrove

(Did some searches but couldn't find answers to these basic questions.)


move.l #$0,d0

moveq #$0,d0


1. The Coldfire reference manual lists the execution times for both instructions as one cycle yet I know moveq is quicker.  Why is that?  I'm assuming it is because the move.l instruction length is 48-bits and moveq is only 16-bits. Would someone be kind enough to explain what is going on here?


2. What is the relation of a one cycle execution time to the number of core clocks?  I thought it was one to one but a 16-bit instruction must take more than one clock to make it through the Instruction Fetch Pipeline (IFP) and Operand Execution Pipeline (OEP).  Is the key here that the IFP time is not counted due to execution time assumption #1 (OEP loaded with opword & extension words)?


3. How many clocks does the 48-bit instruction "move.l #$0, d0" really take?


4. I would have expected the time difference between these two instructions to be listed in a table somewhere but couldn't find such a table in the programmer's or device's reference manuals.  Did I miss it?


T3li.MsoNormal, div.MsoNormal {mso-style-parent:""; margin:0in; margin-bottom:.0001pt; mso-pagination:widow-orphan; font-size:12.0pt; font-family:"Times New Roman"; mso-fareast-font-family:"Times New Roman";} @page Section1 {size:8.5in 11.0in; margin:1.0in 1.25in 1.0in 1.25in; mso-header-margin:.5in; mso-footer-margin:.5in; mso-paper-source:0;} div.Section1 {page:smileyfrustrated:ection1;} -->