ECE 290_ LAB #8 Solutions - 12290 130 10(initially This is...

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ECE 290: LAB #8 Solutions Lab 8 Solutions Part 2 a) b) circle6 ADDMI ADRS, #imm11 circle6 ADDMR ADRS, SR c) Page 1 of 4 Mallard ECE 290: Computer Engineering I - Spring 2007 - ECE 290: LAB #8 Solutions 4/20/2007 https://mallard.cites.uiuc.edu/ECE290/material.cgi?SessionID=mding3_1070420_040634...
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d) Simulation Results: Page 2 of 4 Mallard ECE 290: Computer Engineering I - Spring 2007 - ECE 290: LAB #8 Solutions 4/20/2007 https://mallard.cites.uiuc.edu/ECE290/material.cgi?SessionID=mding3_1070420_040634...
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Part 3 a) Of course, there is more than one way to do this. Congrats if your progam used fewer instructions than mine. The explanations next to each entry weren't required - they're there to help show what's going on. b) Of course during operation, location 12291 will change. 12288 ADD R0, R0, #3 R0 <- 3, R0 will be used as a counter 12289 ADDMI 130, #-1 M[M[12290]] <- M[M[12290]]-1 This is where the memory decrement happens
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Unformatted text preview: 12290 130 10 (initially) This is the first memory location to decrement. Each time through the loop this value gets incremented 12291 LD R1, #-2 R1<-M[12290] 12292 ADD R1, R1, #1 R1<-R1+1 (M[12290]+1) 12293 ST R1, #-4 M[12290]<-R1 (M[12290]+1) 12294 ADD R0, R0, #-1 R0<-R0-1, decrement the counter 12295 BRzp #-7 If counter is not negative yet, loop to instruction at 12289 Page 3 of 4 Mallard ECE 290: Computer Engineering I - Spring 2007 - ECE 290: LAB #8 Solutions 4/20/2007 https://mallard.cites.uiuc.edu/ECE290/material.cgi?SessionID=mding3_1070420_040634. .. Simulation Results: Mallard Copyright © 1995-2000 Course Content Copyright© 1995-2007 Donna J. Brown. All rights reserved. Page 4 of 4 Mallard ECE 290: Computer Engineering I - Spring 2007 - ECE 290: LAB #8 Solutions 4/20/2007 https://mallard.cites.uiuc.edu/ECE290/material.cgi?SessionID=mding3_1070420_040634. .....
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