Instructions:
You are encouraged to work on the problem set in groups and turn
in one problem set for the entire group. Remember to put all
your names on the solution sheet. Also, remember to put the name
of the TA and the time for the discussion section you would like the problem
set turned back to you. Show your work.
Bob Computer just bought a fancy new graphics display
for his LC-3. In order to test out how fast it is, he rewrote the OUT
trap
handler so it would not check the DSR
before outputting. Sadly he discovered
that his display was not fast enough to keep up with the speed at which the LC-3
was writing to the DDR
. How was he able to tell?
Bob also rewrote the handler for GETC
, but when he typed
ABCD
into the keyboard, the following values were input:
AAAAAAAAAAAAAAAAABBBBBBBBBBBBBBBBBBBBCCCCCCCCCCCCCCCCCCCDDDDDDDDDDDDDDDDDDDD
What did Bob do wrong?
The following program does not do anything useful. However, being an “electronic idiot,” the LC-3 will still execute it.
.ORIG x3000
LD R0, Addr1
LEA R1, Addr1
LDI R2, Addr1
LDR R3, R0, #-6
LDR R4, R1, #0
ADD R1, R1, #3
ST R2, #5
STR R1, R0, #3
STI R4, Addr4
HALT
Addr1 .FILL x300B
Addr2 .FILL x000A
Addr3 .BLKW 1
Addr4 .FILL x300D
Addr5 .FILL x300C
.END
Without using the simulator, answer the following questions:
What will the values of registers R0
through R4
be after the LC-3 finishes executing the ADD
instruction?
What will the values of memory locations Addr1
through Addr5
be after the LC-3 finishes executing the HALT
instruction?
The LC-3 has just finished executing a large program. A careful
examination of each clock cycle reveals that the number of executed
store instructions (ST
, STR
, and STI
) is greater than the number of executed load instructions (LD
, LDR
, and LDI
). However, the number of memory write accesses is less than the number of memory read accesses, excluding instruction fetches. How can that be? Be sure to specify which instructions may account for the discrepancy.
Assume that you have the following table in your program:
MASKS .FILL x0001
.FILL x0002
.FILL x0004
.FILL x0008
.FILL x0010
.FILL x0020
.FILL x0040
.FILL x0080
.FILL x0100
.FILL x0200
.FILL x0400
.FILL x0800
.FILL x1000
.FILL x2000
.FILL x4000
.FILL x8000
Write a subroutine SET
in LC-3 assembly language that sets a bit in R0
using the table above. The index of the bit to set is specified in R1
.
Write a similar subroutine CLEAR
that clears the specified bit instead of setting it.
Jane Computer (Bob's adoring wife), not to be outdone
by her husband, decided to rewrite the TRAP x22
handler at a different place in
memory. Consider her implementation below. If a user writes a program that uses this
TRAP
handler to output an array of characters, how many times is the ADD
instruction at the location with label A
executed? Assume that the
user only calls this "new" TRAP x22
once. What is wrong with this TRAP
handler?
Now add the necessary instructions so the TRAP
handler executes properly.
Hint: RET
uses R7
as linkage back to the caller.
; TRAP handler
; Outputs ASCII characters stored in consecutive memory locations.
; R0 points to the first ASCII character before the new TRAP x22 is called.
; The null character (x00) provides a sentinel that terminates the output sequence.
.ORIG x020F
START LDR R1, R0, #0
BRz DONE
ST R0, SAVER0
ADD R0, R1, #0
TRAP x21
LD R0, SAVER0
A ADD R0, R0, #1
BRnzp START
DONE RET
SAVER0 .BLKW #1
.END
How many TRAP
service routines can be implemented in the LC-3? Why?
Why must a RET
instruction be used to return from a TRAP
routine? Why won't
a BRnzp
(unconditional BR
) instruction work instead?
How many accesses to memory are made during the processing of a TRAP
instruction?
Suppose we are writing an algorithm to multiply the elements of an array (unpacked, 16-bit 2's complement numbers), and we are told that a subroutine "mult_all" exists which multiples four values, and returns the product. The mult_all subroutine assumes the source oprands are in R1, R2, R3, R4, and returns the product in R0. For purposes of this assignment, let us assume that the individual values are small enough that the result will always fit in a 16-bit 2's complement register.
Your job: Using this subroutine, write a program to multiply the set of values contained in consecutive locations starting at location x6001. The number of such values is contained in x6000.
Hint: Feel free to include in your program
PTR .FILL x6001
CNT .FILL x6000
.ORIG x3000
JSR A
OUT
BRnzp DONE
A AND R0,R0,#0
ADD R0,R0,#5
JSR B
RET
DONE HALT
ASCII .FILL x0030
B LD R1,ASCII
ADD R0,R0,R1
RET
.END
PUSH V PUSH W PUSH X PUSH Y MUL ADD PUSH Z SUB DIV POP U