Department of Electrical and Computer Engineering

The University of Texas at Austin

EE 306, Fall 2013
Problem Set 5
Due: 11 November, before class
Yale N. Patt, Instructor
TAs: Ben Lin, Mochamad Asri, Ameya Chaudhari, Nikhil Garg, Lauren Guckert
Jack Koenig, Saijel Mokashi, Sruti Nuthalapati, Sparsh Singhai, Jiajun Wang

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.
Please staple your Problem Sets BEFORE coming to class!


  1. Moved from problem set 4

    1. 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?

    2. 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?


  2. Now moved to problem set 6 (Updated 11/06/2013) (Adapted from 6.16) Shown below are the partial contents of memory locations x3000 to x3006.

      15 0
    x3000 0 0 1 0 0 0 0                  
    x3001 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 1
    x3002 1 0 1 1 0 0 0                  
    x3003                                
    x3004 1 1 1 1 0 0 0 0 0 0 1 0 0 1 0 1
    x3005 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0
    x3006                                


    The PC contains the value x3000, and the RUN button is pushed.

    As the program executes, we keep track of all values loaded into the MAR. Such a record is often referred to as an address trace. It is shown below.

    MAR Trace
    x3000
    x3005
    x3001
    x3002
    x3006
    x4001
    x3003
    x0021

    Your job: Fill in the missing bits in memory locations x3000 to x3006.


  3. Moved to problem set 6 (Updated 11/06/2013)

    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 (RET is equivalent to JMP R7).

    
; 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

  4. Moved to problem set 6 (Updated 11/06/2013) (Adapted from 9.2)

    1. How many TRAP service routines can be implemented in the LC-3? Why?

    2. Why must a RET instruction be used to return from a TRAP routine? Why won't a BRnzp (unconditional BR) instruction work instead?

    3. How many accesses to memory are made during the processing of a TRAP instruction?


  5. 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

    1. Write a subroutine CLEAR in LC-3 assembly language that clears a bit in R0 using the table above. The index of the bit to clear is specified in R1. R0 and R1 are inputs to the subroutine.

    2. Write a similar subroutine SET that sets the specified bit instead of clearing it.

      3. Hint: You should remember to save and restore any registers your subroutine uses (the "callee save" convention). Use the RET instruction as the last instruction in your subroutine (R7 contains the address of where in the caller to return to.)


  6. 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 multiplies four values, and returns the product. The mult_all subroutine assumes the source operands 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. Store your result at location x7000. Assume there is at least one value in the array(i.e., M[x6000] is greater than 0).

      Hint: Feel free to include in your program 

      
PTR	.FILL x6001
CNT	.FILL x6000


  7. (Adapted from 9.13)
    The following program is supposed to print the number 5 on the screen. It does not work. Why? Answer in no more than ten words, please. 
    
	.ORIG 	x3000
	JSR	A
	OUT			;TRAP  x21
	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

  8. Moved to problem set 6 (Updated 11/06/2013) (Adapted from 8.15)
    1. What does the following LC-3 program do? 
      
        .ORIG  x3000
        LD R3 , A
        STI R3, KBSR
AGAIN   LD R0,B
        TRAP X21
        BRnzp AGAIN
A       .FILL X4000
B       .FILL X0032
KBSR    .FILL XFE00
        .END
    2. If someone strikes a key, the program will be interrupted and the keyboard interrupt service routine will be executed as shown below. What does the keyboard interrupt service routine do? 
      
        .ORIG X1000
        LDI R0,KBDR
        TRAP X21
        TRAP X21
        RTI
KBDR    .FILL XFE02
        .END
    3. Finally, suppose the program of part a started executing, and someone sitting at the keyboard struck a key. What would you see on the screen?

  9. (Adapted from 8.16)
    What does the following LC-3 program do? 
    
        .ORIG X3000
        LD R0,ASCII
        LD R1,NEG
AGAIN   LDI R2,DSR
        BRzp AGAIN
        STI R0,DDR
        ADD R0,R0,#1
        ADD R2,R0,R1
        BRnp AGAIN
        HALT
ASCII   .FILL X0041
NEG     .FILL XFFB6
DSR     .FILL XFE04
DDR     .FILL XFE06
        .END

  10. (Adapted from 9.5)
    The following LC-3 program is assembled and then executed. There are no assemble time or run-time errors. What is the output of this program? Assume all registers are initialized to 0 before the program executes. 
    
        .ORIG X3000
        ST R0, X3007
        LEA R0, LABEL
        TRAP X22
        TRAP x25
LABEL   .STRINGZ "FUNKY"
LABEL2  .STRINGZ "HELLO WORLD"
        .END

  11. The memory locations given below store students' exam scores in form of a linked list. Each node of the linked list uses three memory locations to store

    1. Address of the next node
    2. Starting address of the memory locations where name of the student is stored
    3. Starting address of the memory locations where the his/her exam score is stored

    in the given order. The first node is stored in locations x4000 ~ x4002. The ASCII code x0000 is used as a sentinel to indicate the end of the string. Both the name and exam score are stored as strings.
    Write down the student's name and score in the order that it appears in the list. 
    
    Address         Contents
    x4000           x4016
    x4001           x4003
    x4002           x4008
    x4003           x004D
    x4004           x0061
    x4005           x0072
    x4006           x0063
    x4007           x0000
    x4008           x0039
    x4009           x0030
    x400A           x0000
    x400B           x0000
    x400C           x4019
    x400D           x401E
    x400E           x004A
    x400F           x0061
    x4010           x0063
    x4011           x006B
    x4012           x0000
    x4013           x0031
    x4014           x0038
    x4015           x0000
    x4016           x400B
    x4017           x400E
    X4018           x4013
    x4019           x004D
    x401A           x0069
    x401B           x006B
    x401C           x0065
    x401D           x0000
    x401E           x0037
    x401F           x0036
    x4020           x0000

    1. The program below counts the number of zeros in a 16-bit word. Fill in the missing blanks below to make it work.
                  .ORIG x3000
            AND   R0, R0, #0
            LD    R1, SIXTEEN
            LD    R2, WORD
A           BRn   B
            ________________		
B           ________________
            BRz   C
            ________________	
            BR    A	; note: BR = BRnzp
C           ST    R0, RESULT
            HALT

SIXTEEN     .FILL #16
WORD        .BLKW #1
RESULT      .BLKW #1
            .END
    2. After you have the correct answer above, what one instruction can you change (without adding any instructions) that will make the program count the number of ones instead?

  13. Added on 11/05/2013 Suppose we use the unused opcode 1101 to specify a new instruction. We will require 3 states after decode (state 32) to complete the job. The control signals required to carry out the work of the new instruction are shown below. All control signals not shown below are 0.
    1. What does this new instruction do?
    2. There are two different formats for specifying the operands of this instruction. Fill them out below.