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. From PS4.
   (Adapted from 6.14) Consider the following machine language program:

   	AND R2, R2, #0	R2 <- 0
   LOOP	ADD R1, R1, #-3	R1 <- R1-3
   	BRn END	End when R1 is negative
   	ADD R2, R2, #1	R2 <- R2+1
   	BRnzp LOOP
   END	HALT

   What are the possible initial values of R1 that cause the final value in R2 to be 3?

   For R2 to contain the value 3, the BRn must not have intiated a branch for 3 consecutive times. Therefore, R1 wasn't negative after the instruction ADD R1, R1, #-3 was exectued 3 times and was negative after the instruction was executed 4 times. That is, R1-3x3 = R1-9 >= 0 and R1-3x4 = R1-12 < 0. Solving the inequalities yields, 9 <= R1 < 12. Since a register contains integers, R1 could have been 9, 10, or 11.




2. From PS4.
   (Adapted from 7.16) Assume a sequence of nonnegative integers is stored in consecutive memory locations, one integer per memory location, starting at location x4000. Each integer has a value between 0 and 30,000 (decimal). The sequence terminates with the value -1 (i.e., xFFFF).
   1. Create the symbol table entries generated by the assembler when translating the following routine into machine code:

      	.ORIG x3000
      	AND R4, R4, #0
      	AND R3, R3, #0
      	LD R0, NUMBERS
      LOOP	LDR R1, R0, #0
      	NOT R2, R1
      	BRz DONE
      	AND R2, R1, #1
      	BRz L1
      	ADD R4, R4, #1
      	BRnzp NEXT
      L1	ADD R3, R3, #1
      NEXT	ADD R0, R0, #1
      	BRnzp LOOP
      DONE	TRAP x25
      NUMBERS	.FILL x4000
      	.END

      Symbol Table

      Label	Memory Address
      LOOP	x3003
      L1	x300A
      NEXT	x300B
      DONE	x300D
      NUMBERS	x300E

      
      
   2. What does the above program do?

      The instruction AND R2, R1, #1 performs a bit mask (x0001) to decide whether the least significant bit of the value is 0 or 1. The LSB of a number is used to determine whether the integer was even or odd. For example, numbers with a zero LSB are: 0000 (#0), 0010 (#2), 0100 (#4), 0110 (#6), which are all even. Hence, R3 counts the amount of even numbers in the list and R4 counts the amount of odd numbers.




3. From PS4.
   Below is a segment of LC-3 machine language program.

   	ADD R2, R1, #0
   HERE	ADD R3, R2, #-1
   	AND R3, R3, R2
   	BRz END
   	ADD R2, R2, #1
   	BRnzp HERE
   END	HALT

   If the data in R1 is an unsigned integer larger than 1, what does the program do? (Hint: what is the relationship between the resulting integer in R2 and the original integer in R1?)

   The program finds out the smallest power of 2 which is larger than or equal to the unsigned integer in R1.




4. From PS4.
   (Adapted from 7.18) The following LC-3 program compares two character strings of the same length. The source strings are in the .STRINGZ form. The first string starts at memory location x4000, and the second string starts at memory location x4100. If the strings are the same, the program terminates with the value 1 in R5; otherwise the program terminates with the value 0 in R5. Insert one instruction each at (a), (b), and (c) that will complete the program. Note: The memory location immediately following each string contains x0000.

   	.ORIG x3000
   	LD R1, FIRST
   	LD R2, SECOND
   	AND R0, R0, #0
   LOOP	LDR R3, R1, #0	; (a)
   	LDR R4, R2, #0
   	BRz NEXT
   	ADD R1, R1, #1
   	ADD R2, R2, #1
   	NOT R4, R4	; (b)
   	ADD R4, R4, #1	; (c)
   	ADD R3, R3, R4
   	BRz LOOP
   	AND R5, R5, #0
   	BRnzp DONE
   NEXT	AND R5, R5, #0
   	ADD R5, R5, #1
   DONE	TRAP x25
   FIRST	.FILL x4000
   SECOND	.FILL x4100
   	.END




5. From PS4.
   The data at memory address x3500 is a bit vector with each bit representing whether a certain power plant in the area is generating electricity (bit = 1) or not (bit = 0). The program counts the number of power plants that generate electricity and stores the result at x3501. However, the program contains a mistake which prevents it from correctly counting the number of electricity generating (operational) power plants. Identify it and explain how to fix it.

   	.ORIG x3000
   	AND R0, R0, #0
   	LD R1, NUMBITS
   	LDI R2, VECTOR
   	ADD R3, R0, #1
   CHECK	AND R4, R2, R3
   	BRz NOTOPER
   	ADD R0, R0, #1
   NOTOPER	ADD R3, R3, R3
   	ADD R1, R1, #-1
   	BRp CHECK
   	LD R2, VECTOR	<- missing instruction
   	STR R0, R2, #1
   	TRAP x25
   NUMBITS	.FILL #16
   VECTOR	.FILL x3500
   	.END

   R2 contains the bit vector, and not the address at which the bit vector is contained. The instruction LDI R2, VECTOR loaded the value at x3500 into R2 since the value at the memory address labeled as VECTOR was used as the address from which to load. The store instruction STR R0, R2, #1 uses the value of R2 to evaluate an address. However, R2 must be modified to contain an address for an STR instruction to work. Thus, LD R2, VECTOR is an additional required instruction.




6. From PS4.
   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:

   1. What will the values of registers R0 through R4 be after the LC-3 finishes executing the ADD instruction?

   R0 - x300B
   R1 - x300D
   R2 - x000A
   R3 - x1263
   R4 - x300B

   2. What will the values of memory locations Addr1 through Addr5 be after the LC-3 finishes executing the HALT instruction?

   Addr1 - x300B
   Addr2 - x000A
   Addr3 - x000A
   Addr4 - x300B
   Addr5 - x300D




7. From PS4.
   

   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?

      Some of the characters written to the DDR weren't being output to the screen.

   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?

      The handler didn't check the KBSR before inputting the character.




8. From PS4. Added 10/26/2011.
   (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	0	0	0	0	0	0	1	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	0	0	0	0	0	0	0	1	1
   x3003	1	1	1	1	0	0	0	0	0	0	1	0	0	0	0	1
   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	0	1	0	0	0	0	0	0	0	0	0	0	0	0	0	1

   
   
   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.




9. 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.

   
   
   Solution: If sequence starting at location in R0 is null, instruction at label A is never executed. However, if there is a string of characters at R0, the instruction at label A is executed an infinite number of times. (Why?) because the RET at END will always go back to LD R0,SAVER0.
   TRAP handler needs to save the registers prior to using them within the handler. i.e.R1 in this case. R7 must saved before using TRAP x21 and restored afterwards.

   
   ; 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    ST R7,SAVER7 
   	 ST R1,SAVER1 
   	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	 LD R7,SAVER7 
   	 LD R1,SAVER1 
   	RET
    
   SAVER0  .BLKW #1
    SAVER7 .BLKW #1
    SAVER1 .BLKW #1
           .END
   




10. (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?

    Solution:
    
    a. 256 TRAP service routines can be implemented. x0000- x00FF
    
    b.
    RET stores the value of PC (before execution of the service routine) in R7 so that it can return control to the original program after execution of the service routine. A BRnzp would not work because:
    • the TRAP routine may not be reached by a 9 bit offset.
    • if TRAP is called multiple times, the computer would not know which LABEL to go to (can change every time).
    
    c. 2 memory accesses are made during TRAP instruction
    1st access:- instruction in fetch
    2nd access:- trap vector table to get address of TRAP service routine