 A zeroaddress machine is a stackbased machine where all operations are
done by using values stored on the operand stack. For this problem, you may
assume that the ISA allows the following operations:
PUSH M  pushes the value stored at memory location M onto the operand stack.
POP M  pops the operand stack and stores the value into memory location M.
OP  Pops two values off the operand stack and performs the binary operation OP
on the two values. The result is pushed back onto the operand stack.
Note: OP can be ADD, SUB, MUL, or DIV for parts a and b of this problem.
Note: See the stack machine supplemental handout for help on this problem.
 Draw a picture of the stack after each of the instructions below are executed.
What is the minimum number of memory locations that have to be used on the stack
for the purposes of this program? Also write an arithmetic equation expressing
u in terms of v, w, x, y, and z. The values u, v, w, x, y, and z are stored in
memory locations U, V, W, X, W, and Z.
PUSH V
PUSH W
PUSH X
PUSH Y
MUL
ADD
PUSH Z
SUB
DIV
POP U
 Write the assembly language code for a zeroaddress machine
(using the same type of instructions from part a) for calculating
the expression below. The values a, b, c, d, and e are stored in
memory locations A, B, C, D, and E.
e = ((a * ((b  c) + d))/(a + c))
 Do Problem 9.13 on page 244 in the textbook.
 Do Problem 6.16 on page 175 in the textbook.
 Consider the following LC3 assembly language program. Assumming that the memory locations DATA get filled before the program executes, what is the relationship between the final values at DATA and the initial values at DATA?
.ORIG x3000
LEA R0, DATA
AND R1, R1, #0
ADD R1, R1, #9
LOOP1 ADD R2, R0, #0
ADD R3, R1, #0
LOOP2 JSR SUB1
ADD R4, R4, #0
BRzp LABEL
JSR SUB2
LABEL ADD R2, R2, #1
ADD R3, R3, #1
BRp LOOP2
ADD R1, R1, #1
BRp LOOP1
HALT
DATA .BLKW #10
SUB1 LDR R5, R2, #0
NOT R5, R5
ADD R5, R5, #1
LDR R6, R2, #1
ADD R4, R5, R6
RET
SUB2 LDR R4, R2, #0
LDR R5, R2, #1
STR R4, R2, #1
STR R5, R2, #0
RET
.END
 During the initiation of the interrupt service routine, the N, Z, and P condition codes are saved on the stack.
By means of a simple example show how incorrect results would be generated if the condition codes were not saved.
Also, clearly describe the steps required for properly handling an interrupt.

 The program below counts the number of zeros in a 16bit 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
C ST R0, RESULT
HALT
SIXTEEN .FILL #16
WORD .BLKW #1
RESULT .BLKW #1
.END
 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?
 (Updated 12/8/06) Fill in the missing blanks so that the subroutine below implements a stack multiply. That is it pops the top two elements off the stack,
multiplies them, and pushes the result back on the stack. You can assume that the two numbers will be nonnegative integers
(greater than or equal to zero) and that their product will not produce an overflow. Also assume that the stack has been properly
initialized, the PUSH and POP subroutines have been written for you and work just as described in class, and that the stack will not
overflow or underflow.
Note: All blanks must be filled for the program to operate correctly.
MUL _______________
ST R0, SAVER0
ST R1, SAVER1
ST R2, SAVER2
ST R5, SAVER5
AND R2, R2, #0
JSR POP
ADD R1, R0, #0
JSR POP
ADD R1, R1, #0
_______________
JSR POP
AGAIN ADD R2, R2, R0
_______________
BRp AGAIN
DONE ADD R0, R2, #0
JSR PUSH
_______________
LD R0, SAVER0
LD R1, SAVER1
LD R2, SAVER2
LD R5, SAVER5
RET
 The program below calculates the closest integer greater than or equal to the square root of the number stored in NUM,
and prints it to the screen. That is, if the number stored in NUM is 25, "5" will be printed to the screen. If the number
stored in NUM is 26, "6" will be printed to the screen. Fill in the blanks below to make the program work.
Note: Assume that the value stored at NUM will be between 0 an 81.
.ORIG x3000
AND R2, R2, #0
LD R3, NUM
BRz OUTPUT
NOT R3, R3
ADD R3, R3, #1
OUTLOOP ADD R2, R2, #1
_______________
AND R1, R1, #0
INLOOP ADD R1, R1, R2
ADD R0, R0, #1
BRp INLOOP
_______________
BRn OUTLOOP
OUTPUT LD R0, ZERO
_______________
TRAP x21
HALT
NUM .BLKW 1
ZERO .FILL x30
.END
 The figure below shows the part of the LC3 data path that deals with memory and I/O.
Note the signals labeled A through F. A is the memory enable signal, if it is 1 memory is enabled, if
it is 0, memory is disabled. B, C, and D are the load enable signals for the Device Registers. If the load
enable signal is 1, the register is loaded with a value, otherwise it is not. E is the 16bit output of INMUX, and
F is the 2bit select line for INMUX.
The initial values of some of the processor registers and
the I/O registers, and some memory locations are as follows:
R0 = x0000
PC = x3000
 KBSR = x8000
KBDR = x0061
DSR = x8000
DDR = x0031
 M[x3009] = xFE00
M[x300A] = xFE02
M[x300B] = xFE04
M[x300C] = xFE06

During the entire instruction cycle, memory is accessed
between one and three times (why?). The following table lists two consecutive
instructions to be executed on the LC3. Complete the table with the values
that each signal or register takes right after each of the memory accesses
performed by the instruction. If an instruction does not require three memory accesses, draw a line accross the unused accesses.
To help you get started, we have filled some of the values for you.
PC 
Instruction 
Access 
MAR 
A 
B 
C 
D 
E[15:0] 
F[1] 
F[0] 
MDR 
x3000  LD R0, x9  1  x3000      x2009    
2          
3          
x3001  LDR R0, R0, #0  1          
2          
3          
 Note: This problem is NOT easy. In fact, it took me a while to solve it, and I am
supposed to be an expert on 306 material. So, if you are struggling to pass this
course, I suggest you ignore it. On the other hand, if you are a hot shot and think
no problem is beyond you, then by all means go for it. We put it on the problem set
to keep some of the hot shots out of mischief. We would not put it on the final,
because we think it is too difficult to put on the exam.
A programmer wrote this program to do something useful. He, however, forgot to comment his code,
and now can't remember what the program is supposed to do. Your job is to save him the trouble and
figure it out for him. In 20 words or fewer tell us what valuable information the program below provides
about the value stored in memory location INPUT. Assume that there is a nonzero value at location INPUT before the program is executed.
HINT: When testing different values of INPUT pay attention to their bit
patterns. How does the bit pattern correspond to the RESULT?
.ORIG x3000
LD R0, INPUT
AND R3, R3, #0
LEA R6, MASKS
LD R1, COUNT
LOOP LDR R2, R6, #0
ADD R3, R3, R3
AND R5, R0, R2
BRz SKIP
ADD R3, R3, #1
ADD R0, R5, #0
SKIP ADD R6, R6, #1
ADD R1, R1, #1
BRp LOOP
ST R3, RESULT
HALT
COUNT .FILL #4
MASKS .FILL 0xFF00
.FILL 0xF0F0
.FILL 0xCCCC
.FILL 0xAAAA
INPUT .BLKW 1
RESULT .BLKW 1
.END