## Department of Electrical and Computer Engineering

### The University of Texas at Austin

EE 306, Fall 2006
Problem Set 6
Due: Not to be turned in
Yale N. Patt, Instructor
TAs: Aseem Bathla, Cameron Davison, Lisa de la Fuente, Phillip Duran, Jose Joao,
Jasveen Kaur, Rustam Miftakhutdinov, Veynu Narasiman, Nady Obeid, Poorna Samanta

Note: This problem set is unusually long, and is not to be turned in. We have put it together and handed it out to give you some challenging examples to help you prepare for the final exam.
1. A zero-address machine is a stack-based 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.

1. 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
PUSH Z
SUB
DIV
POP U
```
2. Write the assembly language code for a zero-address 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))

2. Do Problem 9.13 on page 244 in the textbook.

3. Do Problem 6.16 on page 175 in the textbook.

4. Consider the following LC-3 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
LOOP2   JSR     SUB1
BRzp    LABEL
JSR     SUB2
BRp     LOOP2
BRp     LOOP1
HALT
DATA    .BLKW   #10
SUB1    LDR     R5, R2, #0
NOT     R5, R5
LDR     R6, R2, #1
RET
SUB2    LDR     R4, R2, #0
LDR     R5, R2, #1
STR     R4, R2, #1
STR     R5, R2, #0
RET
.END
```
5. 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.

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
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?

6. (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 non-negative 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
JSR POP
_______________
JSR POP
_______________
BRp AGAIN
JSR PUSH
_______________
LD R0, SAVER0
LD R1, SAVER1
LD R2, SAVER2
LD R5, SAVER5
RET
```
7. 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
_______________
AND R1, R1, #0
BRp INLOOP
_______________
BRn OUTLOOP
OUTPUT   LD R0, ZERO
_______________
TRAP x21
HALT
NUM      .BLKW 1
ZERO     .FILL x30
.END
```
8. The figure below shows the part of the LC-3 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 16-bit output of INMUX, and F is the 2-bit 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 LC-3. 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

9. 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 non-zero 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
LD R1, COUNT
LOOP          LDR R2, R6, #0
AND R5, R0, R2
BRz SKIP