Final Review Questions (without Answers)

There will be ONLY one online exam including both lectures and labs on Tuesday, August 24, 2021 at 9:00 AM. This is a online exam via Canvas/Zoom.

Review

The final exam covers topics after the lab 3 and is NOT comprehensive.

Overall, the exam may include material from the following sources:

• This review
• labs 4 through 7

I will not ask any questions which were not somehow covered by one of the above three sources.

Questions

1. What is wrong with the following code, if anything?

     .equ Exit, 0x11
     .equ Open, 0x66
     .equ Close, 0x68
     .equ Read_Int, 0x6C
   
     .data
   filename:
     .asciz "myFile.txt"
   
     .text
     .global _start
   _start:
     ;; open the file
     ldr r0, =filename
     mov r1, #0
     swi Open
   
     ;; read an integer from it
     swi Read_Int
   
     ;; close the file
     swi Close
   
     ;; exit the program
     swi Exit
     .end
   

2. What is wrong with the following code, if anything?

     .equ Write_Int, 0x6B
   
     .text
     .global _start
   _start:
     ;; print out 42
     mov r0, #1
     mov r1, #42
     swi Write_Int
   

3. Write ARM assembly code which will read two integers from the file myFile.txt and print them out.

4. Consider the following code, which sets up a .data section:

     .data
   label1:
     .asciz "Hi"
   label2:
     .word 1, 2
   label3:
     .asciz "Bye"
   

   Assuming the .data section starts at address 0, how does this look in memory? Use the following table as a template.

   Value
   Index	0	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20

5. Convert the following Java/C-like code into ARM assembly. The names of the variables reflect which registers must be used for the ARM assembly.

   if (r0 >= 5) {
     r1 = r6;
   } else {
     r2 = r7;
   }
   

6. Convert the following Java/C-like code into ARM assembly. Use branch intructions instead of conditional execution. The names of the variables reflect which registers must be used for the ARM assembly.

   if (r5 < r6) {
     r2 = r3;  
     print_string("Less");
   } else if (r5 == r6) {
     r3 = r4;	 
     print_string("Equal");
   } else {
     r4 = r5; 
     print_string("Greater"); 
   }
   

7. Convert the following Java/C-like code into ARM assembly. The names of the variables reflect which registers must be used for the ARM assembly.

   for (int r2 = r1; r2 <= 150; r2 += 4) {
     int r3 = (r2 - 1) * (r2 + 1);
     print_int(r3);
     print_char('\n');
   }
   

8. Convert the following Java/C-like code into ARM assembly. The names of the variables reflect which registers must be used for the ARM assembly. Non-register variable names indicate a value that should be stored in memory.

   int[] myArray = new int[]{19, 21, -5, 4};          
   int r2 = 0;
   int r3 = 0;
   do {
     r2 += myArray[r3];
     r3++;
   } while (r3 < 4);
   print_int(r2);
   

9. Convert the following Java/C-like code into ARM assembly. The names of the variables reflect which registers must be used for the ARM assembly. Non-register variable names indicate a value that should be stored in memory.

   int myArray[4] = {19, 21, -5, 4};
   int* r2 = myArray;
   int r3 = 4;
   int r4 = 0;
   do {
     r4 += *r2;
     r2++;
     r3--;
   } while (r3 != 0);
   print_int(r4);
   

10. Convert the following Java/C-like code into ARM assembly. The names of the variables reflect which registers must be used for the ARM assembly.

    if (r2 < r3 && r3 < r4) {
      r5 = r6;
    } else {
      r6 = r5;
    }
    

11. Convert the following Java/C-like code into ARM assembly. The names of the variables reflect which registers must be used for the ARM assembly.

    if (r2 < r3 || r3 < r4) {
      r5 = r6;
    } else {
      r6 = r5;
    }
    

12. Convert the following Java/C-like code into ARM assembly. The names of the variables reflect which registers must be used for the ARM assembly. Non-register variable names indicate a value that should be stored in memory.

    int[] first = new int[]{0, 5, 27, 98};
    int[] second = new int[]{1, 2, 8, 29, 42};
    int[] result = new int[9];
    int r0 = 0;
    int r1 = 0;
    
    while (r0 < 4) {
      result[r0] = first[r0];
      r0++;
    }
    
    while (r1 < 5) {
      result[r0] = second[r1];
      r0++;
      r1++;
    }
    
    for (r2 = 0; r2 < 9; r2++) {
      print_int(result[r2]);
      print_newline();
    }
    

13. What component is shown below?
    
14. What component is shown below?
    
15. What component is shown below?
    
16. Draw the circuit corresponding to the following sum-of-products equation, where !A refers to the negation of variable A, and so on:

    R = !A!B + AB
    

17. Consider the following sum-of-products equation:

    R = !ABC + ABC + A!B!C
    

    Using the above equation, do the following:

    • Write it as a truth table
    • Simplify it using boolean algebra
    • Simplify it using a Karnaugh map
18. Consider the truth table augmented with don't cares, shown below:
    
    

    A	B	C	D	U
    0	0	0	0	1
    0	0	0	1	X
    0	0	1	0	0
    0	0	1	1	1
    0	1	0	0	X
    0	1	0	1	1
    0	1	1	0	0
    0	1	1	1	X
    1	0	0	0	1
    1	0	0	1	0
    1	0	1	0	X
    1	0	1	1	0
    1	1	0	0	1
    1	1	0	1	0
    1	1	1	0	X
    1	1	1	1	0

    Using the above truth table, write out the following:

    1. The unoptimized sum-of-products equation, skipping over don't cares
    2. A Karnaugh map, along with boxes which exploit don't cares where appropriate.
    3. An optimized sum-of-products equation, derived from the Karnaugh map created in the previous step.