CISP317 Practice Exam 2

Prof. Tak Auyeung

Instructions: You may bring any material that is handwritten or printed prior to the examination to help you. You can also bring a calculator if you think it may help you. However, you can only use the calculator for numerical computations only. You cannot let your calculator compile a program, or to communicate with others.

You, as an individual, are expected to do your own work. This means you cannot seek, receive or otherwise acquire any assistance except clarifications from the professor during an examination. Any communication involving the contents of the subject matter or the examination is considered cheating. Do not initiate or accept such communication, or the result of your examination is automatically voided.

New rules, read this! As of 2003.09.22, I no longer deduct points for wrong answers. Each correct answer is worth one point, each wrong answer is worth zero point, and each unanswered question is also worth zero point. This means you should guess and leave no question unanswered.

As a result, I also need to adjust the letter grade assignment break points. For your individual examinuation, ``A'' means at least 94%, ``B'' means at least 74%, ``C'' means at least 54%, ``D'' means at least 34% and ``F'' means below 34%. The break points for the final grade are now 26.83%, 48.5%, 70.17% and 91.83% as minimums for ``D'', ``C'', ``B'' and ``A'', respectively.

Please note that this change does not affect your letter grade at all, it is just a number game to make some people feel better about guessing.

Make sure you write down you name on the upper right corner first, otherwise I cannot give points to anonymous students!

The baseline is XX, there are YY questions.

1. Assume the following byte sequence is at label s1 and s2 (bytes represented in hexadecimal digits):

   s1: 61 62 63 64 00 66 67 68
   s2: 31 32 33 00 35 36 37 38
   

   What does the following code do?

      ldi   r26,low(s1)
      ldi   r27,high(s1)
      ldi   r28,low(s2)
      ldi   r29,high(s2)
      ldi   r16,2
   
   l1:
      ld    r17,X+
      st    Y+,r17
      cpi   r17,0
      breq  l2
      dec   r16
      brne  l1
      ldi   r17,0
      st    Y+,r17
   
   l2:
     nop
   

   1. s1: 61 62 63 64 00 66 67 68
      s2: 61 62 63 64 00 36 37 68
      

   2. s1: 61 62 63 64 00 66 67 68
      s2: 61 62 33 00 35 36 37 38
      

   3. s1: 61 62 63 64 00 66 67 68
      s2: 61 62 63 00 35 36 37 38
      

   4. s1: 31 32 33 00 00 66 67 68
      s2: 31 32 33 00 35 36 37 38
      

   5. s1: 61 62 63 64 00 66 67 68
      s2: 61 62 00 00 35 36 37 38
      

2. What is the result of the following instructions? Assume r0 has a value of 0x59 and r1 has a value of 0x31:

      rol   r0
      rol   r1
   

   1. r0=0xb2, r1=0x62
   2. r0=0xb?, r1=0x62, ? depends on the value of the C flag before the first instruction
   3. r0=0xb2, r1=0x6?, ? depends on the value of the C flag before the first instruction
   4. r0=0x24, r1=0x98
   5. r0=0x?4, r1=0x98, ? depends on the value of the C flag before the first instruction
3. What is the minimum for ? to guarantee that r0 becomes zero at the nop instruction?

      ldi   r16,?
   l1:
      cpi   r16,0
      breq  l2
      lsl   r0
      dec   r16
      rjmp  l1
   l2:
      nop
   

   1. 0
   2. 7
   3. 8
   4. 255
   5. 256
4. What is the value of r0 after these instruction? Assume r0=0xbd before these instructions:

      lsr   r0
      lsr   r0
      lsr   r0
      lsr   r0
      lsl   r0
      lsl   r0
      lsl   r0
      lsl   r0
   

   1. 0xbd
   2. 0x0b
   3. 0xd0
   4. 0xb0
   5. cannot be determined because it depends on the initial value of the C flag
5. Assume r16 contains the least significant byte of a number $n$, and r17 contains the most significant byte of the same number. What is the value of r18 after the following instructions?

      
      ldi   r20,3
   l1:
      lsr   r17
      ror   r16
      ror   r18
      dec   r20
      brne  l1
      ldi   r20,5
   l2:
      lsr   r18
      dec   r20
      brne  l2
   

   1. 3
   2. 5
   3. the dividend of $n \div 5$
   4. the remainder of $n \div 8$
   5. the dividend of $n \div 32$
6. What happens when the following code is executed? The inc instruction affects the N flag.

      sub   r0,r0
   l1:
      inc   r0
      brlo  l1
      nop
   

   1. this code never gets to the nop instruction
   2. this code does not assemble
   3. when this code executes nop, r0 has a value of 0x00
   4. when this code executes nop, r0 has a value of 0x80
   5. when this code executes nop, r0 has a value of 0xff
7. As 8-bit signed numbers, which of the following represents the least value?
   1. $00001010_2$
   2. $00100011_2$
   3. $11001010_2$
   4. $00001011_2$
   5. $00110000_2$
8. Which single instruction replaces the following instruction sequence without any change of behavior?

   breq   l1
   brne   l1
   

   1. nop
   2. rjmp l1
   3. brcc l1
   4. brcs l1
   5. breq l1
9. What is the 16-bit value stored in r0 (LSB) and r1 (MSB) after the following code executes? Assume the following byte sequence (in hexadecimal digits) is at label m1:

   m1: 01 00 00 10 02 00 00 20 00 01
   

   The code to analyze is as follows:

       ldi   r26,low(m1)
      ldi   r27,high(m1)
      sub   r0,r0
      sub   r1,r1
      ldi   r16,4
   
   l1:
      ld    r2,X+
      add   r0,r2
      ld    r2,X+
      adc   r1,r2
      dec   r16
      brne  l1
   

   (Reminder: LSB is the rightmost byte, MSB is the leftmost one.)

   1. 0x3021
   2. 0x1202
   3. 0x3013
   4. 0x0100
   5. 0x3103
10. Which of the following instruction sequences end up at label l1 if and only if r16 has a value of 0? Assume the instruction following the sequences is not at label l1.

    1.    cpi   r16,0
          breq  l1
       

    2.    cp    r16,r16
          breq  l1
       

    3.    cpi   r16,0
          brne  l1
       

    4.    cp    r16,r16
          brne  l1
       

    5.    cpi   r16,0
          brcc  l1
       

11. Assume a block of memory is allocated as follows:

    blk1: .byte 40
    blk2:
    

    What does the following code do?

       ldi   r26,low(blk1)
       ldi   r27,high(blk1)
       ldi   r16,low(blk2)
       ldi   r17,high(blk2)
       sub   r0,r0
    
    l1:
       cp    r26,r16
       cpc   r27,r17
       brcc  l2
       st    X+,r0
       rjmp  l1
    
    l2:
      nop
    

    1. initializes all 40 allocated bytes to zero
    2. initializes all except the first of the 40 allocated bytes to zero
    3. initializes all except the last of the 40 allocated bytes to zero
    4. never terminates (infinite loop)
    5. the code does not assemble
12. What are the values of flags after the following instruction, assuming r0=0x54 and r1=0x5d?

       cp  r0,r1
    

    1. N=1, Z=0, C=0
    2. N=1, Z=0, C=1
    3. N=0, Z=1, C=0
    4. N=0, Z=0, C=1
    5. N=0, Z=1, C=1