Grade Distributions
Exam 2
Mean = 3.88
Median = 3.8
X X
X X X X
X X X X
X X X X X
X X X X X X X
X X X X X X X X X X X
X X X X X X X X X X X
_______X_X_X_X_X_X_X_X_X_X_X_X___X___________
0 . 1 . 2 . 3 . 4 . 5 . 6 . 7 . 8 . 9 . 10
Exams 1 and 2
Mean = 10.15 X
Median = 10.1 X X
X X X X
X X X X X X X X
X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X X X
___________X_X_X_X_X_X_X_X_X_X_X_X_X_X_______
0 . 2 . 4 . 6 . 8 . 10. 12. 14. 16. 18. 20
Homework 1 to 9
X
Mean = 23.90 X
Median = 25.8 X
X X
X X
X X
X X X
X X X X X
X X X X X
X X X X X X X
X X X X X X X
X X X X X X X X X
_____________________________X_X_X___X___X_X_X_X_X_X_X_X_X_
0 . 2 . 4 . 6 . 8 . 10. 12. 14. 16. 18. 20. 22. 24. 26.
Machine Problems 1 to 4
Mean = 11.70 X
Median = 12.0 X X X
X X X
X X X
X X X X X
X X X X X X X
X X X X X X X X
X X X X X X X X X X X
___X_X_______X_X_X_X_X_X_X_X_X_X_X_X_X_X_X_X_
0 . 2 . 4 . 6 . 8 . 10. 12. 14. 16. 18. 20
Total Scores
mean = 45.58 X
median = 48.1 X
X X
X X
X X
X X X X X X X X
X X X X X X X X X X
X X X X X X X X X X X X X X
___X_______X_X_X_X_X_X_X_X_X_X_X_X_X_X_X_X_X_X___
20. 24. 28. 32. 36. 40. 44. 48. 52. 56. 60. 62
D C B A
If this was the end of the semester, the grade scale might look something
like the above. The D-C division is largely decided by the fact that those
with overall scores below 33 have generally failed to demonstrate competence
in programming.
Note: We will only count the top 10 homework scores, no
matter how many are assigned.
Solutions and Commentary
HAWK EMULATOR
/------------------CPU------------------\ /----MEMORY----\
PC: 00000000 R8: 00000050 FFFFFC: --
PSW: 0000FF03 R1: 000010B2 R9: 00000009 FFFFFE: --
NZVC: 0 0 1 1 R2: 00010080 RA: 00000000 ->000000: LIL R2,#010070
R3: 16710189 RB: 00000000 000004: JSR R1,#00032C
R4: 00084F70 RC: 00000000 000008: LIL R1,#00109E
R5: 00000050 RD: 00000000 00000C: JSRS R1,R1
R6: 00000000 RE: 00000000 00000E: BR #000000
R7: FF000000 RF: 00000000 000010: CPUSET R2,#3
**HALTED** r(run) s(step) q(quit) ?(help)
Bus Trap. Trap PC = #00001022
Trap MA = #16710188 invalid addr
+0000000C: F1 A2 25 PLOTIT: STORES R1,R2
+0000000E: F8 22 0004 26 STORE R8,R2,X
+00000012: F9 22 0008 27 STORE R9,R2,Y
+00000016: F2 62 000C 28 ADDI R2,R2,ARSIZE
+0000001A: F3 D8 29 LOADS R3,R8
+0000001C: F4 D9 30 LOADS R4,R9
+0000001E: 93 01 31 SR R3,1
+00000020: 94 01 32 SR R4,1
+00000022: F8 D3 33 LOADS R8,R3
+00000024: F9 D4 34 LOADS R9,R4
+00000026: E1 +000000 35 LIL R1,PUTAT
- Give the line number of the instruction that directly caused the trap.
___33_____________
- Which register held the value that directly caused this trap?
___R3______________________
The LOADS R3,R8 on line 33 is at address 2216,
and R3 agrees with the trap MA.
Most of the class got this right.
- The programmer used LOADS when they should have used what?
___MOVE_________________
Every single LOADS in the above code shoul have been MOVE!
Only 1/4 of the class got this.
- EXTB R3,R4,R0 does the same thing as
MOVE R3,R4 followed by TRUNC R3,_8___________
Only 1/5 of the class got this.
MOVESL R1,R3,5
ADDSL R3,R1,3
|
- The code to the right multiplies by _40____________
1/3 got this, 1/4 got 288, 1/9 got 297. The wrong answers came from looking
at shift counts without close attention to the use of R1.
_ADDSL___R3,R2,2_
_SL______R3,3____
|
- Can you multiply by the same without using R1?
(answer to the right)
1/2 the class were able to multiply by their answer to the previous question.
FUNC: ; expects __R3_=_x_______________
; __R4_=_y_______________
; uses __R3,_R4,_R5_=_z_______
; returns __R3___________________
TESTR R4
BZS LABEL2 ; _if_(y_!=_0)_{______
LABEL1: ; ___do_{_____________
MOVE R5,R3
AND R5,R4 ; _____z_=_x_&_y______
EQU R3,R4
NOT R3 ; _____x_=_x_^_y______
MOVESL R4,R5,1 ; _____y_=_z_<<_1_____
BZR LABEL1 ; ___}_while_(y_!=_0)_
LABEL2: ; _}__________________
JUMPS R1 ; _return_x___________
- In the appropriate spaces above, identify and name (briefly and concisely)
the registers used above for parameters, return
values, and temporary values.
1/10 did well here.
Many students said that R4 was the return value, despite a) the fact
that our convention is that R3 is always the return value and b) the
fact that this code can only return when R4 is zero.
- In the appropriate spaces above, document all the if statements and loops.
(use a C or Java-like style of comment, with appropriate indenting)
Only 1/20 did well here. Most used if-goto style comments, or worse,
comments that suggested calls to labels instead of just branches. Those
who commented the loop as a loop rarely indicated the end of loop with a closing
brace. Hardly anyone used indenting to set off the loop body.
- In the appropriate spaces above, document all the assignment statements.
(use C or Java-ish notation and indenting)
Perhaps 1/6 did well here. Many gave verbose English instead of boiling their
comments down to assignment statements. Most either did not understand
the EQU instruction or ignored it. Few correctly understood that
the combination of EQU and NOT used her is an exclusive or,
which is ^ in C, C++, Java and Python.
Many used && when they should have used &.
Most correctly understood the shift instruction.
- If the code above is called with parameters
310 and 510 what does it return?
__8_______
Only 1/20 got this. Here is a table of successive register values (in binary)
through the computation:
| x | y
|
|---|
| 0011 | 0101
|
| 0110 | 0010
|
| 0100 | 0100
|
| 0000 | 1000
|
| 1000 | 0000
|
- How many iterations does it take to reach that value?
__4______________________________
1/11 did well here. Wild guesses seemed common.
- In the space below, draw the logic circuit that describes a bit slice of
what the above code does to each bit of the two primary variables during
each iteration:
2 did well here. Most of the logic diagrams had an and gate and
not gate with interconnections unrelated to the code given. Most
only had one output, despite the explicit statement in the question
that the circuit should update "the two primary variables." Many had input
or output labels unrelated to either the registers used in the code or to
names used for those variables in the comments written for problem 7.
| _xi_ | _yi_
| _xi_ | yi+1
|
|---|
|
|
| _0__ | _0__
| _0__ | _0__ |
|
|
| _0__ | _1__
| _1__ | _0__ |
|
|
| _1__ | _0__
| _1__ | _0__ |
|
|
| _1__ | _1__
| _0__ | _1__
|
- Label the columns in the truth table to the right to match the inputs
and outputs of the logic circuit.
1/4 gave labels that matched the logic circuit drawn in answer to problem 12.
1/4 earned partial credit, usually by adding spurious outputs.
1/6 gave answers with no discernible connection to the logic diagram.
1/5 left it blank.
- Fill in the values on the input side of the truth table.
5/7 gave the standard arrangement of input values for a 2-input truth table.
1/6 gave no answer. The most common error leading to partial credit
was to give the values in some kind of nonstandard order. Those who claimed
the circuit had more than 2 inputs but only gave 4 rows earned no credit.
- Fill in the values on the output side of the truth table so they
properly describe the circuit you have drawn for problem 12.
1/3 gave output values that corresponded to what their answer to problem 12.
1/3 earned no credit, typically by filling in outputs with no discernible
relationship to the logic diagram.
would produce.
- What does FUNC do? ___It_adds__________________________
2 got this.
Most gave wild guesses. Greatest common divisor? Least common multiple?
Definitely not.
; Fibonacci code to test ADD and SUB functions
;RETAD = 0;
I = 4;
ARSIZE = 8;
FIB:
; expects R3 = i
STORES R1,R2
ADDSI R2,ARSIZE
CMPI R3,1
BLE FIBQ ; if (i > 1) { // recursive case
STORE R3,R2,I-ARSIZE ; -- save i
LIS R4,1 ; -- param
JSR R1,SUB ; -- sub( i, 1 )
JSR R1,FIB ; -- fib( i - 1 )
LOAD R4,R2,I-ARSIZE
STORE R3,R2,I-ARSIZE ; -- save fib( i - 1 )
MOVE R3,R4 ; -- param i
LIS R4,2 ; -- param
JSR R1,SUB ; -- sub( i, 2 )
JSR R1,FIB ; -- fib( i - 2 )
LOAD R4,R2,I-ARSIZE ; -- param fib( i - 1 )
JSR R1,ADD ; i = add( fib( i - 2 ), fib( i - 1 ) )
FIBQ: ; }
LOADS PC,R2 ; return i;
- When called from a test program, the above code fails by returning to
location zero, sometimes after several recursive calls. What is the error?
___The_AR_gets_pushed_at_the_start,_but_never_popped_at_the_end____________
1/5 did well. 1/9 understood that the problem involved something about
the activation record. 1/2 gave answers unrelated to the problem, many of
which could not possibly lead to the symptom described. 1/11 left it blank.
- Why does the code use I-ARSIZE to access its local variable
instead of just I?
___Because_the_access_is_after_the_AR_is_pushed____________________________
1/6 got this. 2/7 understood that the issue involved something about
the activation record. Almost all of the rest gave answers that didn't
seem to connect with the issue.
- The local variable I in the activation record is used for the
parameter i, but it is also used for something else. What?
___It_is_used_to_save_the_result_of_fib(i-1)_during_the_call_to_fib(i-2)___
1/9 got this, 1/9 earned good partial credit for saying it had something
to do with recursive calls. 1/8 earned a bit of credit for saying something
about needing to save something to compute the sum. 1/2 gave answers that
didn't seem to indicate a useful level of understanding of the code.
- If ARSIZE was larger, what code in the above would need changing?
___ADDSI_R2,ARSIZE_would_change_to_ADDI_R2,R2,ARSIZE_______________________
1/4 got this. 1/8 gave strong hints for strong partial credit. Most of the
remainder earned no credit. The most common useless answer was that if
ARSIZE was bigger, the definition of
ARSIZE would need to be changed. (Yes, but that is trivially implied
by the question.)
For the curious: The code used for problems 7 to 16 is the add function called
by the code for problems 17 to 20. The entire program is
here.