Exam 1: MidtermSolutions and Commentary
Part of
the homework for 22C:60 (CS:2630), Spring 2013
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Mean = 7.39 X
Median = 7.8 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__
0 . 1 . 2 . 3 . 4 . 5 . 6 . 7 . 8 . 9 . 10.
Mean = 8.29 XXXX
Median = 10.0 XXXX
XXXX
XXXX
XXXX
X X XXXX
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.
Mean = 12.34 X
Median = 13.4 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__
. 3 . 4 . 5 . 6 . 7 . 8 . 9 . 10. 11. 12. 13. 14. 15. 16. 17. 18.
Mean = 27.57 X
Median = 29.3 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_______
4 . 6 . 8 . 10. 12. 14. 16. 18. 20. 22. 24. 26. 28. 30. 32. 34. 36. 38. 40
Midterm Grade Scale F | D | C | B | A |
Note: There were two versions of this exam that differed in Problems 1, 2 and 4.
Bit Pattern x 110010 001001 100101 010111 110011 Unsigned decimal ___50___ ____9___ ___37___ ___23___ ___51___ 1's complement of x _001101_ _110110_ _011010_ _101000_ _001100_ (binary) 2's complement of x _001110_ _110111_ _011011_ _101001_ _001101_ x as a 1's comp num __-13___ ____9___ __-26___ ___23___ __-12___ (signed decimal) x as a 2's comp num __-14___ ____9___ __-27___ ___23___ __-13___
1 in 4 did perfect work here. A number of students made various clerical errors. About 1 in 4 had serious problems with signed number representations. The most common error in this category involved the mindless negation of all signed values, even if the sign was positive, but many students gave apparently random results.
The solution presented above is version B of the exam. In the original, one entry in each row and column was pre-solved so that students had one last chance to learn how to do it correctly while working the exam.
. = 0 || Address 3 2 1 0
A: H #4321 || |----|----|----|----|
B: B #65,#87 || 000000: | 87 | 65 | 43 | 21 |
; solved to here || |----|----|----|----|
ALIGN 2 || 000004: | | 0D | 08 | 42 |
B "B",C || |----|----|----|----|
B #D || 000008: | 00 | 0A | 00 | 00 |
ALIGN 4 || |----|----|----|----|
C: H A,#A || 00000C: | 0C | 08 | 02 | 00 |
ALIGN 4 || |----|----|----|----|
D: B A,B,C,D || 000010: | | | | |
END || |----|----|----|----|
|| 000014: | | | | |
|| |----|----|----|----|
|| 000018: | | | | |
|| |----|----|----|----|
|| 00001C: | | | | |
|----|----|----|----|
1 in 5 did perfect work here. 1 in 8 gave answers that were total nonsense, and 1 in 6 gave answers that earned almost no credit. There were two major classes of problems: First, students who did not understand labels. For example students looking at the above code associated the value #4321 with A instead of the address zero. Second, some students did not understand the ALIGN directive. For example, some students assumed that forced all successive lines to be aligned, as opposed to inserting a one-time padding to force the indicated alignment.
The solution presented above is version A of the exam. In both versions, the solution to (bytes 00 to 03 of the answer) was given on the exam, so that students had a last-minute chance to learn something about how the assembler worked.
Addr: Val || . = #1000
1000: ___A2F1___ ||
|| FUNCT:
1002: ___3420___ || STORES R1,R2
|| CMP R3,R4
1004: ___0306___ || BLE FUNXT
|| ADD R3,R3,R4
1006: ___3433___ || SUB R4,R3,R4
|| SUB R3,R3,R4
1008: ___3424___ || FUNXT:
|| LOADS R1,R2
100A: ___3423___ || JUMPS R1
||
100C: ___D2F1___ ||
||
100E: ___B1F0___ ||
||
1010: __________ ||
||
1012: __________ ||
||
1014: __________ ||
1 in 6 gave perfect answers. Small clerical answers were quite common, and 1 in 10 left the entire problem blank. (It was an open book exam, perhaps those students forgot to bring their book? But note, students who knew anything about the Hawk architecture should have been able to fill in the register fields of the instructions without a book, earning roughly half credit). The most common errors had to do with computing the branch displacement for the BLE instruction. 1 in 5 who did the rest of the problem reasonably well could not do this. The second most common problem among those who generally did well was to reverse the byte order. 1 in 12 who did the rest reasonably made this error.
b) It is a subroutine. If FUNCT is called with the integers i in R3 and j in R4, what values are returned? (The answer is very short.) (0.5 points)
__It_sorts_things_so_i_<=_j_(R3<=R4)____
1 in 12 got this. 1 in 15 concluded that the code swaps the values of i and j, earning 0.3 points. 1 in 6 left earned no credit while giving odd and confusing answers. 1 in 10 left this blank.
Admittedly, this code uses a trick for exchanging the values of two variables without using a third, temporary variable. Aside from this, though, the body of the routine is directly based on that from the solution to problem 4 on homework assignment 5, so it should have looked at least marginally familiar.
DATA = 0 ; word 0, points to the string in this node
COUNT = 4 ; word 1, how many array elements
ARRAY = 8 ; word 2, the first array element
; additional words if count requires them
The following program was somewhat optimized before parts of it were carefully cut out and replaced by blanks. Fill in the missing parts. (2 points)
SUBTITLE "Traverse the tree and print in preorder"
; activation record
;RETAD = 0 ; return address
SAVER8 = 4 ; saved copy of R8
SAVER9 = 8 ; saved copy of R9
ARSIZE = 12
TRAVERSE: ; expects R3 = p, a pointer to a tree node
; may use R4-R7
STORES R1,R2 ; -- save return address (blank in A)
STORE R8,R2,SAVER8 ; -- save R8
STORE R9,R2,SAVER9 ; -- save R9
ADDI R2,R2,ARSIZE ; -- push activation record (blank in B)
MOVE R8,R3 ; -- from here on, R8 is p
LOAD R3,R8,DATA ; -- parameter
LIL R1,PUTS
JSRS R1,R1 ; puts( p->data )
LOAD R9,R8,COUNT ; c = p->count
LEA R8,R8,ARRAY ; p = &p->array[0]
TRAVLP: ; for (;;) {
ADDSI R9,-1 ; c = c - 1 (blank in B)
BLE TRAVQT ; if (c <= 0) break (blank in A)
LOADS R3,R8 ; -- parameter
JSR R1,TRAVERSE ; traverse( *p );
ADDSI R8,4 ; p = p + 4 (blank in A)
BR TRAVLP ; } (blank in B)
TRAVQT:
ADDI R2,R2,-ARSIZE ; -- pop activation record (blank in A)
LOAD R8,R2,SAVER8 ; -- restore R8
LOAD R9,R2,SAVER9 ; -- restore R9
LOADS R1,R2 ; -- restore return address (blank in B)
JUMPS R1 ; return
1 in 5 did perfect work here. Branch instructions posed werious problems for 1 in 3 students, while load and store instructions psed serious problems for 1 in 8.
Both versions of the exam were based on the above code, with 4 blank lines added as indicated by the comments to the right. In each case, the instructions omitted were carefully chosen so that filling in the blanks would be equal in difficulty.
The code here is fairly tightly optimized, but the fill-in-the-blanks questions did not involve the optimization. The purpose of this is to avoid providing a template for a solution to Machine Problem 3, since untangling the difference in problem specification between the version solved by this code on the exam and the version assigned for the machine problem is significantly complicated by the optimizations. Nonetheless, those students studied the machine problem before the exam, as was suggested, should have done better on this problem than those who came at it cold. In addition, the size and top-level structure of the code presented on this exam question should serve as a hint about the size of the code that can solve the assigned machine problem.