Assembly Language & System Software
99 Spring
National Chiao-Tung University
Time: 120 minutes
Instructions:
A. There are FIVE pages.
B. There are 13 questions. You must answer all of them.
C. Write your answers on the answer book ONLY.
D. DO NOT TURN OVER the cover page until you are told to do so.
E. CHEATING IS A SERIOUS MATTER. YOU WILL RECEIVE A SCORE OF ZERO and
SERIOUS PENALTY IF YOU ARE CHEATING.
1.
Estimation: Questions 1 to 4: 10 minutes.
[ 4%] Write down the value (hexadecimal format) of AX after each of the following code
segments is executed.
(a)
(b)
(c)
(d)
mov ax, 10h
mov ax, 100
mov ax, 8
mov ax, 0f9h
or ax, 7h
and ax, 1dh
xor ax, 7
test ax, 1
2.
[ 4%] Write down the value (hexadecimal format) of AX after each of the following code
segments is executed. Note: AX is a part of EAX.
(a)
(b)
(c)
(d)
mov eax, 11h
mov eax, 99
mov eax, 0ffedh
mov eax, 0fah
shl eax, 2
shr eax, 16
sal eax, 4
ror eax, 4
3.
[2%] (a) What is the value of ABC? Your answer should be in decimal format.
symbols
WORD
1, 2, 3, 21h, 22h, 23h, 40h, 41, 42, 43
ABC
= ($ - symbols)
[1%] (b) What does the directive “.data?“ indicate ?
4.
[1%] (c) Give one example.
[6%] (each cell 0.25%) Draw and fill out the cells for memory based on the following
declarations. Each cell represents one byte. Little endian is used for storing integers. All
numbers are in hexadecimal format.
V0 BYTE
12h, 13h, 14h
BYTE
“test”, 0
V1 WORD
1024, 1025, 1026
V2 SWORD
-1, -2, -3
V3 DWORD
98765432h
Remark: you must follow the sample of the layout as follows (memory increasing from top to
bottom):
0
1
2
3
4
5
6
7
Estimation: Questions 5 to 6: 10 minutes.
5.
[2%] (a) Write at most three lines of code to combine ax:dx as a single value into eax. Hint: the
high order part of eax is the content of the initial value of ax.
[2%] (b) Write two lines of code to compute the sum of ax, bx and cx and the result is stored in
ax.
[1%] (c) When is zero flag set to one?
[1%] (d) When is overflow flag set to one?
[5%] (e) In the following instruction sequence, show the values of the Zero flag (ZF) and
overflow flag (OF) where they appears:
L1: mov
L2: test
L3: mov
L4: and
L5: mov
L6: add
L7: mov
L8: sub
L9:jmp
6.
al,
al,
al,
al,
al,
al,
al,
al,
L1
00001111b
10h
0bh
5
00001101b
11110011b
10000001b
01111111b
; ZF= ?; OF = ?
; ZF= ?; OF = ?
; ZF= ?; OF = ?
; ZF= ?; OF = ?
; ZF= ?; OF = ?
The following fragment computes a value and the value is stored in EDX.
[5%] (a) What does the fragment do? Describe in words. Assume that all the values can be
stored in a 32-bit register.
[2%] (b) What is the value stored in EDX after the fragment is executed? Your answer should
be in hexadecimal format.
mov ax, 1
xor edx, edx
mov ecx, 5
L1: mov bx, ax
push ax
push edx
mul bx
pop edx
add edx, eax
pop ax
inc ax
Loop L1
Estimation: Questions 7 to 10: 40 minutes.
7.
[8%] The following fragment computes the sum of the signed integers of an array ARRAY and
then displays the sum on the screen. The array has at least one element. The total size, i.e.
number of bytes, of the array is stored in ECX. A signed integer is represented by FOUR
bytes. Fix the syntax or semantic error for each line. You must not change the structure of
the fragment.
.data
L1: ARRAY WORD
10000 DUP(?) ; reserve space for the array
......
L2: mov edi, ARRAY ; store the address of ARRAY to edi
L3: shr ecx, 1
; compute the number of integers in the array
L4: xor ax, ax
;
L5: add ax, DWORD PTR [edi]
L6: add edi, TYPE NEXT
L7: loop L4
L8: call WriteDec
8.
[4%] EAX and EBX store integers. For different types of integers, we need different
instructions for determining the jumping branches after a comparison is performed. Explain the
purpose of each conditional jump:
(a) cmp eax, ebx
(b) cmp eax, ebx
(c) cmp eax, ebx
(d) cmp eax, ebx
ja L1
jge L1
jnl L1
jnbe
9.
[7%] Suppose that we have five SDWORDs, a, b, c, d, and e. Write assembly instructions to
compute: eax = (a-b)*(-e) / (d – b + a)
Note that you must use no more than 7 instructions. Assume that all the numbers (including
the ones from computation) can be stored in a 32-bit register. -1pts per extra instruction.
10. Translate each of the following C++ functions into a function written in assembly language:
[5%] (a)
unsigned int computeTaskOne ( unsigned int a )
{
cout << a << endl; //Hint: endl should be handled.
return a + a;
}
The result should be stored in EAX.
[8%] (b)
void computeTaskThree ( int n, int *arr, int *brr )
{
if ( n <= 0 ) return;
for ( int i = 0; i < n; ++i ) {
arr[i] = brr[i] * brr[i] – 1;
}
}
Estimation: Questions 11 to 13: 40 minutes.
11. A one-dimensional array is used for representing a 2D array which is organized as row-major
order. The numbers of rows and columns are stored in eax, and ebx, respectively. The starting
address of the array is stored in edi. An element of the array can be represented as (r, c), where
r and c are the row index and column index of the element, respectively. The index starts from
0. The data type of elements is DWORD.
Example: write one line of code to store the element (1, 2) to ecx. mov [edi+ebx*4+2*4], ecx
[1%] (a) write one line of code to store the element (0, 0) to ecx.
[1%] (b) write one line of code to store the element (0, ebx-1) to ecx
[4%] (c) write at most four lines of code to store the element (ecx, edx) to esi
[8%] (d) write at most 8 instructions to compute the sum of the edx –th row to ebx. -1pts per
extra instruction.
Hint: Draw a EAX x EBX two dimensional array for easy calculation.
12. [12%] Translate the SIC/XE program into object code (a) – (l).
Register Code: S : 4; T: 5;
X : 1;
Location
Object code (Hints)
--------------------------------------------------------FINAL START 0
(La)0000
+LDS
LENGTH
(a)
(format 4)
(Lb)0004
STL
RETADR
(b)
(PC-relative)
(Lc)0007
CLEAR S
(c)
(Ld)0009
CLEAR X
(d)
(Le)000B
LDB
#LENGTH
(e)
(imme. addr)
(Lf)000E
CMPR
X, T
(f)
(format 2)
(Lg)0010
JGT
EXIT
(g)
(PC-relative)
(Lh)0013
COMP
#0
(h)
(imme. addr)
(Li)0016
JEQ
EXIT
(i)
(PC-relative)
(Lj)0019
ADDR
S, X
(j)
(Lk)001B
LDL
RETADR
(k)
(PC-relative)
(Ll)001E
EXIT
RSUB
(l)
(format 3)
(Lm)0021
LENGTH WORD
41
(Ln)0024
RETADR RESW
2
13. [6%] A SIC/XE program is given as follows:
Location
Object code
--------------------------------------------------------OBJP
START 1012
1012
CLEAR B
B430
1014
CLEAR T
B450
1016
CLEAR F
B460
1018
RETADR
RESW
2
no object code
101E
DATA
RESW
2
no object code
1024
LENGTH
WORD
32
20 00 00
Translate the SIC/XE program to an object program. You can use at most TWO text records.
Hint: make sure the number of columns is correct. A column represents a half-byte (4-bit).