Assembly Language Programming By Ytha Yu, Charles Marut Chap 10 ( Arrays and Addressing Modes)

Outline
• One‐Dimensional Arrays
• Addressing Modes
• Two‐Dimensional Arrays
• Based Indexed Addressing Mode
2

One‐Dimensional Array A
3
• One‐Dimensional Array is an
ordered list of elements, all of
same type
• DB & DW pseudo‐ops to
declare byte and word sized
arrays

Arrays
W DW 1000,40,29887,329
Address of the array variable is called base address of array
Offset Address Symbolic Address Contents
0300h W 1000d
0302h W +2 40d
0304h W +4 29887d
0306h W +6 329d
4

The DUP Operator
• The DUP (duplicate) is used to define arrays
whose elements share a common initial value.
• repeat_count DUP (value)
• GAMMA DW 100 DUP (0)
• DELTA DB 212 DUP (?)
• LINE DB 5, 4, 3 DUP (2, 3 DUP (0), 1)
• LINE DB 5,4,2,0,0,0,1,2,0,0,0,1,2,0,0,0,1
5
DUP may be
nested

One‐Dimensional Array A
W DW 10, 20, 30, 40, 50, 60
Offset address Symbolic address Decimal address
0200h W 10
0202h W + 2h 20
0204h W + 4h 30
0206h W + 6h 40
0208h W + 8h 50
020Ah W + Ah 60
6

Location of Array Elements
• The address of an array element can be
specified by adding a constant to the base
address
– If A is an array
– S is the size of every element in bytes
Position Location
1 A
2 A + 1 x S
3 A + 2 x S
.
.
N A + (N‐1) x S

Example: Exchange 10th and 25th element of an array
• W[10] is located at W + 9x2 = W + 18
• W[25] is located at W + 24x2 = W + 48
MOV AX, W + 18
XCHG W+48, AX
MOV W+18, AX

Addressing Modes
• The way an operand is specified
• register mode: an operand is a register.
• immediate mode: an operand is a constant.
• direct mode: an operand is a variable.
– MOV AX, 0
– ADD ALPHA, AX
9

ADDITIONAL ADDRESSING MODES
Four additional addressing modes to address
memory operands indirectly
1. Register Indirect Mode
2. Based
3. Indexed
4. Based Indexed

Register Indirect Mode
Offset address of the operand is contained in a
register. Register acts like a pointer to the
memory location
• [register]
• The register is BX, SI, DI, or BP.
11
the operand’s
segment number
is contained in DS
the operand’s
segment number
is contained in SS

Suppose that SI contains 0100h, and
the word at 0100h contains 1234h.
• MOV AX, [SI] ; AX = 1234h
The CPU
1. examines SI and obtains the offset address 100h,
2. uses the address DS:0100h to obtain the value
1234h, and
3. moves 1234h to AX.
• MOV AX, SI ; AX = 0100h
12

Suppose that
BX contains 1000h Offset 1000h contains 1BACh
SI contains 2000h Offset 2000h contains 20FEh
DI contains 3000h Offset 3000h contains 031Dh
where the above  offsets are in the data segment addressed by DS.
13
Tell which of the following instructions are legal.
If legal, give the source offset address and the
result or number moved.
a. MOV  BX, [BX]
b. MOV  CX, [SI]
c. MOV  BX, [AX]
d. ADD  [SI], [DI]
e. INC  [DI]
Source offset Result
1000h 1BACh
2000h 20FEh
illegal source register
illegal memory‐memory addition
3000h 031Eh

Write some code to sum in AX the elements of the 10‐element
array W defined by
W DW 10,20,30,40,50,60,70,80,90,100
14
The idea is to set a pointer to the base of the array, and let it
move up the array, summing elements as it goes.
XOR AX, AX; AX holds sum
LEA SI, W ; SI points to array W
MOV CX, 10 ; CX has number of elements
ADDNOS:
ADD AX, [SI] ; sum = sum + element
ADD SI, 2 ; move pointer to
; the next element
LOOP ADDNOS ; loop until done

Home Assignment
• Write a procedure to reverse an array of N
words

Based and Indexed Addressing Mode
• Operand’s offset address is obtained by adding a number
called displacement to the contents of a register
• Displacement can be of following forms;
– Offset address of a variable
– A constant (+ve or ‐ve)
– Offset address of a variable plus or minus a constant
• If A is a variable;
16
A
‐2
A + 4

• [register + displacement]
• [displacement + register]
• [register] + displacement
• displacement + [register]
• displacement[register]
• The register is SI, DI, BX, or BP.
17

Suppose W is a word array; BX contains 4
MOV AX, W[BX]
Will move the element at address W + 4 to AX (the third element
of array)
MOV AX, [W + BX]
MOV AX, [BX + W]
MOV AX, W + [BX]
MOV AX, [BX] + W
18

Rework the last example by using
based mode.
XOR AX, AX ; AX holds sum
XOR BX, BX ; clear base register
MOV CX, 10 ; CX has number of elements
ADDNOS:
ADD AX, W[BX] ; sum = sum + element
ADD BX, 2 ; index next element
LOOP ADDNOS ; loop until done
19

Suppose that ALPHA is declared as
ALPHA DW 0123H, 0456h, 0789h, 0ABCDh
in the segment addressed by DS.
Suppose also that
BX contains 2 Offset 0002 contains 1084h
SI contains 4 Offset 0004 contains 2BACh
DI contains 1
20
Tell which of the following instructions are legal. If legal, give the
source offset address and the result or number moved.
Source offset Number moved
a. MOV  AX, [ALPHA+BX] ALPHA+2 0456h
b. MOV  BX, [BX+2] 2+2 = 4 2BACh
c. MOV  CX, ALPHA[SI] ALPHA+4 0789h
d. MOV  AX, –2[SI] –2+4 = 2 1084h
e. MOV  BX, [ALPHA+3+DI] ALPHA+4 0789h
f. MOV  AX, [BX] 2 Illegal form of source operand
g. ADD  BX, [ALPHA+AX] Illegal source register

PTR OPERATOR
• Operands of instruction must be of the same type
MOV AX, 1 ; legal word instruction
MOV BH, 5; legal byte instruction
MOV [BX], 1 ; illegal cant interpret whether destination is a
;byte operand pointed by bx or a word
• For destination to be byte
MOV BYTE PTR [BX], 1
• For destination to be word
MOV WORD PTR [BX], 1

Using PTR to Override a type
type PTR address_expression
If we delare
DOLLARS DB 1AH
CENTS DB 52H
MOV AX, DOLLARS ; ILLEGAL
MOV AX, WORD PTR DOLLARS; AL= DOLLARS, AH = CENTS
;will move 521AH to AX

Label pseudo op
• To get around problem of type conflict;
MONEY LABEL WORD
DOLLARS DB 1AH
CENTS DB 52H
• Declares MONEY as a word variable and
components DOLLARS and CENTS as byte
• MOV AX, MONEY ; LEGAL
• MOV AL, DOLLARS ; LEGAL
• MOV AH, CENTS ; LEGAL

SEGMENT OVERRIDE
• BX, SI, DI specify offset relative to DS
• Possible to specify offset relative to other
segments
Segment_register : [pointer_register]
MOV AX, ES:[SI]
CAN also be used with based and indexed
addressing modes

Accessing the STACK
• When BP specifies the offset address, SS supplies
the segment number
• BP may be used to access elements of the stack
• Move top three elements of stack to AX, BX, CX
without changing the stack
MOV BP, SP
MOV AX, [BP]
MOV BX, [BP + 2]
MOV CX, [BP + 4]

Two‐Dimensional Array B
26
• An array of arrays
• A One Dimensional Array whose elements are
also One Dimensional Array
• Arranged as rows and cloumns

How Two‐Dimensional Array are stored
Suppose array B has
10, 20, 30, 40 in the first row,
50, 60, 70, 80 in the 2nd row,
&
90, 100, 110, 120 in the 3rd row

Row‐Major Order
B DW 10, 20, 30, 40
DW 50, 60, 70, 80
DW 90, 100, 110, 120
Used when elements in a row are to be
processed together sequentially
28

Column‐Major Order
B DW 10, 50, 90
DW 20, 60, 100
DW 30, 70, 110
DW 40, 80, 120
Used when elements in a column are to be
processed together sequentially
29

Locating an element in an array
• Consider M xN array stored in row major order
• Size of element is S
• To find location of A[I,j]
• Find where the row i begins
• Location of jth element in that row
– Row 1 begins at A
– Row 2 begins at A + N x S
– Row 3 begins at A + 2 x N x S
– Row i begins at A + (i‐1) x N x S
jth element in a row is located at (j‐1) x S

Final Result
Location of A[i,j]
A + { (i‐1) x N + (j ‐ 1)} x S
• For column major ordered array;
Location of A[i,j]
A + { (i‐1) + (j ‐ 1) x M} x S

Based Indexed Addressing Mode
• Offset address of the operand is the sum of;
– Contents of a base register (BX or BP)
– Contents of an index register (DI or SI)
– Optionally a variable's offset address
– Optionally a constant (+ve or ‐ve)

• variable [base_register][index_register]
• [base_register + index_register + variable + constant]
• variable [base_register + index_register + constant]
• constant [base_register + index_register + variable]
33

• W is a word variable
• BX contains 2
• Si contains 4
MOV AX, W[BX][SI]
Will move contents of W + 6 to AX
MOV AX, [W + BX + SI]
MOV AX, W[BX + SI]
34

• M is a 5x7 word array stored in row major
order, write some code to;
– Clear row 3
– Clear column 4
1. Clear row 3
For an MxN array; Row i begins at A + (I‐1) x N x S
Thus in a 5x7 array, row 3 begins at; A + (3‐1) x7x2 = A + 28;
MOV BX, 28
XOR SI, SI
MOV CX, 7
CLEAR:
MOV A[BX][SI], 0
ADD SI,2
LOOP CLEAR

• M is a 5x7 word array stored in row major
order, write some code to;
– Clear row 3
– Clear column 4
2. Clear column 4
For an MxN array; column j begins at A + (j‐1) S
Column 4 begins at A + (4‐1) x 2 = A + 6;
Since A is a 7 column array stored in row major order, to get to the next
element in column 4 we need to add 7x 2 = 14 ;
MOV SI, 6
XOR BX, BX
MOV CX, 5
CLEAR:
MOV A[BX][SI], 0
ADD BX,14
LOOP CLEAR

An application: Average Test Scores
sum[j] = 0
i = 1
FOR 5 times DO
sum[j] = sum[j] + score[i, j]
i = i + 1
END_FOR
37

Assembly Language Programming By Ytha Yu, Charles Marut Chap 10 ( Arrays and Addressing Modes)

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Assembly Language Programming By Ytha Yu, Charles Marut Chap 10 ( Arrays and Addressing Modes)