1. Assembly Language Lecture 9
Multiplication and Division Instructions

2. Signed vs. Unsigned Multiplication
Signed multiplication and unsigned multiplication lead to different results. Consider two bytes:
and
Signed multiplication of these two numbers gives:
(-1) * (1) = (-1)
Unsigned multiplication of these two numbers gives:
(255) * (1) = (255)
REMEMBER:
If you multiply an 8-bit number by an 8-bit number, you will get a 16-bit number!
If you multiply a 16-bit number by a 16-bit number, you will get a 32-bit number!

3. Signed vs. Unsigned Multiplication
Thus there is one instruction to multiply signed (IMUL)and another instruction to multiply unsigned (MUL).
These instructions execute one way if you're working with bytes and they execute in a different way if you're working with words.

4. MUL and IMUL Instructions
Unsigned Multiplication
Syntax: MUL source
Signed Multiplication
IMUL source
Byte Form
The multiplier is contained in the source field and the multiplicand is assumed to be in AL. The 16-bit product will be in AX. The source may be a byte register or a memory byte. It cannot be a constant.
Word Form
The multiplier is contained in the source field and the multiplicand is assumed to be in AX. The 32 bit product will be split over two registers. The most significant 16 bits will be in DX and the least significant bits will be in AX (This is referred to as DX:AX). The source may be a word register or a memory word. It cannot be a constant.

5. MUL and IMUL Instructions
Effect of MUL/IMUL on the status flags:
SF,ZF,AF,PF : undefined
CF / OF :
If the product is too big to fit in the lower half of the destination (AL for byte multiplication and AX for word multiplication), the CF/OF will be set to 1.
If the upper half of the result on a MUL is a zero or the upper half of the result on an IMUL is the sign extension of the lower half, the CF/OF will be set to 0.

6. Examples Example 1 Example 2 Before the instruction executes:
DX:???? AX: BX: 0010 (16) CF/OF:?
IMUL BX multiplies the contents of AX by BX
placing the 32 bit answer in DX:AX
After the instruction executes:
DX:0000 AX: 0020 (32) BX: CF/OF:0 (Result fits in AX)
Example 2
DX:???? AX: 0003 BX: 7000 (28672) CF/OF:?
IMUL BX multiplies the contents of AX by BX
DX:0001 AX: BX: CF/OF:1
(Result [86016] does not fit in AX)

7. Examples Example 3 Example 4 Before: AL: 30 (48) BL: 02 CF/OF:?
IMUL BL multiplies the contents of AL
by BL placing the 16 bit answer in AX
After:
AX: 0060 (96) BL: 02 CF/OF:0 (Result fits in AL)
Example 4
AL: 80 (-128) BL: CF/OF:?
IMUL BL multiplies the contents of AL by BL
placing the 16 bit answer in AX
AX: FF00(-256) BL: 02 CF/OF:1
(High 8 bits are not the signed extension of the lower half)

8. After: AX: 00E0 (224) BL: 02 CF/OF:0 (High 8 bits are zero)
Examples (cont’d)
Example 5
Before: AL: 80 (-128) BL: FF (-1) CF/OF:?
IMUL BL multiplies the contents of AL by BL placing
the 16 bit answer in AX
After: AX: 0080 (128) BL: FF CF/OF:1
(High 8 bits are not the signed sense signed extension of the lower half)
Example 6
Before: AL: 70 (112) BL: CF/OF:?
MUL BL multiplies the contents of AL by BL placing
After: AX: 00E0 (224) BL: CF/OF:0 (High 8 bits are zero)

9. Examples (cont’d) Example 7 Before: AL: A0 (160) BL: 02 CF/OF:?
MUL BL multiplies the contents of AL by BL
placing the 16 bit answer in AX
After:
AX: 0140 (320) BX: 02 CF/OF:1
(High 8 bits are not zero)

10. Problem 1
Translate into ASSMBLER the following high-level assignment statement: ANSWER = 13 * A + B * C - D
where ANSWER, A, and D are words in memory and B and C are bytes in memory. Assume no overflow and use signed multiplication.
MOV AX,13 ;move constant to reg
IMUL A ;assume A*13 in AX
MOV ANSWER,AX ;ANSWER=13*A
MOV AL,B
IMUL C ;assume B*C in AX
ADD ANSWER,AX ;ANSWER=13*A+B*C
MOV BX,D ;Put D in register so you can ;subtract it from ANSWER
SUB ANSWER,BX ;ANSWER=13*A+B*C-D

11. Problem 2
Write a procedure to calculate N! Remember N! = 1*2*3*...* N-1 * N For example, 4! = 1*2*3*4 = 24 0! and 1! are defined to be 1 and the factorial for a negative number is undefined and should be returned as 0.

12. Solution
; compute N! ; input : CX = N ;output: AX = N! MOV AX,1 ; AX holds product TOP: MUL CX ; product = product X term LOOP TOP

13. Signed and Unsigned Division
When any kind of division is performed, two results are obtained - the quotient and the remainder.
The effect of the division instructions on the flags register is that all status flags are undefined after a divide has taken place !!!

14. Signed and Unsigned Division
Unsigned Division Instruction
Syntax: DIV divisor
Signed Division Instruction
Syntax: IDIV divisor
These instructions divide 8 (or 16) bits into 16 (or 32) bits. The quotient and remainder will be the same size as the divisor.
Byte Form
The divisor is an 8-bit register or memory byte. The 16-bit dividend is assumed to be in AX. After the division the 8 bit quotient is in AL and the 8-bit remainder is in AH. The divisor cannot be a constant!

15. Signed and Unsigned Division
Word Form
The divisor is 16-bit register or memory word. The 32 bit dividend is assumed to be in DX:AX. After the division, the 16 bit quotient is in AX and the 16-bit remainder is in DX. The divisor cannot be a constant!
The divide instruction does not set the status flags to any particular settings.
If the quotient will not fit into AL or AX, the system will display "DIVIDE OVERFLOW".

16. Signed and Unsigned Division
The sign of the quotient will be determined by the laws of algebra. The sign of the remainder will be the sign of the original dividend.
That is:
QUOTIENT REMAINDER
+/+ = /+ = +
+/- = /- = +
-/+ = /+ = -
-/- = /- = -

17. Signed and Unsigned Division
To set to divide one word by another word
1. Place the dividend in AX
2. For DIV, clear DX (XOR DX,DX will clear DX)
3. For IDIV, extend the sign of the value in AX to DX by issuing the CWD instruction (Convert a Word to a Doubleword).
To set up to divide one byte by another byte
1. Place the dividend in AL
2. For DIV, clear AH (XOR AH,AH will clear AH)
3. For IDIV, extend the sign of the value in AL to AH by issuing
the CBW instructions (Convert a Byte to a Word).

18. CWD Instruction
Converts the signed 16-bit number in AX into a signed 32-bit number in DX:AX
NOTE: Use this instruction only with signed arithmetic!!
Syntax: CWD
If bit 15 of AX is a 1, the instruction will place FFFF in DX.
If bit 15 of AX is a 0, the instruction will place 0000 in DX.
No flags are affected by this instruction.
EXAMPLES:
Before:
DX:???? AX:7000 (28672)
CWD converts the signed 16-bit number in AX to a
32-bit signed number in DX:AX
After:
DX:0000 AX:7000 (28672)

19. CWD Instruction Example 2: Before the instruction executes:
DX:???? AX:8000 (-32768)
CWD converts the signed 16-bit number in AX to a
32-bit signed number in DX:AX
After the instruction executes:
DX:FFFF AX:8000 (-32768)

20. CBW Instruction
This instruction converts the signed 8-bit number in AL into a signed 16-bit number in AX
NOTE: Use this instruction only with signed arithmetic!!
Syntax: CBW
If bit 7 of AL is a 1, the instruction will place FF in AH.
If bit 7 of AL is a 0, the instruction will place 00 in AH.
No flags are affected by this instruction.
Examples:
Before: AX:??40 (64)
CBW converts the signed 8-bit number in AL to a
16-bit signed number in AX
After: AX:0040 (64)

21. CBW Instruction Example 2 Before: AX:??FC (-4)
CBW converts the signed 8-bit number in AL
to a 16-bit signed number in AX
After: AX:FFFC (-4)

22. IDIV BX divides the contents of DX:AX by BX placing
DIV and IDIV Examples
Example 1
MOV AX,9
CWD
Thus, before the instruction executes:
DX:0000 AX: BX: 0002
IDIV BX divides the contents of DX:AX by BX placing
the quotient in AX and the remainder in DX
After the instruction executes:
DX:0001 AX: BX: 0002

23. IDIV BX divides the contents of DX:AX by BX placing
DIV and IDIV Examples
Example 2
MOV AX,9
CWD
Thus, before the instruction executes:
DX:0000 AX: BX: FFFE (-2)
IDIV BX divides the contents of DX:AX by BX placing
the quotient in AX and the remainder in DX
After the instruction executes:
DX:0001 AX: FFFC (-4) BX: FFFE

24. IDIV BL divides the contents of AX by BL placing the
DIV and IDIV Examples
Example 3
Before the instruction executes:
AX: 0A00 (2560) BL: 02
IDIV BL divides the contents of AX by BL placing the
quotient in AL and the remainder in AH
As the instruction executes, the system will display on the screen:
DIVIDE OVERFLOW
because the quotient (1280 or 500h) will not fit into AL

25. DIV BX divides the contents of DX:AX by BX placing
DIV and IDIV Examples
Example 4
MOV AX,9
XOR DX,DX
Thus, before the instruction executes:
DX:0000 AX: BX: FFFE (65534)
DIV BX divides the contents of DX:AX by BX placing
the quotient in AX and the remainder in DX
After the instruction executes:
DX:0009 AX: BX: FFFE

26. DIV BL divides the contents of AX by BL placing the
DIV and IDIV Examples
Example 5
Before the instruction executes:
AX: FFFF (65535) BL: 02
DIV BL divides the contents of AX by BL placing the
quotient in AL and the remainder in AH
As the instruction executes, the system will display on the screen:
DIVIDE OVERFLOW
because the quotient (32767 or 7FFFh) will not fit into AL

27. CODE : DIV and IDIV Examples Example 6 Divide -1250 by 7
AX: BX: 7
CODE :
MOV AX, ; AX gets dividend
CWD ;Extend sign to DX
MOV BX, ;BX has divisor
IDIV BX ;AX gets quotient, DX has remainder

28. CODE : DIV and IDIV Examples Example 7
Divide the signed value of the byte variable XBYTE by -7
AL: XBYTE BL: -7
CODE :
MOV AL, XBYTE ;AL has dividend
CBW ;Extend sign to AH
MOV BL, ;BL has divisor
IDIV BL ;AL has quotient, AH has remainder

29. Problem
Translate into ASSMBLER the following high-level assignment statement:
ANSWER = 130 / A + B / C - D / 7
where ANSWER, A, and D are words in memory and B and C are bytes in memory. Assume no overflow and use signed multiplication.
MOV AX,130 ;move constant to reg
CWD ;expand to DX:AX
IDIV A ;assume QUOTIENT 130/A in AX
MOV ANSWER,AX ;ANSWER=130/A
MOV AL,B
CBW ;expand to AX
IDIV C ;assume quotient B/C in AX

30. CBW ;convert byte quotient in AL to a word
Problem (cont’d)
CBW ;convert byte quotient in AL to a word
ADD ANSWER,AX ;ANSWER=130/A + B/C
MOV AX,D
MOV BL, ;put constant in reg
IDIV BL ;assume quotient D/7 in AL
CBW ;expand to AX so you can SUB from ANSWER
SUB ANSWER,AX ;ANSWER=130/A + B/C - D/7