Chapter 9: Multiplication….. Saj Alam and Samya Zain

1

Chapter 9: Multiplication…..Saj Alam and Samya Zain

Monday

November 6, 2006

2

Simple Multiplication Examples

53

10

530

53

55

265

+265

3915

53 22

106

+106 1166

3

Multiplication

A B = CA= Multiplicand nA= Number of digits of Multiplicand (A)

B = Multiplier nB= Number of digits of Multiplier (B)

C = product nC nA + nB

4

nA= 2

nB= 2

nC= 4

53

55

265

+265

3915

nC = nA + nB

Example 2:

53 nA= 2

10 nB= 2

530 nC= 3

nC < nA + nB

Example 1:

5

Binary Number System

System Digits: 0 and 1

Bit (short for binary digit): A single binary digit

LSB (least significant bit): The rightmost bit MSB (most significant bit): The leftmost bit

53 = 110101

MSB LSB

6

Binary To Decimal Conversion

=

=

=

=

=

=

=

7

Binary Multiplication

Example 1:

Example 2:

Example 3:

LSBMSB

addition

multiplication

8

110101 10110

1101010

100111110

100111110

= Multiplicand(MC) ; nA= 2

= Multiplier(MP) ; nB= 2

= product ; nC nA + nB

53

22

106

+106

1166

53 = 11010122 = 10110

Decimal Binary

21029282726 25 24 23 22 21 20

1024 +128+8+4+ 2 =1166Binary weight

10010001110product

000000Bit0 : start product

110101 Bit1 : MC shifted left

110101Bit2 : MC shifted left



9

Left Shift Algorithm (LSA)

Step1:Write the Multiplicand(MC) and the Multiplier(MP) with LSB lined up.

Step2: Look at LSB of Multiplier. IF LSBMP = 0 put {00…nA} as initial product

IF LSBMP = 1 copy Bit0 of MC as initial product

Step3: Look at the next LSB bit (Bit1MP) of Multiplier. IF Bit1MP = 0 put {00…nA} as product 2 IF Bit1MP = 1 Shift left and copy Bit1 of MC as product 2

Step4:Repeat STEP 3 until all bits in the MP are used up.

Step5:Check the number of digits or bits in the MC and MP. nC nA + nB

10

Right Shift Algorithm (RSA)Decimal Multiplication

11

111111 1234

3777774

444444 shift right one digit

000000initial product

444444+

3333330+

3777774shift right one digit+ 22222200

2599997425999974shift right one digit

111111000137110974 final product

12

11000001111 final product

101111 100001

= Multiplicand(MC)

= Multiplier(MP)

47 =33 =

Decimal Binary 25 24 23 22 2120 Binary weight

000000 initial product

1011110101111shift right one digit

000000

000000 00101111Add & shift right one digit

Add & shift right one digit 000101111000000

Add & shift right one digit 0000101111

101111 211210 29 28 27 26 252423222120

47 33 = 1551

1024 + 512 ++8+4+ 2+1= 1551

00000000000101111Add & shift right one digit

13

Right Shift Algorithm (RSA)Step1:Write the Multiplicand(MC) and the Multiplier(MP) with LSB lined up.

Step2: Put initial product {000…nA}

Step3a: Look at LSB of Multiplier. IF LSBMP = 0 put {00…nA} as product 2

IF LSBMP = 1 copy Bit0 of MC as product 2

Step3b: add initial product and product 2

Step3c: Shift the resulting sum one digit to right.

Step4a: Look at the next LSB bit (Bit1MP) of Multiplier. IF Bit1MP = 0 put {00…nA} as product 3 IF Bit1MP = 1 copy Bit1 of MC as product 3

Step4b: add product2 and product 3

Step4c: Shift the resulting sum one digit to right.

Step5: Repeat Steps 4a, 4b and 4c until all bits in the MP are used up.

Step6:Check the number of digits or bits in the MC and MP. nC nA + nB

14

111 33

Try yourself:

15

Right Shift Algorithm (RSA)for 16- Bit Multiplication

Wednesday

November 8, 2006

16

16- Bit Multiplication

AX

DXBXCX

CF

Multiplicand = FFFFMultiplier = FFFF

AX Multiplier BX Multiplicant DX Bit count CX CF

00 FF10H

CFFF00

FF FF 00

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AX SUB AX, AX ; AX ; CF Multiplier BX MOV BX , MPLYMultiplicant DX MOV DX, MCANDBit count CX MOV CX, 10H ; number of bits CF

Is CX = 0 ?

Examine LSB of MPLIER RCR BX, 1 ; Examine LSB of MPLIER

Add MCND to AXNO

YES ADD AX, DX ; add MCAND to product

Bit = 0 JNC SHIFT ; skip addition step if bit = 0 NEXT:

Shift the product right by 1 bit.

Re-examine the next bit.Decrement counter

SHIFT:RCR AX, 1 ; shift the product 1 bit rightRCR BX, 1 ; & examine next bit of MPLYDEC CX ; change counter down by 1

YESNO

JCXZ DONE ; if CX = 0, all bits are doneJMP NEXT ; otherwise go on to next bit

Store final product in memory

DONE:MOV PRODL, BX ; store low wordMOV PRODH, AX ; store high wordRET

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AX SUB AX, AX ; AX ; CF

Multiplicand = FF DX MOV DX, MCAND

Multiplier = 3 BX MOV BX , MPLY

Bit count CX MOV CX, 10H ; number of bits

03

16HF0 0F

000000 0000

CF 0000 0000

1111 0000 0000 11110000 0000 0000 10110000 0000 0001 0000

AXDX

BXCX

00

RCR BX, 1 ; Examine LSB of MPLIER

0000 0000CF

0000 00001111 0000 0000 1111

0000 0000 0000 01010000 0000 0001 0000

AXDX

BXCX

1 No change

No changeNo change

03

F0 0F

1111 0000CF

0000 11111111 0000 0000 1111

0000 0000 0000 01010000 0000 0001 0000

AXDX

BXCX

0

NEXT:JNC SHIFT ; skip addition step if bit = 0 ADD AX, DX ; Since CF = 1 & put in AX

No change

No change

Note: CF is reset

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0111 1000CF

0000 01111111 0000 0000 1111

0000 0000 0000 01010000 0000 0001 0000

AXDX

BXCX

1

SHIFT:RCR AX, 1 ; shift the product 1 bit right

0111 1000CF

0000 01111111 0000 0000 1111

1000 0000 0000 00100000 0000 0000 1111

AXDX

BXCX

1

15H

DEC CX ; change counter down by 1

No change

No changeNo change

Put CF = 0 and Right shift all values

No change

No change

No change

No change

No change No change

Note: CF did not change

RCR BX, 1 ; & examine next bit of MPLY (BX)

0111 1000CF

0000 01111111 0000 0000 1111

1000 0000 0000 00100000 0000 0001 0000

AXDX

BXCX

1

16H

As CX 0 :JCXZ DONE ; if CX = 0, all bits are doneJMP NEXT ; otherwise go on to next bitJNC SHIFT ; skip addition step if bit = 0& CF = 1: ADD AX, DX

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0110 1000CF

0001 01101111 0000 0000 1111

1000 0000 0000 00100000 0000 0000 1111

AXDX

BXCX

1

15H

ADD result in AX:

No change

No changeNo change

0110 1000 0001 0110 1CarryOverFlow

0111 1000CF

0000 01111111 0000 0000 1111

1000 0000 0000 00100000 0000 0000 1111

AXDX

BXCX

1

15H

Adding:

No change

1011 0100CF

0000 10111111 0000 0000 1111

0100 0000 0000 00010000 0000 0000 1111

AXDX

BXCX

0

15H

SHIFT:RCR AX, 1 ; shift the product 1 bit rightRCR BX, 1 ; & examine next bit of MPLY

DEC CX ; change counter down by 1JCXZ DONE ; if CX = 0, all bits are doneJMP NEXT ; otherwise go on to next bit

21

1011 0100CF

0000 10111111 0000 0000 1111

1010 0000 0000 00000000 0000 0000 1110

AXDX

BXCX

1

14H

No change No change

As CX 0 & CF = 0:SHIFT:RCR AX, 1 ; shift the product 1 bit rightRCR BX, 1 ; & examine next bit of MPLY

1011 0100CF

0000 10111111 0000 0000 1111

0100 0000 0000 00010000 0000 0000 1110

AXDX

BXCX

0

14H

DEC CX ; change counter down by 1

No change

No change No change

0101 1010 0000 0101

AND SO ON untill the counter CX comes to 0H……..

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;*****************************************************************************

TITLE MAIN PROGRAM MUL16(UNSIGNED)

;-----------------------------------------------------------------------------

; Author: M. Sajjad Alam (University at Albany)

;-----------------------------------------------------------------------------

; This program multiplies two 16-bit numbers. The numbers to be multiplied are stored in the data segment at MPLY and MCND. Space is reserved at

; PRODL and PRODH for the product. The program multiplies unsigned numbers using the right shift algorithm.

; The largest two numbers that can be multiplied are: FFFFh and FFFFh or in decimal 65535 and 65535. For the purpose of testing the program, we suggest

; you test the program for the following examples:

;FFFFh x 10h = FFFF0h, FFFFh x 100h = FFFF00h, FFFFh x 1000h = FFFF000h

; FFFFh x Fh = EFFF1h, FFFFh x FFh = FEFF01h, FFFFh x FFFh = FFEF001h

; FFFFh x FFFFh = FFFE0001

; INPUT: DX <- Multiplicand

; BX <- Multiplier

; CX <- Bit Counter = 10h = 16

; OUTPUT: AX <- High word of product

; BX <- Low Word of product

;*****************************************************************************

SSEG SEGMENT PARA STACK 'STACK' ; Start of Stack Segment

DB 64 DUP('STACK') ; Reserve 64 bytes

SSEG ENDS ; End of Stack Segment

;*****************************************************************************

DSEG SEGMENT PARA PUBLIC 'DATA' ; Start of data segment

; Enter here the data words

MCND DW 0FFFFH ; Store multiplicand

MPLR DW 0FFFH ; Store multiplier

; Reserve two words for the product

PRODL DW ? ; Low byte of Product

PRODH DW ? ; High byte of Product

DSEG ENDS ; End of data segment

;*****************************************************************************

CSEG SEGMENT PARA PUBLIC 'CODE' ; Start of Code Segment

ASSUME CS:CSEG, DS:DSEG, SS:SSEG

MUL16 PROC FAR

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; Set up the Stack with proper values to return to DOS or DEBUG

PUSH DS ; Save DS on the stack

SUB AX,AX ; Load 0 into AX

PUSH AX ; Save 0 on the stack

; Initialize the Data Segment registers

MOV AX,DSEG ; Store Address DSEG in AX

MOV DS,AX ; Transfer AX to Data Seg Reg

;----------------------------------------------------------------------------

; Programming task starts here

START:

SUB AX,AX ; Store 0 in AX and clear CF

MOV BX,MPLR ; Store multiplier in BX

MOV DX,MCND ; Store multiplicand in DX

MOV CX,16 ; Store # of bits in CX

RCR BX,1 ; Rotate BX one bit right

NEXT:

JNC SHIFT ; If CF=1, skip the adding

ADD AX,DX ; Add multipicand to AX

SHIFT: ; Perform right shift of prod

RCR AX,1 ; Right shift of AX thru CF

RCR BX,1 ; Right shift CF into BX

DEC CX ; Decrease bit count by 1

JCXZ DONE ; Done if bit count is 0

JMP NEXT ; Otherwise loop for next bit

DONE: ; Multiplication is finished

MOV PRODL,BX ; Store low word at PRODL

MOV PRODH,AX ; Store high word at PRODH

RET ; return to debug

MUL16 ENDP

CSEG ENDS

END MUL16

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****************************************************************MUL16 MULTIPLICATION OF TWO 16-BIT NUMBERSAuthor: M. Sajjad AlamDate: November 4, 1996Debug session showing the multiplication of 0FFh by 0FFFFh----------------------------------------------------------------C:\PHY353\MUL16>debug mul16_m.exe -u 1B5E:0000 1E PUSH DS 1B5E:0001 2BC0 SUB AX,AX 1B5E:0003 50 PUSH AX 1B5E:0004 B85D1B MOV AX,1B5D 1B5E:0007 8ED8 MOV DS,AX 1B5E:0009 2BC0 SUB AX,AX 1B5E:000B 8B1E0000 MOV BX,[0000] 1B5E:000F 8B160200 MOV DX,[0002] 1B5E:0013 B91000 MOV CX,0010 1B5E:0016 D1DB RCR BX,1 1B5E:0018 7302 JNB 001C 1B5E:001A 03C2 ADD AX,DX 1B5E:001C D1D8 RCR AX,1 1B5E:001E D1DB RCR BX,1 1B5E:0020 49 DEC CX 1B5E:0021 E302 JCXZ 0025 1B5E:0023 EBF3 JMP 0018 1B5E:0025 891E0400 MOV [0004],BX 1B5E:0029 A30600 MOV [0006],AX 1B5E:002C CB RETF ------------------------------------------------------------

-g9AX=1B5D BX=0000 CX=017D DX=0000 SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0009 NV UP EI PL ZR NA PE NC 1B5E:0009 2BC0 SUB AX,AX

We want to get past the point where the data segment register is loadedand the data segment is defined. At offset 0009, we have already movedthe data segment offset value 1B5D into the data segment register DS.-----------------------------------------------------------------------------------------

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----------------------------------------------------------------------We can now do a dump of the data segement and identify the location of the data words corresponding to multiplicand and multiplier.--------------------------------------------------------------------------------------------------------------------------------------------Note the mulplicand 0FFFH was the first data word in the data segmentand now present at location 1B5D:0000 . The multiplier 0FFH which is the second data word in the data segment is 0FFH which is equivalent to00 FF. In the memory the lower byte FFH is loaded first, followed by the high byte 00H as can be seen at location 1B5D:0003 and 1B5D:0004,respectively.-----------------------------------------------------------------------dds:00 1B5D:0000 FF FF FF 00 00 00 00 00-00 00 00 00 00 00 00 00 ................ 1B5D:0010 1E 2B C0 50 B8 5D 1B 8E-D8 2B C0 8B 1E 02 00 8B .+.P.]...+...... 1B5D:0020 16 00 00 B9 10 00 D1 DB-73 02 03 C2 D1 D8 D1 DB ........s....... 1B5D:0030 49 E3 02 EB F3 89 1E 04-00 A3 06 00 CB 13 A1 06 I............... 1B5D:0040 3C 8B 16 08 3C A3 1A 41-89 16 1C 41 C6 06 B2 10 <...<..A...A.... 1B5D:0050 FF 5E 8B E5 5D C3 55 8B-EC 56 8B 5E 04 8B 76 06 .^..].U..V.^..v. 1B5D:0060 8B 04 39 07 73 06 B8 FF-FF 5E 5D C3 8B 76 06 8B ..9.s....^]..v.. 1B5D:0070 04 39 07 76 07 B8 01 00-5E 5D C3 90 2B C0 5E 5D .9.v....^]..+.^]

1B5D:00031B5D:0000

----------------------------------------------------------------------A look at the general purpose registers shows that they have data fromprevious operations sitting in them at this stage.-----------------------------------------------------------------------r AX=1B5D BX=0000 CX=017D DX=0000 SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0009 NV UP EI PL ZR NA PE NC ----------------------------------------------------------------------

We will now carry on the necessary steps so that we have the registers AX, BX, CX, and DX loaded with the multiplier and multiplicand.---------------------------------------------------------------------- 1B5E:000B 8B1E0200 MOV BX,[0002] DS:0002=00FF -t AX=0000 BX=00FF CX=017D DX=0000 SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=000F NV UP EI PL ZR NA PE NC 1B5E:000F 8B160000 MOV DX,[0000] DS:0000=FFFF -t AX=0000 BX=00FF CX=017D DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0013 NV UP EI PL ZR NA PE NC 1B5E:0013 B91000 MOV CX,0010 -t AX=0000 BX=00FF CX=0010 DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0016 NV UP EI PL ZR NA PE NC

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------------------------------------------------------------------------------NoteNote: now the AX=0000, the initial product. BX contains the multiplier and DX contains the multiplicand. The register CX is loaded with the counter 10Hor 16, indicating the number of shifts needed to complete the multiplication.We are now ready to shift the multiplier right into the carry bit to examineif the bit is a '1' or '0'.------------------------------------------------------------------------------AX=0000 BX=00FF CX=0010 DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0016 NV UP EI PL ZR NA PE NC NC 1B5E:0016 D1DB RCR BX,1 -t AX=0000 BX=007F CX=0010 DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0018 NV UP EI PL ZR NA PE CY CY 1B5E:0018 7302 JNB 001C ----------------------------------------------------------------------------Note when RCR BX,1 is executed, the 1 in the LSB (least significant bit) of register BX is shifted to the carry flag so that is now read CY. BX now reads 007FH which is exactly what you expect if you shift 0000 0000 1111 1111 right by one bit. The '0' from the 9th bit shifts into the 8th bit position, the '1' from the 5th bit position moves into the 4th bit position and the '1‘ from the 1st bit moves into the carry flag. In the meantime, the 0 from the carry flag moves into the 16th bit. We have 007FH left in register BX. Note:Note: the 1st bit of the multiplier moved into the carry flag is a '1' so we have to add the multiplicand in register DX to the intermediate product in AX. The instruction at offset 0018 when executed changes the next instruction counter IP to offset 001A. -------------------------------------------------------------------------- -t AX=0000 BX=007F CX=0010 DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=001A NV UP EI PL ZR NA PE CYCY 1B5E:001A 03C2 ADD AX,DX -t AX=FFFF BX=007F CX=0010 DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=001C NV UP EI NG NZ NA PE NCNC 1B5E:001C D1D8 RCR AX,1

0000 0000 0111 1111

--------------------------------------------------------------------------Register AX now contains the sum of the multiplicand in register DX andthe initial product 0000. Note :Note : the carry flag has been reset to NC or '0'.We are ready to shift the intermediate product right by one bit into the register BX. ---------------------------------------------------------------------------tAX=7FFF BX=007F CX=0010 DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=001E OV UP EI NG NZ NA PE CY CY

When we executed RCR AX,1, the LSB of AX was moved into the carry flag, mchanging it to CY. Also note the '0' from the carry flag has been stuffed into the MSB (most significant bit) of AX so that the highest nibble in AX is 7 instead of F. Further the LSB bit of BX or the next mutliplier bit is presented into the carry flag leaving as CY or "1".

27

-tAX=7FFF BX=007F CX=0010 DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=001E OV UP EI NG NZ NA PE CYCY1B5E:001E D1DB RCR BX,1 ----------------------------------------------------------------- -------When we executed RCR BX,1, the LSB of BX is shifted into the carry flagand the LSB of AX sitting in the carry flag is shifted to the highest bitof the register BX. This completes the shift of the register combinationAX BX right by one bit.----------------------------------------------------------------- --------t AX=7FFF BX=803F CX=0010 DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0020 OV UP EI NG NZ NA PE CYCY 1B5E:0020 49 DEC CX -------------------------------------------------------------------------- When we executed RCR BX,1, the LSB of BX is shifted into the carry flagand the LSB of AX sitting in the carry flag is shifted to the highest bitof the register BX. This completes the shift of the register combinationAX BX right by one bit.----------------------------------------------------------------- --------t AX=7FFF BX=803F CX=000F DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0021 NV UP EI PL NZ AC PE CYCY 1B5E:0021 E302 JCXZ 0025 -t AX=7FFF BX=803F CX=000F DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0023 NV UP EI PL NZ AC PE CY CY 1B5E:0023 EBF3 JMP 0018 --------------------------------------------------------------------------We have completed examining the first bit of the multiplier so we decreasethe CX counter by 1 so that CX = 000FH. Since this is not '0', the jump to end of the looping is skipped and the instruction pointer IP is set to the next location, given by offset 0018.---------------------------------------------------------------------------t AX=7FFF BX=803F CX=000F DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0018 NV UP EI PL NZ AC PE CYCY 1B5E:0018 7302 JNB 001C

----------------------------------------------------------------------------We are back at the point where are ready to check the carry flag carries a '0' or '1' from the next bit of the mutliplier in register BX. In the mean-time, the intermediate product has been shifted one bit to the right. And we are into the second loop already. We want to shorten our work, so we jump to the end of loop.

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----------------------------------------------------------------------------We are back at the point where are ready to check the carry flag carries a '0' or '1' from the next bit of the mutliplier in register BX. In the meantime, the intermediate product has been shifted one bit to the right. And we are into the second loop already. We want to shorten our work, so we jump to the end of loop.-----------------------------------------------------------------------------g21 AX=BFFF BX=401F CX=000E DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0021 NV UP EI PL NZ NA PO CYCY 1B5E:0021 E302 JCXZ 0025 -t AX=BFFF BX=401F CX=000E DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0023 NV UP EI PL NZ NA PO CYCY 1B5E:0023 EBF3 JMP 0018 -t AX=BFFF BX=401F CX=000E DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0018 NV UP EI PL NZ NA PO CYCY 1B5E:0018 7302 JNB 001C ----------------------------------------------------------------------------Above, we have completed the multiplication by the second bit. The counter CX has been deceremented by 1 to 000EH. We are ready to start examining the third bit of the multiplier which is present as CY or '1'. We do not wish todemonstrate all the identical loop for the remaining bits so we wil jump to the part where multipplication by all bits have been executed. The counter CX=0 at this stage.----------------------------------------------------------------------------

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-g25 AX=00FE BX=FF01 CX=0000 DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0025 NV UP EI PL ZR NA PE NC 1B5E:0025 891E0400 MOV [0004],BX DS:0004=0000 -t AX=00FE BX=FF01 CX=0000 DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=0029 NV UP EI PL ZR NA PE NC 1B5E:0029 A30600 MOV [0006],AX DS:0006=0000 -t AX=00FE BX=FF01 CX=0000 DX=FFFF SP=013C BP=0000 SI=0000 DI=0000 DS=1B5D ES=1B39 SS=1B49 CS=1B5E IP=002C NV UP EI PL ZR NA PE NC 1B5E:002C CB RETF ----------------------------------------------------------------------------We executed all instructions up to code offset 0025H. Here the counter CX=0 signifies that 16 shifts have taken place and mutliplication by the 16 bits in the multiplier have been completed. The product should be in registers AX AX BX BX = 00FE 00FE FF01FF01 which is indeed what the product of FFFFH x FF should be. Below, we show where the product is stored in the data segment. The low word of the product in register BX is loaded first at the location 1B5D:0004 and 1B5D:005 with the low byte being loaded before the higer byte.Again, the higher word of the product in register AX=00FE is loaded at location 1B5D:0006 and 1B5D:007. -----------------------------------------------------------------------------dds:00 1B5D:0000 FF FF FF 00 01 FF01 FF FE 00FE 00-00 00 00 00 00 00 00 00 ................----------- product stored here 1B5D:0010 1E 2B C0 50 B8 5D 1B 8E-D8 2B C0 8B 1E 02 00 8B .+.P.]...+...... 1B5D:0020 16 00 00 B9 10 00 D1 DB-73 02 03 C2 D1 D8 D1 DB ........s....... 1B5D:0030 49 E3 02 EB F3 89 1E 04-00 A3 06 00 CB 0F 2B C0 I.............+. 1B5D:0040 50 57 E8 7B 93 83 C4 04-89 46 F8 EB 06 8B 5E F8 PW.{.....F....^. 1B5D:0050 C6 07 00 FF 46 F8 8B 46-F8 A3 06 3A 56 E8 D9 93 ....F..F...:V... 1B5D:0060 83 C4 02 0B C0 76 0B 56-FF 36 E0 11 E8 7B 93 83 .....v.V.6...{.. 1B5D:0070 C4 04 57 B8 19 40 50 E8-70 93 83 C4 04 B8 01 00 [email protected].......

AX: 1B5D:0006 & 1B5D:0007 BX :1B5D:0004 & 1B5D:0005

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BACK UP SLIDES

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110101 10110 000000 110101 1101010 110101 100111110 000000 100111110 110101 10010001110

= Multiplicand ; MC

= Multiplier ; MP53 =22 =

Bit0 : start product

21029282726 25 24 23 22 21 20

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And Truth Table

Or Truth Table

Exclusive Or Truth Table

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AND Gate: a logic circuit designed to compare TRUE-FALSE (or on-off or one-zero) inputs and pass a resultant TRUE signal only when all the inputs are TRUE.

Binary: having two components or possible states.

Binary code: a system for representing things by combinations of two symbols, such as one and zero, TRUE and FALSE, or the presence or absence of voltage.

Binary number system: a number system that uses two as its base and expresses numbers as strings of zeros and ones.

Bit: the smallest unit of information in a computer, equivalent to a single zero or one. The word "bit" is a contraction of binary digit.

Boolean algebra: a method for expressing the logical relationships between entities such as propositions or on-off computer circuit; invented by the 19th Century English mathematician George Boole.

Byte: a sequence of bits, usually eight, treated as a unit for computation or storage.

Carry: the digit added to the next column in a an additional problem when the sum of the numbers in a column equals or exceeds the number base.

Kilobyte: 1,024 bytes (1,024 being one K, or two to the 10th power); often used as a measure of memory capacity.

Logic Gate: a circuit that accepts one or more than one input and always produces a single predicatable output.

OR gate: a circuit designed to compare binary TRUE-FALSE (or on-off or one-zero) inputs and pass a resultant TRUE signal if any input is TRUE.

XOR gate: a circuit designed to compare binary TRUE-FALSE (or on-off or one-zero) inputs and pass a resultant TRUE signal if a single input is TRUE, otherwise the output is FALSE.

1 Bit = Binary digit1 Byte = 8 Bits

1 Kilobyte = 1,000 Bytes 1 Megabyte = 1,000 Kilobytes1 Gigabyte = 1,000 Megabytes1 Terabyte = 1,000 Gigabyte1 Pet byte = 1,000 Terabytes1 Exabyte = 1,000 Petabytes1 Zettabyte = 1,000 Exabytes

1 Yottabyte = 1,000 Zettabytes

Chapter 9: Multiplication….. Saj Alam and Samya Zain

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Transcript of Chapter 9: Multiplication….. Saj Alam and Samya Zain