programb10 - mov ch, 04h ; Count of digits to be displayed...

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Appendix B  Explanation : Consider that a byte of data is present in the AL register and second byte of data is present in the BL register. We have to multiply the byte in AL with the byte in BL. Using MUL instruction, multiply the contents of two registers. The multiplication of two 8 bit numbers may result into a 16 bit number. So result is stored in AX register. The MSB is stored in AH and LSB in AL. For example : AL = 09 H 09 H BL = 02 H × 02 H 0012 H  Algorithm : Step I : Initialise the data segment. Step II : Get the first number in AL register. Step III : Get the second number in BL register. Corel - 3 Step IV : Multiply the two numbers. Step V : Display the result. Step VI : Stop  Flowchart : Refer flowchart 10.  Program : .model small .data Flowchart 10 a db 09H b db 02H .code mov ax, @data ; Initialize data section mov ds, ax mov ah, 0 mov al, a ; Load number1 in al mov bl, b ; Load number2 in bl mul bl ; multiply numbers and result in ax
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Unformatted text preview: mov ch, 04h ; Count of digits to be displayed mov cl, 04h ; Count to roll by 4 bits mov bx, ax ; Result in reg bx l2: rol bx, cl ; roll bl so that msb comes to lsb mov dl, bl ; load dl with data to be displayed and dl, 0fH ; get only lsb cmp dl, 09 ; check if digit is 0-9 or letter A-F Microprocessors & Applications (UPTU) Appendix B jbe l4 add dl, 07 ; if letter add 37H else only add 30H l4: add dl, 30H mov ah, 02 ; Function 2 under INT 21H (Display character) int 21H dec ch ; Decrement Count jnz l2 mov ah, 4cH ; Terminate Program int 21H end Result : C:\>tasm 8bit-mul.asm Turbo Assembler Version 3.0 Copyright (c) 1988, 1991 Borland International Assembling file: 8bit-mul.asm Error messages: None Warning messages: None Passes: 1 Remaining memory: 438k C:\>tlink 8bit-mul.obj Turbo Link Version 3.0 Copyright (c) 1987, 1990 Borland International Warning: No stack C:\>8bit-mul 0012...
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programb10 - mov ch, 04h ; Count of digits to be displayed...

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