Эмулятор процессора Intel 8086 версия 4.05 (ENG) / asm_files / 0_sample_add_huge_n

Assembler Source Code a:link {text-decoration: none} a:visited {text-decoration: none} a:hover {text-decoration: none; color:#ff0000}     ; 0_sample_add_huge_numbers.asm name "huge" ; this example shows how to add huge unpacked bcd numbers. ; this allows to over come the 16 bit and even 32 bit limitation. ; because 32 digit decimal value holds over 100 bits! ; with some effort the number of digits can be increased. org 100h ; skip data: jmp code ; the number of digits in numbers: ; it's important to reserve 0 as most significant digit, to avoid overflow. ; so if you need to operate with 250 digit values, you need to declare len = 251 len equ 32 ; every decimal digit is stored in a separate byte. ; first number is: 423454612361234512344535179521 num1 db 0,0,4,2,3,4,5,4,6,1,2,3,6,1,2,3,4,5,1,2,3,4,4,5,3,5,1,7,9,5,2,1 ; second number is: 712378847771981123513137882498 num2 db 0,0,7,1,2,3,7,8,8,4,7,7,7,1,9,8,1,1,2,3,5,1,3,1,3,7,8,8,2,4,9,8 ; we will calculate this: ; sum = num1 + num2 ; 423454612361234512344535179521 + 712378847771981123513137882498 = ; = 1135833460133215635857673062019 sum db len dup(0) ; declare array to keep the result. ; you may check the result on paper, or click Start , then Run, then type "calc" and hit enter key. code: nop ; entry point (nop does nothing, it's nope). ; digit pointer: xor bx, bx ; setup the loop: mov cx, len mov bx, len-1 ; point to lest significant digit! next_digit: ; add digits: mov al, num1[bx] adc al, num2[bx] ; this is a very useful instruction that ; adjusts the value of addition ; to be string compatible aaa ; aaa stands for ascii add adjust. ; --- algorithm behind aaa --- ; if low nibble of al > 9 or af = 1 then: ; al = al + 6 ; ah = ah + 1 ; af = 1 ; cf = 1 ; else ; af = 0 ; cf = 0 ; ; in both cases: clear the high nibble of al. ; --- end of aaa logic --- ; store result: mov sum[bx], al ; point to next digit: dec bx loop next_digit ; include carry in result (if any): adc sum[bx], 0 ; print out the result: mov cx, len ; start printing from most significant digit: mov bx, 0 print_d: mov al, sum[bx] ; convert to ascii char: or al, 30h mov ah, 0eh int 10h inc bx loop print_d ; wait for any key press: mov ah, 0 int 16h ret ; stop

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