32-bit calculator in assembly 8086

Question

CALCULATOR 32 bit

Can anyone help me with my 32-bit calculator in MASM32. I think adding and subtracting is fine, but I cannot print the number in decimal;

0002FFFF - 10005 = 1fffa

In MEMORY: 0001 0FFFA

PRINT: 165530 in decimal form

DATA_HERE SEGMENT mult1 dw 0002H dw 0FFFFH mult2 dw 0001H dw 0005H ans dw 0,0 DATA_HERE ENDS STACK_HERE SEGMENT STACK DW 40 DUP(0) STACK_HERE ENDS CODE_HERE SEGMENT ASSUME CS:CODE_HERE, DS:DATA_HERE, SS: STACK_HERE INICIO: MOV AX,DATA_HERE MOV DS,AX ADD: MOV AX,mult1+2 ; take lower 16-bit of NUM1; take ADD AX,mult2+2 ; AX = AX + lower 16-bit of NUM2 MOV ans+2,AX ; Store lower 16-bit result at ans MOV AX,mult1 ; take higher 16-bit of NUM1 in AX; ADC AX,mult2 ; AX = AX + NUM2 + CF (add with carry) MOV ans,AX ; Store higher 16-bit result at ans SUBTRACT: MOV AX,mult1+2 ; take lower 16-bit of NUM1 in AX ; SUB AX,mult2+2 ; AX = AX - lower 16-bit of NUM2 MOV ans+2,AX ; Store lower 16-bit result at ans MOV AX,mult1 ; take higher 16-bit of NUM1 in AX; SUB AX,mult2 ; AX = AX - NUM2 MOV ans,AX ; Store higher 16-bit result at ans xor si,si mov si,0 ciclo: mov AX, ans[si]; call display ; print AX add si, 2 cmp si, 2 JLE ciclo mov ax,4C00h int 21h display proc ;push CX ;Beginning of procedure MOV BX, 10 ;Initializes divisor ;MOV DX, 0000H ;Clears DX MOV CX, 0000H ;Clears CX ;Splitting process starts here .Dloop1: MOV DX, 0000H ;Clears DX during jump DIV BX ;Divides AX by BX PUSH DX ;Pushes DX(remainder) to stack INC CX ;Increments counter to track the number of digits CMP AX, 0 ;Checks if there is still something in AX to divide JNE .Dloop1 ;Jumps if AX is not zero .Dloop2: POP DX ;Pops from stack to DX ADD DX, 30H ;Converts to it ASCII equivalent MOV AH, 02H INT 21H ;calls DOS to display character LOOP .Dloop2 ;Loops till CX equals zero ;pop CX RET ;returns control display ENDP CODE_HERE ENDS END INICIO 

Answer 1

First you print the bottom 16 bit as decimal FFFA → 65530 and 1 from the second word.

1 65530.

If you have a 32-bit processor, you can perform the separation using the EAX / EDX registers.

If not, it's harder. Then you should simulate 32-bit division. This is not fun ;-).

Here is a hint: http://en.wikipedia.org/wiki/Division_algorithm (yes, you have to handle the bit)

If you are writing a C program that divides a long value by 10L, compile it for your processor (16 bits) and decompose the output file, you will see how they do it. This is just an idea. I have not tried it myself. You must state that your long value is large enough so that the compiler cannot optimize .; -)

Answer 2

The following code done with EMU8086 gets a 32-bit number in two separated words, then both of them are converted to a string. The code uses the largest 32-bit number 0FFFF FFFFh, you can change it for any other number, from 0 to FFFF FFFFh, it will work (comments will help you understand what is happening in the code):

 .model small .stack 100h .data num_low dw ? ;32 BIT num_hig dw ? ;NUMBER. buf db 12 dup('$') ;NUMBER CONVERTED TO STRING. .code start: ;INITIALIZE DATA SEGMENT. mov ax, @data mov ds, ax ;STORE BIG 32 BIT NUMBER = 4294967295 = 0FFFF FFFFh. mov num_low, 0FFFFh mov num_hig, 0FFFFh ;CONVERT 32 BIT NUMBER TO STRING. mov si, offset buf call number2string32bit ;DISPLAY STRING (32 BIT NUMBER CONVERTED). mov ah, 9 mov dx, offset buf int 21h ;WAIT FOR ANY KEY. mov ah, 7 int 21h ;FINISH PROGRAM. mov ax, 4c00h int 21h ;------------------------------------------ ;CONVERT 32 BIT NUMBER IN STRING. ;ALGORITHM : EXTRACT DIGITS ONE BY ONE DIVIDING ;NUMBER BY 10, STORING REMAINDERS IN STACK, THEN ;EXTRACT THEM IN REVERSE ORDER TO BUILD STRING. ;PARAMETERS : num_low = LOW WORD OF 32 BIT NUMBER. ; num_hig = HIGH WORD OF 32 BIT NUMBER. ; SI = POINTING WHERE TO STORE STRING. number2string32bit proc mov bx, 10 ;DIVIDE NUMBER BY 10 TO EXTRACT DIGITS. mov cx, 0 ;DIGITS COUNTER. NECESSARY TO POP REMAINDERS. extracting: ;DIVIDE HIGHER WORD OF 32 BIT NUMBER. mov dx, 0 ;DX NOT NECESSARY FOR THE HIGH WORD. mov ax, num_hig div bx mov num_hig, ax ;HIGHER WORD OF RESULT. ;DIVIDE LOWER WORD OF 32 BIT NUMBER. ;VERY IMPORTANT : PREVIOUS DX IS NECESSARY FOR THE LOW WORD. mov ax, num_low div bx mov num_low, ax ;LOWER WORD OF RESULT. push dx ;STORE REMAINDER (EXTRACTED DIGIT). inc cx ;INCREASE DIGIT COUNTER. ;CHECK END OF PROCESS. cmp ax, 0 ;IF LOWER WORD IS jne extracting ;NOT ZERO, REPEAT. ;NOW RETRIEVE PUSHED DIGITS. THERE ARE CX DIGITS STORED IN STACK. poping: pop dx ;GET STORED DIGIT. add dl, 48 ;CONVERT DIGIT TO CHARACTER. mov [ si ], dl ;STORE CHARACTER IN STRING. inc si ;POSITION FOR NEXT CHARACTER. loop poping ;CX--. IF ( CX > 0 ) REPEAT. ret number2string32bit endp ;------------------------------------------ end start