This program is supposed to be in assembly language. Please use basic MASM only.

Question

This program is supposed to be in assembly language. Please use basic MASM only. Use Kip Irvine's Assembly Language for x86 Processors (7th edition).

Write a procedure (MultiplyBit) that multiplies any unsigned 16-bit integer by AX, using ONLY shifting and rotation. Ask the user for two numbers, and using a procedure (DisplayProduct) the results as follows: For example, n = 5 and m-3 5*3=15

Explanation / Answer

;------------ ; muldemo.asm ; This is a demo program for emu 8086 ; ; This demo reads two unsigned 16-bit ; integer values (words) from keyboard and ; prints their product. ; ; In this demo the function readint is defined ; to read a 16-bit unsigned integer value (word). ; If the number entered by user is too big, ; it prints an error message. ; ; In this program we use our defined function ; printud to print the product. ; ; printud prints the value of AX register as ; a 16-bit unsigned integer (word). ; ; The main program of this demo gets two words ; from key board and saves them in value1 and ; value2 respectively, then it calculates the ; product. If the product has overflow (dx > 0) ; then an overflow message is printed. Otherwise, ; the value of product is printed. ; We have used various prompts and messages to ; support a user-friendly interaction. ; multi-segment executable file template. data segment ; add your data here! ; Memory variables value1 dw 24 value2 dw 35 value3 dw ? prompt1 db "Enter two integers: $" product db "Product = $" interror db 13, 10, "Number is too big!", 13, 10, '$' mulerror db 13, 10, "Multiplication overflow!", 13, 10, '$' ; endl : string of return + new-line endl db 13, 10, '$' ; a string to keep prompt visible pkey db "press any key...$" ends stack segment dw 128 dup(0) ends code segment ;-------------------- ; MAIN PROGRAM start: ; set segment registers: mov ax, data mov ds, ax mov es, ax ;-------------------- ; add your code here ; Print prompt1 lea dx, prompt1 mov ah, 9 int 21h ; Read first int, if it is too large, error message. call readint cmp dx, 0 jne L1001 ; Save the number in value1 mov value1, ax ; Read the second int, if it is too large, error message. call readint cmp dx, 0 jne L1001 ; Save the number in value2 mov value2, ax ; print a return/new-line char. lea dx, endl mov ah, 9 int 21h ; perform the multiplication mov ax, value1 mul value2 ; If the result is too big, error message. cmp dx, 0 jne L1002 ; save the product in value3 mov value3, ax ; print the message lea dx, product mov ah, 9 int 21h ; print the product result mov ax, value3 call printud ; print return/new-line lea dx, endl mov ah, 9 int 21h ; goto end jmp L1000 ; print number too large error L1002: lea dx, mulerror mov ah, 9 int 21h jmp L1000 ; print multiplication overflow error L1001: lea dx, interror mov ah, 9 int 21h L1000: ;-------------------- lea dx, pkey mov ah, 9 int 21h ; output string at ds:dx ; wait for any key.... mov ah, 1 int 21h mov ax, 4c00h ; exit to operating system. int 21h ;-------------------- ; Function printd ; prints the value of AX register in signed ; decimal format. ; ; This function uses a recursive algorithm to print ; The value in AX register. For example, to print the ; value 3187, this function call itself (printd) ; for 318, then prints 7. ; If the value to print is less than 10, then it is ; printed and recursion terminates. ; If the value is negative, a - is printed then ; printd is called for the negate of value. printd proc ; preserve used registers push ax push bx push cx push dx ; if negative value, print - and call again with -value cmp ax, 0 jge L1 mov bx, ax ; print - mov dl, '-' mov ah, 2 int 21h ; call with -AX mov ax, bx neg ax call printd jmp L3 L1: ; divide ax by 10 ; ( (dx=0:)ax / cx(= 10) ) mov dx, 0 mov cx, 10 div cx ; if quotient is zero, then print remainder cmp ax, 0 jne L2 ; DX contains the remainder, but since DX < 10; ; actually DL contains it. In order to print it ; In ASCII format, we should add '0' to it. ; For example, the ascii code of 5 is 53, ; and the ascii code of '0' is 48. In order to ; print 5, we add '0' to it to make it '5'. add dl, '0' mov ah, 2 int 21h jmp L3 L2: ; if the quotient is not zero, we first call ; printd again for the quotient, and then we ; print the remainder. ; call printd for quotient: call printd ; print the remainder add dl, '0' mov ah, 2 int 21h L3: ; recover used registers pop dx pop cx pop bx pop ax ret printd endp ;-------------------- ; Function printud ; prints the value of AX register in unsigned ; decimal format. ; ; This function uses a recursive algorithm to print ; The value in AX register. For example, to print the ; value 3187, this function call itself (printud) ; for 318, then prints 7. ; If the value to print is less than 10, then it is ; printed and recursion terminates. ; The comments are exactly like printd. We just dropped ; the code for negative case. There is no negative case ; for unsigned integer: -1 is assumed 65535. printud proc push ax push bx push cx push dx mov dx, 0 mov cx, 10 div cx cmp ax, 0 jne L4 add dl, '0' mov ah, 2 int 21h jmp L5 L4: call printud add dl, '0' mov ah, 2 int 21h L5: pop dx pop cx pop bx pop ax ret printud endp ;-------------------- ; Function readint ; ; Reads an integer from keyboard. ; IF OK, AX contains the value (DX = 0) ; Otherwise, DX > 0 and number is too large. readint proc ; preserve the used registers push bx push cx ; bx contains the number entered so far. ; It is initially 0 mov bx, 0 ; dx indicates if there is overflow. mov dx, 0 L12: ; reads a character from keyboard mov ah, 1 int 21h ; Checks if it is a valid numeric key. cmp al, '0' jb L11 cmp al, '9' ja L11 ; if it is valid, convers fro ascii to integer, sub al, '0' mov ah, 0 ; multiples the number read so far by 10 and adds ; the new value to it xchg ax, bx mov cx, 10 mul cx ; If there is a overflow, exits the function ; with dx > 0 cmp dx, 0 jne L11 ; Otherwise, saves the number so far in bx add bx, ax ; Goes for another key. jmp L12 L11: ; If for any reason input terminates, ; that is, either user enters a white ; separator or the number is too large, ; we are here: move the number read so far ; to ax and returns. mov ax, bx ; recover used registers . pop cx pop bx ret readint endp ;-------------------- ends end start ; set entry point and stop the assembler.

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