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Division Algorithm Assignment Solution using Assembly Language


Assembly Programming Exercises

Exercise 1

Write a program (div.asm) to perform a positive integer long-division algorithm. Your program will have two inputs: the dividend and divisor and have two outputs: the quotient and remainder.

For simplicity, assume, you will be given only positive values and the divisor will be always greater than zero. HINT: For division, use the repeated subtraction method i.e. subtract the divisor from the dividend till your result becomes zero. In some cases, you will have remainder though.

To grade the program, your inputs and outputs are as follows:

Register R1 = Divisor

Register R6 = Dividend

Register R2 = Quotient

Register R3 = Remainder.

Example: R1 = 50, R6 = 105. Therefore 105/50, R2 = 2 and R3 = 5.

Exercise 2

Write a program that takes the string Lastname, firstname. The string is located at location FLNAME. To initialize the constant string we use the.STRINGZ directive which initializes locations with ASCII value of the character in the sequence. Note that.STRINGZ also null terminate the string. Your program will display the string as-is, then your program must display the first name first followed by a space followed by the last name.

Change the characters after.STRINGZ instruction in the file to test different cases.

You need to turn in your .asm code and a screen print of the output. Here is a sample of the output:

The original full name is:

Mansour, Khaled

The modified full name is:

Khaled Mansour

Solution:

1.

.ORIG x3000 ; Program start address LD R1, DIVISOR ; Load divisor LD R6, DIVIDND ; Load dividend ; At the start: ; R1 = Divisor ; R6 = Dividend ; At the end: ; R2 = Quotient ; R3 = Remainder AND R2, R2, #0 ; Start quotient to zero ADD R3, R6, #0 ; Copy dividend, it will be the initial remainder DIVLOOP NOT R5, R1 ; take ones complement of divisor ADD R5, R5, #1 ; add 1 to get the two's complement (negative) ADD R5, R5, R3 ; subtract remainder - divisor BRn DONE ; if result is negative, end division ADD R3, R5, #0 ; Copy result to current remainder ADD R2, R2, #1 ; increment quotient BR DIVLOOP ; repeat division loop DONE BR DONE ; Infinite loop to stop program DIVISOR .FILL #50 ; Divisor DIVIDND .FILL #105 ; Dividend .END

Output:

LC 3 Simulator

2.

.ORIG x3000 ; Program start address LEA R0, ORIGMSG ; Load original string address PUTS ; print original string message LD R0, LF ; load line feed character OUT ; jump to next line LEA R0, FLNAME ; Load fullname string address PUTS ; print fullname string LD R0, LF ; load line feed character OUT ; jump to next line LEA R1, FLNAME ; Load fullname string address LD R2, NCOMMA ; load negative of comma FNDLOOP LDR R3, R1, #0 ; Load character from string ADD R4, R3, R2 ; compare with comma BRz DISPLAY ; if equal, print strings ADD R1, R1, #1 ; advance to next char in string BR FNDLOOP ; repeat loop DISPLAY STR R4, R1, #0 ; save null to replace comma and end string LEA R0, MODMSG ; Load modified string address PUTS ; print modified string message LD R0, LF ; load line feed character OUT ; jump to next line ADD R0, R1, #2 ; point to space after the comma and the space ; which is the start of the first name PUTS ; print first name string LD R0, SPACE ; load space character OUT ; display space LEA R0, FLNAME ; Load last name string address PUTS ; print last name string LD R0, LF ; load line feed character OUT ; jump to next line DONE BR DONE ; Infinite loop to stop program FLNAME .STRINGZ "Mansour, Khaled" ; Original full name ORIGMSG .STRINGZ "The original fullname is:" ; Original message MODMSG .STRINGZ "The modified fullname is:" ; Modified message NCOMMA .FILL #-44 ; negative of ascii value for comma ',' SPACE .FILL 32 ; space character LF .FILL 10 ; line feed to jump to next line .END

Output:

LC 3 Console