Question

Consider a 9-story building with a basement. The first floor is given the number 1, the second is given the number 2 and so on. The basement is given the number 0. You will write a program in ASSEMBLY CODE to simulate an elevator operated throughout the whole building.

Write the following subroutines that will work with the main program.

DELAY: generates a software delay of 1 second. OPEN: displays a ‘Door open’ message on the screen after a delay of 1 second. CLOSE: displays a ‘Door close’ message on the screen after a delay of 1 second. UP: performs the following procedure: Displays a ‘Going UP’ message on the screen. Goes up one floor at a time (1 second delay between each floor), until the elevator reaches the destination. At each floor, prints that number (current floor number) on the screen. After the elevator reaches the destination floor, stores this number as the current floor number in memory. DOWN: performs the following procedure: Displays a ‘Going DOWN’ message on the screen. Goes down one floor at a time (1 second delay between each floor), until the elevator reaches the destination. At each floor, prints that number (current floor number) on the screen. After the elevator reaches the destination floor, stores this number as the current floor number in memory.

Basically, a number is inputted. You are currently on floor one. The number inputted is the floor you are heading to. After initializing your current floor (1), you call an open subroutine, and wait for the number to be inputted. Then the close subroutine is called. After this, depending on what floor number was picked, the UP or DOWN subroutine is called until the floor is reached, and the STAY function is called, once you reach the floor. The floor numbers print along with the going UP or going DOWN functions.

Answer 1

The below program produces a delay of 1 second. here the subrouting makes a delay of 1 microseconds.

it's called 4 times back to back and the entire cycle is repeated 250 times.

Hence the result is like as follows:

A delay of 4 times called x 1 microsecond x 250 = 1000 micro seconds.

ie, 1 second is produced.

Now go to the program code;

DELAY1: MOV R5, #250D

LABEL: ACALL DELAY

ACALL DELAY

ACALL DELAY

ACALL DELAY

DJNZ R5, LABEL

RET

DELAY: MOV R6, #250D

MOV R7, #250D

LOOP1: DJNZ R6, LOOP1

LOOP2: DJNZ R7, LOOP1

RET

NOTE

IN R5, THE DELAY CALLED 4TIMES BACK. ITS NOTED THAT IN R6 AND R7 THE ENTIRE CYCLE PROCESS IS SPECIFIED AS 250 TIMES (250D).

PLEASE LISTEN THE LOOP AS FOLLOWS:

DELAY: MOV R6, #250D

MOV R7, #250D

LABEL1: DJNZ R6, LABEL1

LABEL2: DJNZ R7, LABEL2

RET