CSC103 Assembly Language Lab 2012

--D. Thiebaut 07:15, 20 February 2012 (EST)

---

This lab deals with assembly language and the computer simulator. Work on this lab with a partner. It's easier to figure out some of these questions with another person.

Preparation

Start

• Point your browser to the simulator and start it (third button from the bottom)

Editing Program

• Click on the top bar at the top of the simulator applet, which says computer
• In the edit window that opens up, enter this simple program:

• Enter a simple program in the window:

start: lod-c  1
		sto    var1
		hlt

@10
var1:  data

Translating/loading into RAM

• Click on translate
• Observe that your program has been translated, or coded into numbers. Numbers are the only thing the processor can understand.
• If the top right button says Integer, click on it so that it shows Instructions. Observe that numbers can mean instructions as well as numbers for the processor.
• Click on the top right button and see how the memory can be seen many different ways. The real contents of the memory is when you ask to see it in binary. That's what inside the memory of a computer at any given time. Bits. But because these bits are used in a code, they can mean different things.

Execute the Progam

Single Stepping

• Click on the button that says "Set PC = 0" to make sure the processor will execute the program starting at 0
• Click on the Cycle button and observe that the number 1 is going into the AC register and then into memory, at Address 10.

Running the program

• If you want to run the program full speed, reset the PC register to 0 and click Run instead of Cycle.
• First, change the program so that it loads a number different from 1 into AC:

        lod-c    55

• Translate
• set PC = 0
• Run
• Observe that 55 will be loaded up from memory into AC, and then stored into the variable var1, at Address 10.

You now know the basics of programming in assembly language.

Problem 1

• Using the steps you have just gone through, write a program that initializes 3 variables called var1, var2, and var3 to the value 23.

Problem 2

• Try this program out:

start: lod    var1
		sto    var2
		hlt

@10
var1:  55
var2:	data

• Run it.
• Observe the result, and in particular what happens to var2.
• Add two new variables to the data section: var3, and var4.
• Modify the program so that it stores the contents of var1 into var2, var3, and var4.
• Check that it works
• Modify the number in var1 and restart the program. Does the program put the new value in the other 3 variables?

Problem 3

• Try this program out:

start: lod    var1
		add-c 	1
		sto    var1
		hlt

@10
var1:  55
var2:	 66
var3:   3

• Run the program
• What does it do? What's the result of the computation?
• Once you've figured it out, modify the program so that var1 originally starts with the value 5 in it, and then end with the value 7.

• Once that works, modify the program so that it adds one to all three variables.

Problem 4

• Try this new program out:

start: lod    var1
loop:	add-c	1
		sto    var1
		jmp	   loop
		hlt

@10
var1:  55
var2:	 55

• Run the program. Study it.
• Modify it so that as the program runs, var1 increases by 1 every step through the loop, and var2 decreases by one through the loop. In other words, when the program starts, var1 and var2 are 55. Then, a few cycles later, var1 is 56 and var2 is 54. Then a few cycles later, var1 is 57, and var2 is 53, etc.

Solutions

P1

; Solution for Problem 1
; D. Thiebaut
start: lod-c 23
		sto var1
		sto	var2
		sto var3
		hlt

@10
var1:	0
var2:	0
var3:	0

P2

; Solution for Problem 2
; D. Thiebaut
start: lod	var1
		sto var2
		sto var3
		sto var4
		hlt

@10
var1:	55
var2:	0
var3:	0
var4:  0

P3

; Solution for Problem 3
; D. Thiebaut
start: lod    var1
		add-c 	1
		sto    var1
		lod		var2
		inc
		sto		var2
		lod		var3
		add-c	1
		sto		var3
		hlt

@10
var1:  55
var2:	 66
var3:   3

P4

; Solution for Problem 4
; D. Thiebaut
; First initialize both variables with 55

start:	lod-c  55
	sto		var1
	sto		var2

; now loop and increment var1 and decrement var2
; the loop is endless.
loop:	lod    		var1
    	add-c		1
	sto    		var1
	lod		var2
	dec
	sto		var2
	jmp		loop
	hlt

@15
var1:  	0
var2:  	0