Description
I. Topic Overview:
For assembly language-programs to carry out useful tasks, there must be a way to make decisions. In this lab, students will familiarize themselves with how decisions can be made with the jump instruction. The jump instructions transfer control to another part of the program. This transfer can be unconditional or can depend on a particular combination of status flag settings. After introducing the jump instructions, we’ll use them to implement high-level language decision structures.
II. Lesson Fit:
There is prerequisite to this lab. Students must have a basic idea on the following concepts:
a. Flag register
b. Some basic operations such as MOV, ADD, SUB, MUL and DIV
c. Basic I/O operations
d. Character encoding using ASCII
III. Learning Outcome:
a. Control flow of the program
b. Write conditional statements in assembly
IV. Anticipated Challenges and Possible Solutions
i. Step by step simulation
Solutions:
V. Acceptance and Evaluation
Code: 50%
Viva: 50% VI. Activity Detail a. Hour: 1 Discussion: Jump instruction
Conditional Jumps:
if (x>5){
// code
}
In the line if (x>5), a comparison is done between the content of x and 5. The decision whether to execute the enclosed code or not depends on the result of the comparison. In assembly the comparison is done by a piece of code called CMP. The syntax of CMP is CMP destination, source. The comparison is done by destination – source. The result is not saved anywhere but affects the flags. The result of subtraction can be 0, AX > BX or AX< BX. Below is a piece of code.
MOV AX, first_number
MOV BX, second_number
CMP AX, BX
The comparison is done in the third line. Now there could be 5 possibilities AX==BX, AX> BX, AX< BX, AX>= BX, AX<= BX. We will take the decision based on one of these options. This decision is performed by “jump” instruction denote as “j”.
Condition Jump Instruction Explanation
AX == BX JE jump if destination and source are equal
AX > BX JG jump if destination > source
AX < BX JL jump if destination < source
AX >= BX JGE jump if destination >= source
AX <= BX JLE jump if destination <= source
But jump where???
Jump in that line which we want to execute when one of the above conditions is satisfied. How will we get the line number? We do not have to worry about the line numbers because we will name the line(s) ourselves.
int x = 5;
if (x > 4) {
//code
} mov ax, 5 cmp ax, 4 JG My_Line
My_Line:
;code
Node the declaration of the line name ends with a colon (:).
Please go through page number 96 of the book to know about more conditional jumps.
Unconditional jump
For skipping a portion of code, we use/ need unconditional jump. There is no comparison, it is just a jump from one line to the one we want. The instruction is called JMP and its syntax is JMP destination line_name.
mov ax,5 mov bx,8 jmp my_line mov ah,4 mov dl,6 my_line mov dl,7
In the above code, when the 3rd line is executing the program jumps to the line named my_line without executing the codes in between, This is how codes are skipped.
Branching Structures
In high-level languages, branching structures enable a program to take different paths, depending on conditions. In this section, we’ll look at three structures.
IF-THEN :
The condition is an expression that is true or false. If it is true, the true-branch statements are executed. If it is false, nothing is done, and the program goes on to whatever follows.
Example: Replace the number in AX by its absolute value Solution: A pseudocode algorithm is:
IF AX < 0
THEN replace AX by -AX
END_IF
It can be coded as follows:
The condition AX < 0 is expressed by CMP AX,0. If AX is not less than 0, there is nothing to do, so we use a JNL (jump if not less) to jump around the NEG AX. If condition AX < 0 is true, the program goes on to execute NEG AX.
Problem Task: Task 01 (Page 8)
b. Hour: 2
Discussion: Branching Structures (Cont.) IF-THEN-ELSE:
The jmp instruction comes in use when if – else condition is employed. Let us see the use of conditional and unconditional jumps together in a program where we will find the greater of 2 numbers.
JAVA
System.out.println(“enter the first number”); int x = k.nextInt();
System.out.println(“enter the second number”); int y = k.nextInt();
if (x>y){
System.out.println(x+” is greater”);
} else{
System.out.println(y+” is greater”);
}
Assembly:
data segment
; add your data here!
pkey db “press any key…$” a db “Enter first number$” b db “Enter second number$” c db ” is larger$”
ends stack segment dw 128 dup(0)
ends code segment start:
; set segment registers:
mov ax, data mov ds, ax mov es, ax
lea dx, a ;Print a mov ah,9 int 21h mov ah,1 int 21h
mov bl,al ;move input to bl
; add your code here
lea dx, b ;Print b mov ah,9 int 21h mov ah,1 int 21h
mov cl,al ;move input to cl
cmp cl,bl ;compare the two inputs jg line1
end
lea dx, c mov ah,9 int 21h jmp e
mov dl,bl mov ah,2 int 21h
Discuss why was the „jmp e‟ statement crucial in this program.
CASE
A CASE is a multiway branch structure that tests a register, variable, or expression for particular values or a range of values. The general form is as follows:
CASE expression
Values _ 1: statements_1 Values _ 2: statements_2
..
..
Values_n: statements_n
END_CASE
In this structure, expression is tested; if its value is a member of the set values_1, then statements_1 are executed. We assume that sets values_l, … , values_n are disjoint
Problem Task: Task 02 – 09 (Page 8)
c. Hour: 3
Discussion:
Check progress while the students carry on with the rest of the tasks.
Problem Task: Task 10-14 (Page 9)
VII. Home Tasks: All the unfinished lab tasks.
Lab 4 Activity List
Task 01
Take a number in AX, and if it‟s a negative number, replace it by 5.
Task 02
Suppose AL and BL contain extended ASCII characters. Display the one that comes first in the character sequence.
Task 03
If AX contains a negative number, put -1 in BX; if AX contains 0, put 0 In BX; if AX contains a positive number, put 1 in BX.
Task 04
If AL contains 1 or 3, display “o”; if AL contains 2 or 4 display “e”. Task 05
Read a character, and if it’s an uppercase letter, display it.
Task 06
Read a character. If it’s “y” or “Y”, display it; otherwise, terminate the program.
Task 07
Write an assembly program to check whether a number is even or odd.
Task 08
Write a program to input any alphabet and check whether it is vowel or consonant.
Task 09
Write a program to check whether a number is divisible by 5 and 11 or not.
Task 10
Write a program to find the maximum and minimum between three numbers.
Sample execution:
User input : 2 3 4
Output: Maximum number is 4
Minimum number is 1
Task 11
Write a program that takes as input all sides of a triangle and check whether triangle is valid or not. If the sides form a triangle, print “Y”, otherwise print “N”.
Task 12
Write a program that takes a digit as an input and outputs the following. If the digit is within 0-3, it prints “i”, If it‟s within 4-6, it prints “k”, If it‟s within 7-9, it prints “l” and if it‟s 10, it prints “m”.
Task 13
Sample execution:
User Input: 3
Task 14
Write a case to print the total number of days in a month.
Sample execution:
User Input : 3
Output: 31 days
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