Question

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11. How does a program determine whether or not to jump based off of a comparison?(In assembly language)

12. Left shift 15 by 3

Answer 1

11.

Conditional execution in assembly language is defined by a few branching and looping instructions. These instructions can change the progression of control in a program.

• Conditional jump
• Unconditional jump

Syntax :      CMP destination, source

The CMP instruction compares two operands. It is generally used in conditional execution. This instruction basically subtracts one operand from the other for comparing whether the operands are equal or not. It does not disturb the destination or source operands. It is used along with the conditional jump instruction for decision making. CMP is often used for comparing whether a counter value has reached the number of times a loop needs to be run. Consider the following typical condition

Unconditional Jump - The JMP instruction provides a label name where the flow of control is transferred immediately.

Syntax :    JMP label

Example:

MOV  AX, 00    ; Initializing AX to 0
MOV  BX, 00    ; Initializing BX to 0
MOV  CX, 01    ; Initializing CX to 1
L20:
ADD  AX, 01    ; Increment AX
ADD  BX, AX    ; Add AX to BX
SHL  CX, 1     ; shift left CX, this in turn doubles the CX value
JMP  L20       ; repeats the statements

Conditional Jump -  If some specified condition is satisfied in conditional jump, the control flow is transferred to a target instruction. 

Example:

INC     EDX
CMP     EDX, 10 ; Compares whether the counter has reached 10
JLE     LP1     ; If it is less than or equal to 10, then jump to LP1

LP1: ....

Assembly code:

include 'emu8086.inc'
.model small
.stack 100h

.data


.code

        main proc
                
                mov dl,5
                mov bl,7
                
                cmp dl,bl
                
                je show
                        print 'Both are not same'
                        mov ah,04h
                        int 21h
                        
                        show:
                        print 'Both are same'
                        mov ah,04h
                        int 21h
                
                
        main endp

end main

output:

12.

In a left shift the binary value shifts to left by 1 place. Actually each shift is a 2 multiple of the value. So the 15 (00001111) multipled by 8 (2*2*2) , will calculate 120 (01111000) as the value whose binary form is the left shift of 15 by 3.

Syntax Binary form Value

x=15; 00001111 15

x= x>>3; 01111000 15 *(2*2*2) = 120

left shift of 15 (00001111) by 3 is 01111000 .