Question

What mathematical problem is the above assembly code performing?

for example if the answer was working out what 2 + 2 is, then you would type:

2 + 2

Answer 1

Code for Add two numbers in Assembly Language

DATA SEGMENT
NUM1 DB ?
NUM2 DB ?
RESULT DB ?
MSG1 DB 10,13,"ENTER FIRST NUMBER TO ADD : $"
MSG2 DB 10,13,"ENTER SECOND NUMBER TO ADD : $"
MSG3 DB 10,13,"RESULT OF ADDITION IS : $"
ENDS
CODE SEGMENT
ASSUME DS:DATA CS:CODE
START:
MOV AX,DATA
MOV DS,AX
LEA DX,MSG1
MOV AH,9
INT 21H
MOV AH,1
INT 21H
SUB AL,30H
MOV NUM1,AL
LEA DX,MSG2
MOV AH,9
INT 21H
MOV AH,1
INT 21H
SUB AL,30H
MOV NUM2,AL
ADD AL,NUM1
MOV RESULT,AL
MOV AH,0
AAA
ADD AH,30H
ADD AL,30H
MOV BX,AX
LEA DX,MSG3
MOV AH,9
INT 21H
MOV AH,2
MOV DL,BH
INT 21H
MOV AH,2
MOV DL,BL
INT 21H
MOV AH,4CH
INT 21H
ENDS
END START

In this Assembly Language Programming, A single program is divided into four Segments which are 1. Data Segment, 2. Code Segment, 3. Stack Segment, and 4. Extra Segment. Now, from these one is compulsory i.e. Code Segment if at all you don’t need variable(s) for your program.if you need variable(s) for your program you will need two Segments i.e. Code Segment and Data Segment.

Next Line –CODE SEGMENT

CODE SEGMENT is the starting point of the Code Segment in a Program and CODE is the name given to this segment and SEGMENT is the keyword for defining Segments, Where we can write the coding of the program.

Next Line – ASSUME DS:DATA CS:CODE

In this Assembly Language Programming, their are Different Registers present for Different Purpose So we have to assume DATA is the name given to Data Segment register and CODE is the name given to Code Segment register (SS,ES are used in the same way as CS,DS )

Next Line – START:

START is the label used to show the starting point of the code which is written in the Code Segment. : is used to define a label as in C programming.

Next Line – MOV AX,DATA
MOV DS,AX

After Assuming DATA and CODE Segment, Still it is compulsory to initialize Data Segment to DS register. MOV is a keyword to move the second element into the first element. But we cannot move DATA Directly to DS due to MOV commands restriction, Hence we move DATA to AX and then from AX to DS. AX is the first and most important register in the ALU unit. This part is also called INITIALIZATION OF DATA SEGMENT and It is important so that the Data elements or variables in the DATA Segment are made accessible. Other Segments are not needed to be initialized, Only assuming is enhalf.

Next Line – LEA DX,MSG1
MOV AH,9
INT 21H

he above three line code is used to print String or Message present in the character Array till $ symbol which tells the compiler to stop.

Now, lets understand line by line

LEA DX,MSG1 in this LEA stands for LOAD EFFECTIVE ADDRESS and it loads the effective address of second element into the first element. This same code can be interchangeably written as MOV DX, OFFSET MSG1 where OFFSET means effective address and MOV means move second element into the first element.

MOV AH,9
INT 21H

The above two line code is used to PRINT the String or Message of the address present in DX register.

Standard Input and Standard Output related Interrupts are found in INT 21H which is also called as DOS interrupt. It works with the value of AH register, If the Value is 9 or 9h, That means PRINT the String or Message of the address present in DX register.

Next Line – MOV AH,1
INT 21H
SUB AL,30H
MOV NUM1,AL

The above Four line code is used to Read a Character from Console and save the value entered in variable NUM1 in its BCD form. This can be done by subtracting 30H i.e. SUB AL,30H. The value coming from Console is Basically in ASCII form. eg. When you enter 5 we see 35H,So by subtracting 30H we get back to value as 5.

Standard Input and Standard Output related Interrupts are found in INT 21H which is also called as DOS interrupt. It works with the value of AH register, If the Value is 1 or 1h, That means READ a Character from Console, Echo it on screen and save the value entered in AL register.

SUB AL,30H means subtracting 30H from AL.

MOV NUM1,AL means move value in AL register into variable NUM1.

Next Line – LEA DX,MSG2
MOV AH,9
INT 21H

The above two line code is used to PRINT the String or Message of the address present in DX register i.e. for MSG2.

Next Line – MOV AH,1
INT 21H
SUB AL,30H
MOV NUM2,AL

The above Four line code is used to Read a Character from Console and save the value entered in variable NUM2 in its BCD form.

Next Line – ADD AL,NUM1

As we know the programs work only with the instructions in the instruction set. Instruction ADD is used to add to numbers in the following permutations above. REG stands for Registers (Eg. AX, BX, CX, DX ). memory stands for Variable or Address. immediate stands for Numbers or Values. Let us understand the meanings of the above permutations.

First permutation :- REG , memory means Register can be added with memory.

Second permutation :- memory , REG means memory can be added with Register.

Third permutation :- REG, REG means Register can be added with Register.

Fourth permutation :- memory , immediate means memory can be added with immediate.

Fifth permutation :- REG, immediate means Register can be added with immediate.

Note :- In the permutations above it will work only in the order mentioned above and not by interchanging the first to second and second to first.

Now, we have understood part of it to add to number we can write ADD NUM1, NUM2, But there is no permutation for ADD memory, memory, Hence we have to send one number to AL or AX depending on DB or DW. AX Register is called Accumulator. and is used for holding the result of Addition in it After Addition. Now we are taking DB, So we have t0 instruction MOV AL,NUM1 move NUM1 variable value to AL Register. After moving NUM1 to AL, We can Add REG to memory, So we have ADD AL,NUM2 or (We can Add memory to REG , So we have ADD NUM2,AL) Both are allowed as per permutations so use one from two. Now, the Resultant Value is saved in Accumalator AL for DB and AX for DW, So move Resultant value to RESULT variable by instruction MOV RESULT,AL

Next Line – MOV AH,0
AAA

The first line is just to clear the unwanted garbage value present in AH register as the AH register is going to be used later. The above two line code is used to corrects result in AH and AL after addition when working with BCD values . AAA means ASCII Adjust after Addition. AAA Instruction has NO Operands or values attached to it. If we are adding two BCD numbers the Result is saved in AL register in HEXadecimal form. What AAA exactly does is it converts the result into BCD form and first digit is saved in AH register and second digit is saved in AL register.

Next Line – ADD AH,30H
ADD AL,30H

Since the Result of Multiplication is in AH and AL register in BCD form After using AAM. Now we want to print the result on screen or console. the BCD form value will not show us the Result But will print the Corresponding Ascii Codes of the number, Hence By adding 30H to BCD will Convert it to ASCII code which will print the digit (number) on screen.

Next Line – MOV BX,AX

Since the AH register is used again and again. We cannot lose the result in AH and AL register, So to save the to be printed value in BH and BL register. By moving AX to BX.

Next Line – LEA DX,MSG3
MOV AH,9
INT 21H

The above two line code is used to PRINT the String or Message of the address present in DX register i.e. for MSG3.

Next Line – MOV AH,2
MOV DL,BH
INT 21H

The above Three line code is used to Write a Character on Console present in BH register.

Standard Input and Standard Output related Interupts are found in INT 21H which is also called as DOS interrupt. It works with the value of AH register, If the Value is 2 or 2h, That means WRITE a Character on Console present in DL register hence the value to be printed is moved to DL register. Here we are printing BH register.

Next Line – MOV AH,2
MOV DL,BL
INT 21H

The above Three line code is used to Write a Character on Console present in BL register.

Next Line – MOV AH,4CH
INT 21H

The above two line code is used to exit to dos or exit to operating system. Standard Input and Standard Output related Interrupts are found in INT 21H which is also called as DOS interrupt. It works with the value of AH register, If the Value is 4ch, That means Return to Operating System or DOS which is the End of the program.

Next Line – CODE ENDS

CODE ENDS is the End point of the Code Segment in a Program. We can write just ENDS But to differentiate the end of which segment it is of which we have to write the same name given to the Code Segment.

Last Line – END START

END START is the end of the label used to show the ending point of the code which is written in the Code Segment.

Note :- In this Assembly Language Programming, We have Com format and EXE format. We are Learning in EXE format only which simple then COM format to understand and Write. We can write the program in lower or upper case, But i prepare Upper Case.