Converting an All Upper Case String to an All Lower Case String

The concept of the code is to read each character from the memory allocation and converting one by one into it equivalent lower case character. Since the difference between a Lower Case and an Upper Case letter is equivalent to 20H, and since the Upper Case has a lower value that the Lower Case, to convert to lower case, we add the value of the Upper Case Letter to 20H. To traverse through the string, we start at index 2 up to the last. the CL holds the length of the string so that it is possible to determine how many times the process will repeat. We also use the Stack Index to hold the index of the character we are looking for, so everytime we repeat the loop, we increment the SI to proceed with the next index.

Why didn't we use the CX? It is because, the Counter Register counts backwards. Meaning, if CX=09H, it counts from 09H to 01H.

Here is the code.

Data Segment

   STR1 DB 11, 12 DUP('$')

  STR2 DB 0AH, 0DH, '$'

Code Segment

MOV AX, dseg 

MOV DS, AX

LEA DX, STR1 

MOV AH, 0AH

INT 21H

 

LEA DX, STR2

MOV AH, 09H

INT 21H

LEA DX, STR1

MOV SI, DX ; asign the si with DX since it already holds the address of str1

ADD SI, 02H ; increment 2H to start at the third element

MOV CL, STR1[1] ; assign CL with the length of the string

A: MOV AH, 02H

MOV DL, [SI] ; get each character based on SI

ADD DL, 20H ; convert each character

INT 21H

INC SI ; increment the index

LOOP A ; repeat the process

MOV AH, 4CH   

INT 21H