Assembly 8086 program

Question

I am new to assembly language and this is some code that I didn't understand hoping that someone would help with it.

DATA SEGMENT
     VALUES DB 1,2,3,4,5,6,7,8,9
     ITEM DB 6
DATA ENDS
CODE SEGMENT
     ASSUME CS:CODE,DS:DATA
START: MOV AX,DATA
       MOV DX,AX
       LEA SI,VALUES
       MOV AL,ITEM
COMPARE: CMP AL,[SI]
         JZ Found
         INC SI
         LOOP COMPARE
         CLC
         JMP EXIT
Found: STC
EXIT: MOV AH,4CH
      INT 21H
      ENDS
CODE END

This program is supposed to look for number (6) among 1,2,3,4,5,6,7,8,9

I understand how it works in general but I have a few questions:

1. Why did we use CLC and STC ?? I know they put CF into zero and one but why do we use it??
2. Why did we use MOV AH,4CH in after the EXIT label??

3. After the start label why did we say the following:

   MOV AX,DATA
   MOV DX,AX

Why didn't we just say: MOV DX,DATA

Lastly, Could someone suggest a good book to learn assembly??

Solution 1

MOV AX,DATA

MOV DX,AX

This is wrong.

Correct answer is:

MOV AX,DATA   

MOV DS,AX

we cant send data from memory to segment registers(DS) directly . so we are sending through General purpose registers(AX).

Solution 2

The algorithm searches a number in a list of numbers.

If it is found, CF is set. If it is not found CF is cleared.

INT 21H is the MS-DOS service interupt. Function 4Ch ends the program with an error code in AL (which contains the number to be found).

DATA SEGMENT
     VALUES DB 1,2,3,4,5,6,7,8,9
     ITEM DB 6
DATA ENDS
CODE SEGMENT
     ASSUME CS:CODE,DS:DATA
START: MOV AX,DATA
       MOV DX,AX
       LEA SI,VALUES           ; DS:SI points to the VALUES structure
       MOV AL,ITEM
COMPARE: CMP AL,[SI]           ; Compare with number in list
         JZ Found              ; Jump to Found if equal
         INC SI                ; Try next
         LOOP COMPARE          ;
         CLC                   ; Clear CF (not found)
         JMP EXIT              ; Quit
Found: STC                     ; Set CF (found)
EXIT: MOV AH,4CH               ; End program with error code AL = 6.
      INT 21H
      ENDS
CODE END

Solution 3

The LOOP instruction is curious here. This instruction decrements CX and jumps only if CX is not zero. This means that the loop runs CX times, but CX is never set by the program.

CX is likely zero on entry, and the first decrement will make it 65535, so it will actually loop a maximum of 65536 times, searching past the end of the list if the element is not found.

To make it correct, add MOV CX, ITEM - VALUES before the loop start. Since ITEM comes right after VALUES, subtracting their addresses will give the number of bytes (elements) in the list.

Often a label is added to the end of a list to make such calculations more robust.

         VALUES     DB 1,2,3,4,5,6,7,8,9
         VALUES_END LABEL BYTE
         ; ...

         MOV CX, VALUES_END - VALUES
COMPARE: ; ...
         LOOP COMPARE