Write assembly program using array with the contents given:

Question

5 5 5 5 5 5 5 5 5 5

Define 'Array' and 'RESULT'.

Code:
MOV AX, 0
MOV BX, 0

L1:
ADD AX, [ARRAY+BX]
ADD BX, 2
CMP BX, 20
JL L1

MOV [RESULT], AX

Answer 1

This assembly program calculates the sum of the elements in the 'ARRAY' and stores the result in 'RESULT'.

Here is the code:

```assembly
DATA SEGMENT
ARRAY DB 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
RESULT DW ?
DATA ENDS

CODE SEGMENT
ASSUME DS:DATA, CS:CODE

START:
MOV AX, 0
MOV BX, 0

L1:
ADD AX, WORD PTR [ARRAY+BX]
ADD BX, 2
CMP BX, 20
JL L1

MOV WORD PTR [RESULT], AX

; Exit program
MOV AH, 4Ch
INT 21h
CODE ENDS
END START
```

Explanation:
- The program uses the `DATA` segment to define 'ARRAY' and 'RESULT'.
- 'ARRAY' is defined as a byte array with the given contents: `5 5 5 5 5 5 5 5 5 5`.
- 'RESULT' is defined as a word variable, which will store the sum of the elements in 'ARRAY'.
- The `CODE` segment defines the main code section of the program.
- The `START` label marks the beginning of the program.
- Initialize `AX` and `BX` registers with 0, which will be used for calculations.
- The `L1` loop is used to iterate through the elements of 'ARRAY'.
- The element is accessed using the index `BX`, which starts at 0 and increments by 2 with each iteration since each element in 'ARRAY' is a 2-byte value.
- The element value is added to the `AX` register.
- The `BX` register is incremented by 2 to move to the next element.
- The loop continues until the `BX` register reaches 20, which is the size of 'ARRAY' in bytes (10 elements * 2 bytes each).
- After the loop, the sum in `AX` is stored in 'RESULT' using the `MOV` instruction.
- The program exits gracefully with an interrupt call (`INT 21h` and `AH = 4Ch`) to terminate the program.

Note: The program assumes an Intel x86 architecture and uses the DOS interrupt `INT 21h` for program exit.