What is the operation performed by the following assembly language program of 8051?

             CLR        A

             MOV       R1, # 100H

             MOV       R7, # 21H

AGAIN: MOV      @ R1, A

             INC         R1

             DJNC      R7, AGAIN

 Assembly program of 8051:

8051 microcontroller is the CISC based Harvard Architecture and it has peripherals like 32 I/O, timers/counters, serial communication and memories. 

The microcontroller requires a program to perform the operations that require a memory for saving and to read the functions. The 8051 microcontroller consists of RAM and ROM memories to store instructions.

 Step 1: CLR A: It clears the Accumulator value and make all bits set to 0. [A= 0]

 Step 2: MOV  R1, # 100H : The # symbol represents the value being data. This instruction means the value 100H is moved to the register R1.

 [ R1=100]

 Step 3: MOV R7, # 21H : This instruction means the value 21H is moved to the register R7. [ R7=100]

 Step 4: AGAIN: MOV @ R1, A

 Step 5:INC R1

 Step 6: DJNC R7, AGAIN

 In the above instructions, MOV @ R1, A, will copy the contents of memory whose address is in R1 to the accumulator.

INC R1 : it increments the register value by 1.

DJNC R7: It decrements the register R7 by one and jump to the step 4 if the value in R7 is not zero and the loop continues until the value in the R7 decrements to Zero.

From the above mentioned steps, it is clear that the above program clears the 21 RAM locations starting from address 100 H.

Hence option 1 is correct.