Largest and Smallest :- Store 5 Bytes from location 30h ot 34h find the largest and smallest number and store at 40h and 41h respectively

Explanation

This code is written in 8051 Assembly Language. Its primary purpose is to find both the largest and the smallest numbers from a set of five numbers stored in memory.

The program works by using a "running comparison" logic:

1. Initialization: It points to the start of the data (memory address 30H). It takes the very first number and assumes it is both the largest (stored at 40H) and the smallest (stored at 41H).
2. Iteration: It then moves to the next memory location and enters a loop that runs four times (to check the remaining four numbers).
3. Comparison for Largest: For every new number, the program compares it with the current "largest" value stored at 40H. If the new number is greater, the value at 40H is updated.
4. Comparison for Smallest: Similarly, it compares the same number with the current "smallest" value stored at 41H. If the new number is smaller, the value at 41H is updated.
5. Completion: After checking all numbers, the final largest value is found at memory address 40H and the smallest at 41H.

Algorithm

1. Initialize register R0 with the starting memory address of the data (30H).
2. Set a loop counter R2 to 4 (since the first number is handled separately, 4 numbers remain).
3. Load the first number from the memory location pointed to by R0 into the Accumulator.
4. Store this first number into memory location 40H (Initial Largest) and 41H (Initial Smallest).
5. Increment R0 to point to the next data byte.
6. Start Loop:
   • Load the next number into the Accumulator.
   • Check Largest: Compare the Accumulator with the value in 40H.
     • If the Accumulator is greater, move the Accumulator value into 40H.
   • Check Smallest: Compare the Accumulator with the value in 41H.
     • If the Accumulator is smaller, move the Accumulator value into 41H.
   • Increment R0 to point to the next number.
   • Decrement the loop counter R2. If R2 is not zero, repeat the loop.
7. End the program.