PL/SQL Program for Reverse of a Number

Writing a program to reverse a number (for example, turning 12345 into 54321) is a classic programming challenge. It's a great way to learn how to use loops and basic math functions like MOD and TRUNC to manipulate individual digits.

This simple guide will show you the logic and a complete PL/SQL program to solve this problem.

What You Need to Know

To write this program, you will use a few basic PL/SQL concepts:

1. Enabling Output: You must run this command once in your SQL tool to see the printed results:SET SERVEROUTPUT ON;
2. Anonymous Block: We will write our code in a DECLARE...BEGIN...END; block.
3. Variables: We'll need a variable for the number to reverse (v_num), a variable to hold the new reversed number (v_reverse), and a temporary variable to hold the last digit (v_remainder).
4. WHILE Loop: A WHILE loop is perfect for this. We will tell the program to keep running the loop while the original number is still greater than 0.
5. MOD(n, 10): This math function gives us the remainder of a division. When you divide by 10, the remainder is always the last digit (e.g., MOD(123, 10) is 3).
6. TRUNC(n / 10): This math function cuts off the decimal. When you divide a number by 10 and TRUNCate it, you effectively remove the last digit (e.g., TRUNC(123 / 10) is 12).

PL/SQL Program: Reverse a Number

This program will take the number stored in v_num, reverse its digits, and print the result.

PL/SQL Program

SET SERVEROUTPUT ON;

DECLARE
  -- The number we want to reverse
  v_num NUMBER := 12345;
  
  -- A variable to hold the reversed number as we build it
  v_reverse NUMBER := 0;
  
  -- A temporary variable to hold the last digit
  v_remainder NUMBER;

BEGIN

  -- Create a copy of the original number for the output message
  DBMS_OUTPUT.PUT_LINE('Original number: ' || v_num);

  -- Loop as long as the number is greater than 0
  WHILE v_num > 0 LOOP
    
    -- 1. Get the last digit
    -- Example: MOD(12345, 10) = 5
    v_remainder := MOD(v_num, 10);
    
    -- 2. Build the new reversed number
    -- Example: v_reverse = (0 * 10) + 5  => 5
    --          v_reverse = (5 * 10) + 4  => 54
    v_reverse := (v_reverse * 10) + v_remainder;
    
    -- 3. Remove the last digit from the original number
    -- Example: TRUNC(12345 / 10) = 1234
    v_num := TRUNC(v_num / 10);
    
  END LOOP;
  
  -- Print the final result
  DBMS_OUTPUT.PUT_LINE('Reversed number: ' || v_reverse);

END;
/

Result (for n := 12345)

Original number: 12345
Reversed number: 54321

Program Explanation

1. DECLARE section: We create v_num (12345), v_reverse (initialized to 0), and v_remainder.
2. BEGIN section: The logic starts. We print the original number.
3. WHILE v_num > 0 LOOP: The loop begins and will continue to run as long as v_num is positive.

Inside the loop:

• Loop 1:
  • v_num = 12345
  • v_remainder = MOD(12345, 10) = 5
  • v_reverse = (0 * 10) + 5 = 5
  • v_num = TRUNC(12345 / 10) = 1234 (Loop continues)
• Loop 2:
  • v_num = 1234
  • v_remainder = MOD(1234, 10) = 4
  • v_reverse = (5 * 10) + 4 = 54
  • v_num = TRUNC(1234 / 10) = 123 (Loop continues)
• Loop 3:
  • v_num = 123
  • v_remainder = MOD(123, 10) = 3
  • v_reverse = (54 * 10) + 3 = 543
  • v_num = TRUNC(123 / 10) = 12 (Loop continues)
• Loop 4:
  • v_num = 12
  • v_remainder = MOD(12, 10) = 2
  • v_reverse = (543 * 10) + 2 = 5432
  • v_num = TRUNC(12 / 10) = 1 (Loop continues)
• Loop 5:
  • v_num = 1
  • v_remainder = MOD(1, 10) = 1
  • v_reverse = (5432 * 10) + 1 = 54321
  • v_num = TRUNC(1 / 10) = 0 (Loop stops)

4. DBMS_OUTPUT.PUT_LINE(...): The program exits the loop and prints the final value of v_reverse, which is 54321.