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Reverse a number in PL/SQL
  • Difficulty Level : Medium
  • Last Updated : 20 Nov, 2019


Prerequisite PL/SQL introduction

In PL/SQL code groups of commands are arranged within a block. A block group related declarations or statements.
In declare part, we declare variables and between begin and end part, we perform the operations.

Explanation:
Consider the example, input = 12345.

Step 1 : mod(12345,10) = 5
rev:= 0*10 + 5 = 5
num = floor(12345/10) = 1234

Step 2 : mod(1234,10) = 4
rev:= 5*10 + 4 = 54
num = floor(1234/10) = 123



Step 3 : mod(123,10) = 3
rev:= 54*10 + 3 = 543
num = floor(123/10) = 12

Step 4 : mod(12,10) = 2
rev:= 543*10 + 2 = 5432
num = floor(12/10) = 1

Step 5 : mod(1,10) = 1
rev:= 5432*10 + 1 = 54321
num = floor(1/10) = 0
in step 5, num =0 which doesn’t satisfy the while condition and loop terminates.

rev = 54321

More examples:

Input : 123456
Output :654321
Input :87459
Output :95478

Below is the required implementation:

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SET SERVEROUTPUT ON;
DECLARE
-- declare a number 'num' for reading actual input
-- declare another number 'rev' that would be reverse of num
num NUMBER;
rev NUMBER;
  
BEGIN
-- & is used to read input from keyboard
num:=#
-- initialize rev to 0
rev:=0;
-- the loop runs until num is greater than 0
WHILE num>0 LOOP
-- mod function is used to find the modulus/ remainder of num when divided by 10
  
rev:=(rev*10) + mod(num,10);
-- floor function is used to obtain a result which is an integer
num:=floor(num/10);
END LOOP;
DBMS_OUTPUT.PUT_LINE('Reverse of the number is: ' || rev);
END;
/                        
  
-- Program End

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Output:

Enter value for num : 157439
Reverse of the number is: 934751

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