Java/C2/do-while/Gujarati
From Script | Spoken-Tutorial
| Time' | Narration |
| 00:02 | Welcome to the spoken tutorial on While Loop in java. |
| 00:06 | In this tutorial, you will learn About the while loop.
How To use it. |
| 00:12 | For this tutorial we are using
Ubuntu 11.10, JDK 1.6 and Eclipse 3.7 |
| 00:21 | To follow this tutorial, you must have knowledge of relational operators in Java
|
| 00:26 | If not, for relevant tutorials please visit our website as shown. http://spoken-tutorial.org |
| 00:36 | Here is the structure of a while loop. |
| 00:39 | It has two parts. |
| 00:41 | One is the loop running condition and the second is the loop variable.
|
| 00:48 | Let us now look at an example. Switch to Eclipse |
| 00:55 | Here We have the eclipse IDE and the skeleton required for rest of the code
|
| 01:00 | I have created a class WhileDemo and added the main method to it. |
| 01:05 | We shall print numbers from 1 to 10 using a while loop. Type int n = 1
|
| 01:15 | This variable n is going to be our loop variable |
| 01:21 | type while 'in parenthesis n less than or equal to 10 open and close braces |
| 01:33 | This condition is called looping running condition.
|
| 01:37 | It means the loop will run as long as this condition is true.
|
| 01:42 | In our case, it will run as long as the value of n is less than or equal to 10.
|
| 01:47 | And it will stop only when the value of n become greater than 10. |
| 01:53 | Inside the loop, we shall print the value of n |
| 01:58 | System.out.println(n); and then increment n = n + 1;
|
| 02:12 | This way, first 1 is printed and then the value of n becomes 2. |
| 02:18 | Then the loop condition is checked.
|
| 02:21 | Since it is true. 2 is printed and n becomes 3. |
| 02:25 | And so on loop progresses untill 10 is printed after that n becomes 11 and the condition is not true and the loop will stop
|
| 02:37 | So Let us see the code in action. |
| 02:39 | Save and run. |
| 02:47 | As we can see, the numbers from 1 to 10 are printed.
|
| 02:52 | Now we shall print numbers from 50 to 40 |
| 02:58 | So We start with 50. Change n = 1 to n = 50 |
| 03:03 | And We go till 40.
|
| 03:05 | In other words as long as n is greater than or equal to 40. So change the condition to n greater than or equal to 40.
|
| 03:16 | And since we are looping from a bigger number to a smaller number, we have decrement the loop variable.
|
| 03:22 | So Change n=n + 1 to n=n - 1
|
| 03:27 | Save and run. As we can see, the the numbers 50 to 40 have been printed |
| 03:42 | Now we shall print the first 10 multiples of 7. |
| 03:48 | To do that, we start with 7
|
| 03:50 | So change n = 50 to n = 7and then end with 70.
|
| 03:57 | Change the condition to n less than equal to 70
|
| 04:03 | This way, we make sure the loop stops at 70. |
| 04:07 | To get the multiples, we shall increment the loop variable by 7.
|
| 04:12 | So, change n=n - 1 to n=n + 7
|
| 04:18 | This way first 7 is printed and then n becomes 14, 14 is printed and so on untill 70. Save and run |
| 04:33 | As we can see, the first 10 multiples of 7 are printed.
|
| 04:43 | We can also use a while loop to get the sum of digits of a number.
|
| 04:47 | Let us see how to do so. |
| 04:49 | First clear the main method.
|
| 04:54 | int n equal to 13876. This is the number |
| 05:02 | Then int dSum equal to 0The variable dsum with symbolic for digit sum will contain the sum of digits
|
| 05:18 | Type while, n greater than 0 open close parenthesis
|
| 05:27 | The reason for using this condition will be evident in a while. |
| 05:32 | To get the sum of digits, we must first get the digits. |
| 05:36 | To do that we use modulo operator.
|
| 05:40 | Type dSum = dSum + (n % 10) So we get the unit digit and add it to dsum |
| 05:5 | After that we remove the digit by dividing by 10. n = n / 10 |
| 06:08 | So when the loop is run for the first time, dSum will be 6 and n will be 1387. |
| 06:15 | And when the loop is run for the the second time, dSum will be sum of 7 and 6 which is 13, and n will become 138.
|
| 06:22 | So on, as the loop progresses, the digits will be removed from n and finally |
| 06:28 | n become zero.After that the condition n greater than 0 will be false and the loop will stop |
| 06:36 | So let us now add a print statement |
| 06:42 | System.out.println(dSum) |
| 06:51 | Let us see the code and action. Save and run |
| 06:59 | As we can see the sum of digit which is 25 has been printed |
| 07:06 | This way, a while loop, which is one of the most fundamental constructs in programming, can be used. |
| 07:16 | This brings us to the end of the tutorial.
|
| 07:20 | In this tutorial we have learnt
|
| 07:26 | As an assignment for this tutorial solve the following problem.
|
| 07:29 | Given a number, compute its reverse by using a while loop. Example: 19435 => 53491 |
| 07:37 | To know more about the Spoken Tutorial project, watch the video available at the following link, that summarizes the Spoken Tutorial project |
| 07:45 | If you do not have good bandwidth, you can download and watch it. |
| 07:50 | The Spoken Tutorial Project Team, Conducts workshops using spoken tutorials, Gives certificates for those who pass an online test |
| 07:57 | For more details, please write to contact AT spoken HYPHEN tutorial DOT org. |
| 08:03 | Spoken Tutorial Project is a part of the Talk to a Teacher project. |
| 08:07 | It is supported by the National Mission on Education through ICT, MHRD, Government of India. |
| 08:12 | More information on this Mission is available at the following link |
| 08:17 | This tutorial has been contributed by TalentSprint. Thanks for joining.
|
