Difference between revisions of "Java/C2/while-loop/English-timed"
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| − | | In this tutorial, you will learn: * | + | | In this tutorial, you will learn: |
| − | * | + | * About the '''while loop''' |
| + | * How to use it. | ||
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Revision as of 23:18, 25 March 2015
| Time | Narration |
| 00:02 | Welcome to the spoken tutorial on While Loop in java. |
| 00:06 | In this tutorial, you will learn:
|
| 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. [1] |
| 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' becomes 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 until 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 decremented the loop variable. |
| 03:22 | So Change n=n+1 to n=n-1. |
| 03:27 | Save and run. As we can see, 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 = 7 and 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 until 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 0 The 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. |
| 05:52 | 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 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 becomes 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. |
