Java/C2/while-loop/English-timed
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
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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.
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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
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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
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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.
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01:37 | It means the loop will run as long as this condition is true.
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01:42 | In our case, it will run as long as the value of n is less than or equal to 10.
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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;
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02:12 | This way, first 1 is printed and then the value of n becomes 2. |
02:18 | Then the loop condition is checked.
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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
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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.
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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.
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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.
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03:16 | And since we are looping from a bigger number to a smaller number, we have decrement the loop variable.
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03:22 | So Change n=n + 1 to n=n - 1
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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
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03:50 | So change n = 50 to n = 7and then end with 70.
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03:57 | Change the condition to n less than equal to 70
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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.
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04:12 | So, change n=n - 1 to n=n + 7
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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.
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04:43 | We can also use a while loop to get the sum of digits of a number.
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04:47 | Let us see how to do so. |
04:49 | First clear the main method.
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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
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05:18 | Type while, n greater than 0 open close parenthesis
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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.
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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.
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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.
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07:20 | In this tutorial we have learnt
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07:26 | As an assignment for this tutorial solve the following problem.
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07:29 | Given a number, compute its reverse by using a while loop. Example: 19435 => 53491 |
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08:17 | This tutorial has been contributed by TalentSprint. Thanks for joining.
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