Discussion :: Flow Control - Finding the output (Q.No.3)
| Sanjay said: (Jun 23, 2011) | |
| How 2 is printed in third iteration, ans supposed to be 21012 ? | |
| Xyz said: (Aug 23, 2011) | |
| How x-2 matches in the first iteration? I didn't understand. | |
| Derya said: (Dec 13, 2011) | |
| It doesn't, does it! I think this is wrong. case x: System.out.print("0 "); x=2 case x-1: System.out.print("1 "); x=1 case x-2: System.out.print("2 "); x=0 z has 3 values 1,2,3 at each iteration of the loop, so it should only print 1 and 0 |
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| Anjena said: (Jan 14, 2012) | |
| Can you explain about final keyword? | |
| Kundan said: (Feb 29, 2012) | |
| Final keyword used in java to declare a variable as a constant. Value of these types of variable can, t change during run time. | |
| Hemavathi said: (Mar 27, 2012) | |
| I don't understand the output can anyone please explain this? | |
| Raju said: (Apr 27, 2012) | |
| Friends , If we didn't use break keyword after case statement the next Case also automatically printed In first case case3 satisfies so o/p=2 In second case case2 satisfies so o/p=1 2 In third case case3 satisfies so o/p= 0 1 2 Final output is 212012 |
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| Priya said: (Sep 7, 2012) | |
| Great! I understand Raju. Thanks! | |
| Ram said: (Aug 27, 2013) | |
| Is it necessary to declare x as final? | |
| Singh Sahaab said: (Jan 15, 2014) | |
| The answer is option D because there is no break statement encountered between cases. If you'll don't believe then code and run it. | |
| Prashant Chittam said: (Jan 28, 2014) | |
| For those guys who don't understand the code. Why 3rd condition met.. x is static and its 2. So 2-2 is the first condition for switch and for loop z=0 is true which is less than 3. So 2 is getting print. Now z++ is increased to 1 so the value of z=1. Again 2-1 is the second condition for switch, and for loop z=1 is true which is less than 3. So 1 is getting print...with 2 because after printing 1 ..there is no break so that it continue with print 2. Now the last ...z++ is increased to 2 so the value of z=2. Which is our static final variable which satisfies case 1. So it prints 0 with 1 and 2. Because again there is no break so that is continues printing up to end of switch block. Read carefully. |
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| Ali said: (May 22, 2014) | |
| Raju you explained very well and easily. Prasanth x is static but in this time final is important any way you also explained well. Thanks all of you. |
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| Anshu said: (Aug 6, 2014) | |
| Why we check the conditions from last but not from first? | |
| Hardik Patel said: (May 23, 2015) | |
| But you have passed z as an argument in switch case, and in the case you are doing simple function irrespective of z. I don't understand it what's the work of z in x, x-1, x-2. |
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| Sekhar said: (Jun 2, 2015) | |
| Friends am really not getting the point could please explain any one I need a answer regarding this question? | |
| Vijay Yadav said: (Jul 9, 2015) | |
| Friends x is declared as final so value of x never change please give me correct answer. I haven't satisfy by anyone answer. | |
| Rishanth said: (Oct 18, 2015) | |
| If we do not have a break instruction, the next instructions in the loop gets executed! | |
| Rahul said: (Mar 26, 2016) | |
| Nice programming question. | |
| Jayashree said: (Jul 24, 2016) | |
| Guys! what is the first condition? What is the connection between loop, x and also cases? |
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| Rimsha Fazal said: (Dec 28, 2016) | |
| I didn't understand how the conditions are matching in the program. There are simple values given by z in every iteration which is executed and printed what's the roll of x and y then? | |
| Shital said: (Jul 1, 2017) | |
| Sorry but no one explains about final keyword because x is declared as final then how it's value will change. | |
| Bagsari said: (Nov 18, 2017) | |
| I did not understand how final keyword works and how the control works on the switch statement? | |
| Yeshi said: (May 15, 2018) | |
| Hi. 1. The Final Keyword is like constant keyword in C, we can not re-assign the value once it is declared. 2. As per switch statement, when the value of z is 0 than the case with 0 will get executed. ( switch statement compares the value of a variable to the values specified in case of statements then first). So first time 0 is printed. Similarly for other scenarios. 3.Y is just initialized but not used just to make the program look complex. |
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| Khuddus said: (Jul 23, 2019) | |
| Remember, once a match is found all remaining statements are executed until a break statement is encountered. | |
| Vasavi said: (Jul 27, 2019) | |
| Can anyone explain it clearly? | |
| Rhin said: (Sep 4, 2021) | |
| Let me explain, public class Switch2 { final static short x = 2;//it will never change the value of x public static int y = 0; public static void main(String [] args) { for (int z=0; z < 3; z++) { switch (z) { case x: System.out.print("0 "); case x-1: System.out.print("1 "); case x-2: System.out.print("2 "); } } } } Iteraton->1 :z=0->switch(0) goes to case (x-2)-> case(2-2)->case(0): So 2 will be printed . ====================== Iteration->2 z gets incremented ->z=1->switch(1) goes to case(x-1)->case(2-1) ->case(1): prints 1 and there is no break statement so goes to next print statement prints 2. ====================== Iteration->3 z gets incremented ->z=2 ->switch(2) goes to case(x)->case(2)->case(2): prints 0 and there is no break statement so 1 and 2 will gets printed ANd the final output will be->2 1 2 0 1 2. |
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