StefanK Refer to this stackoverflow question if you'd like help understanding the difference between a static block and specific static initializers.
http://stackoverflow.com/questions/9379426/java-when-is-a-static-initialization-block-useful
I found this useful in comparison to the droves of wordy, example-less explanations (I'm looking at you, Oracle Docs.)
sanwong15 Clean and Clear. Perfect. Thanks. Yea, Oracle Docs has a lot to read.
rohillakriti21 Really helpful. Thanks.
nicolas_saavedr1 thanks.
Meaditya I generally avoid coming here to discussions as people post answers! But when all seemed lost and there was no hope left for humanity. You came in my friend! ALL HAil our lord and saviour Stefank!
Purvi97 thanku so much ..... it clears all doubt
rohitsawai2893 Thanks for this link. It helped me to understand this concept in more better way
bharadwaj_poduri thank u bro..now I got the point.
chjlei Thanks!It's very useful and helpful.
sahilmathurcs1 static Scanner in=new Scanner(System.in); static int B=in.nextInt(); static int H=in.nextInt(); static boolean flag=true;
why did we add static? couldn't get.I am new for Java.
k_saikrishna1998 becz since it is a static block all fields(variables) should be static it is similar to using static variables in static methods
vikas343430 thanks
anose001 That explains it.
arshia_362 cool
aakashmhankale1 static boolean flag = true; static int B,H;
static{
Scanner scan = new Scanner(System.in); B = scan.nextInt(); scan.nextLine(); H = scan.nextInt(); scan.close(); if((B<=0 && H<=0)||(B>=0 && H>=0)){ flag = true; }else if((B<=0 && H>=0)||(B>=0 && H<=0)){ flag=false; System.out.println("java.lang.Exception: Breadth and height must be positive"); }}
brandonzhang00 @above No that code is too complicated. The else if statement is not neccessary (just need an else statement). You just need to set flag = true when both are greater than 0. (Lengths can't be negative)
pbhanuchandra You have a error in else if() block
Here is the actual answer
static boolean flag = true; static int B,H; static { Scanner scan = new Scanner(System.in); B = scan.nextInt(); scan.nextLine(); H = scan.nextInt(); scan.close(); if((B<0 && H<0)&&(B>=0 && H>=0)) { flag = true; } else if((B<=0 && H<=0)||(B>=0 && H<=0)||(B<=0 && H>=0)) { flag=false; System.out.println("java.lang.Exception: Breadth and height must be positive"); } }
ashish_patel9981 [deleted]ashish_patel9981 your program is correct but how i am not able to understand the logic behind this if ((B<0 && H<0)&&(B>=0&&H>=0)) suppose :we will give input 1and 3 then it shoud go to false block because B<0 && H<0 =false B>=0&& H>=0=True false && true false should come
akashraj9598 It won't give the correct answer. you need to remove the less than condition from the first if. other than that there is no need of second if else only else is needed.
rohan_bhat72 can you explain how you directly used B,H as variables instead specifying any datatypes for that.
DEVA_ASURA He used int .Read carefully you can find it.
huzaif_sayed It was not passing two test cases so i did some modification
static boolean flag = true; static int B,H; static{ Scanner scan = new Scanner(System.in); B = scan.nextInt(); scan.nextLine(); H = scan.nextInt(); scan.close(); if(B>0 && H>0){ flag = true;
}else{ flag=false; System.out.println("java.lang.Exception: Breadth and height must be positive"); } }
negi_sapna02 thanks..this is giving correct answer. I was missing to update the flag as false in else loop.
moumitasinha17 static boolean flag = true; static int B,H;
static{
Scanner scan = new Scanner(System.in); B = scan.nextInt(); scan.nextLine(); H = scan.nextInt(); scan.close(); if(B>0 && H>0){ flag = true;
}else{ flag=false; System.out.println("java.lang.Exception: Breadth and height must be positive"); } }
muhammadafzal402 static boolean flag = true; static int B,H;
static{ Scanner s=new Scanner(System.in); B=s.nextInt(); s.nextLine(); H=s.nextInt(); s.close(); if(H<=0 || B<=0){ flag=false; System.out.println("java.lang.Exception: Breadth and height must be positive"); } }
manoj_manu static boolean flag = true; static int B,H; static{ Scanner scan = new Scanner(System.in); B = scan.nextInt(); H = scan.nextInt(); scan.close(); if(B<=0||H<=0){ flag=false; System.out.println("java.lang.Exception: Breadth and height must be positive"); }else flag=true; }
#this is the actual output your output fails in two of the given test casesdevendranath966 why we should keep in static block?
mraemiller What's missing from all these explanations is they all focus on "how it works" and not "why you would want to do it" which remains still partiallly a mystery to me. Apart from allowing one to run commands out of sequence - and why would you want to do that? - the best explanation of what it's for that I could find is...
"The advantage of private static methods is that they can be reused later if you need to reinitialize the class variable. So, you kind of get more flexibility with a private static method in comparison to the corresponding static initialization block. This should not mislead that a 'public' static method can't do the same. But, we are talking about a way of initializing a class variable and there is hardly any reason to make such a method 'public'. "
http://www.jusfortechies.com/java/core-java/static-blocks.php
oishik_raybarman [deleted]
Eg0Jemima Remember, you need to either use a try catch block or add "throws (Exception)" to a method name in order to throw a checked exception. In this case, there is no method name for the static initializer block, so you must use a try catch!
See code below:
MyCode{
static Scanner input = new Scanner(System.in); static boolean flag = true; static int B = input.nextInt(); static int H = input.nextInt(); static{ try{ if(B <= 0 || H <= 0){ flag = false; throw new Exception("Breadth and height must be positive"); } }catch(Exception e){ System.out.println(e); } }}
StefanK Great example! :)
palani_05 how multiplication result comes automatically?
piyusman static boolean flag=true;
static int B,H;
static {
Scanner scan = new Scanner(System.in); B = scan.nextInt(); H = scan.nextInt(); try{ if(B<=0 || H<=0){ flag = false; throw new Exception("Breadth and height must be positive"); } }catch(Exception e){ System.out.println(e); }}
kartheeksakella Instead of using or you can use multiplication i.e, if bh<0 the string is printed. This is better because, if we extend the code this format will be better to debug.
sachingoyal0510 we should not use bh<0 condition because if b and h both are negative, then your code will set flag as true which is incorrect
sanchitsrivasta1 we cant use this as if both b & h are negative the condition will get true which should not be happen
anilsingh22 
if both b & h are negative the condition will get false and it throw the exception
tarunbinwal805 why do we need flag in this to true can we use and operator insted of or .................................please help
tanweiyon I found twice repeating "java.lang.Exception:" in your code. Line 17 should be corrected as : throw new Exception("Breadth and height must be positive");
khushboo_bhagch1 Code is working but how you are calculating the area .
marupatlapranay Area is calculated in the main method. Expand the code and you will find it! :)
khushboo_bhagch1 thank you
prasannakumarko1 cool...
marupatlapranay Why can't we put boolean flag = true; and int B,H; in the static block??
Surya07081995 No You can't do that because we need to reference the variables in main method,as main method is non-editable.so the only option you have that while compiling, when JVM loads (.class) file in the memory intialize all those variables at the same time that can be done when you declare variables as static.
static boolean flag=true; static Scanner in=new Scanner(System.in); static int B=in.nextInt(); static int H=in.nextInt(); static { if(B>0 && H>0) flag=true; else { flag=false; System.out.println("java.lang.Exception: Breadth and height must be positive"); } }
Note: variables inside the static block is not static.
abishake1396 why are we using the boolean variable flag?
Kedharnath_Reddy alredy given flag in main function as a conditon for if so we should use flag
yogesh9661 how to access the main class??
palash_rawat Click on arrow you will find the full part of main class.
ruchipandit40326 i have used the same code, still getting errors...
.java:36: error: class, interface, or enum expected public static void main(String[] args){ ^ .java:39: error: class, interface, or enum expected System.out.print(area); ^ .java:40: error: class, interface, or enum expected } ^
I'm not getting this error...plz help me out.
Surya07081995 [deleted]Surya07081995 System.out.print(area); ?????
static boolean flag=true; static Scanner in=new Scanner(System.in); static int B=in.nextInt(); static int H=in.nextInt(); static { if(B>0 && H>0) flag=true; else { flag=false; System.out.println("java.lang.Exception: Breadth and height must be positive"); } }
paste the same code,complied and tested and let me know!
mayank7544 i did the same thing but got an error class,enum or interface expected.pls help me
ashish_patel9981 i also did exactly same but i am getting error
heenaverma66 Do not close class before main. There is closing parenthesis for class after main.
tpirramba Thanks a ton! It Worked for me.
kvkulakarrni9 its showing semicolan expected
prasannakumarko1 thank u
LucianoPe I tried doing it that way but it went wrong, so I had to take the:
Scanner scan = new Scanner(System.in); B = scan.nextInt(); H = scan.nextInt();
out of the block.
Why is it?
Surya07081995 No You can't do that because we need to reference the variables in main method,as main method is non-editable.so the only option you have that while compiling, when JVM loads (.class) file in the memory intialize all those variables at the same time that can be done when you declare variables as static.
LucianoPe Thanks, that helps a lot to understand how it works
nstelme if you hover you cursor at the bottom of the unmodifiable code, you'll see facing both left and right which you can expand. There you'll see the given multiplication
praveenics kindly open main().
jameydogom [deleted]
jagadeshanh Expant main method.
prunns It doesn't come automatically. The code for multiplication to print the area is written in the Hidden Stub in the main function. You can expand and see the hidden code by clicking on the arrow that appears near line number.
jbristow This problem does not require exceptions. It's actually way easier without them.
klkevelson Very true, but it asks us to display a message indicating that an exception has been thrown, which I think implies that we're supposed to use exceptions.
mpottsmail This problem would be way easier without the unrealistic restraints put on it. 6 years of programming and I've never once had to use a static block like this haha
gk01276602145 can you show me the way without exception?tks
yeyushengfan Like this:
public static int B = 1; public static int H = 1; public static boolean flag = true; static { Scanner scan = new Scanner(System.in); B = scan.nextInt(); H = scan.nextInt(); scan.close(); if (B <= 0 || H <= 0) { flag = false; System.out.println("java.lang.Exception: Breadth and height must be positive"); } }
X_coder1794 Your answer is fail to pass the 6th test case.!!
yeyushengfan Sorry. It should be "if (B <= 0 || H <= 0)". You can try again.
nupoor_1013 It is the same thing. And Answer comes by this.
catequil Test 6 is testing for the upper bound of < 100. If you look in the constraints you will see that the number entered must be greater than -100 and less than 100. The test is failing because it tests a height of 100. The crazy thing that is missing is they do not tell you what should display when this situation happens. But what it is expecting is the exception that you get when your height or breadth is negative.
Corner_Kick @catequil Thanks, you are right. But the constrains are actually: -100 <= B <= 100 -100 <= H <= 100
So the 100 should be included. I guess it's a mistake that Testcase 6 fails, when you take the input 100 for valid.
rohitsawai2893 No need to initialize variables B, H and flag.
nupoor_1013 I think we have to initialize flag...
nupoor_1013 thanks for the solution...also..no need to intialise....
daiict_201501067 [deleted]
chetan_ce99 what is the use of flag over here??
123shuklaprasha1 public static int B = 1; public static int H = 1; public static boolean flag = true; static{ }
"can u explain me plzz why have used this?????"
getsadzeg Very true.
palani_05 how multiplication result comes automatically?
Eg0Jemima There is code they give you behind the scenes. If you look before and after the section you can type in your code, there is a Solution class and a main they give you. You can expand these to see their code by clicking the double sided blue arrow on the display.
Hope this helps
Tushcr7 But why use try-catch?
hsinghal26 [deleted]
KudosSan Understood the problem but I felt like it lack explanition. Thank you for the example! Now I understand it clearly. I probably would've gotten it if I had known you could expand the code by clicking on the blue arrows.
dpak211211 why do we use static in every line??
tuyendev because main method is static ? See more : What is static keyword ?
subh_007 [deleted]subh_007 because we are not allowed to add code to the main method, and static blocks is a way in which we can run our code before main() executes and to modify variable through static block we need to make them static because static block cannot access non-static context directly.
bsaharsh85 thanks,this cleared my doubt!!!!
kosuke_ueki91 Mine is much simpler. I just print the string literally without throwing any exception XD. Because it doesn't make sense to throw an exception on a static block. It should be either in the constructor or in a method.
I'm already an experienced developer and I found this while looking places to practice my coding skills. These challenges are too easy for me XD. But they are good to remember the basics.
I've been coding in Python for a year or so. During this period I almost didn't touched Java. That's why I wanted to practice it a little bit.
Sorry, this went totally out of topic... I just felt I had to say it.
subh_007 Can you please send me the link where I can learn about throwing exception in static block is no good..
kosuke_ueki91 [deleted]tuyendev i use try-with-resources to create new Scanner. ( like this try (Scanner scanner = new Scanner(System.in)) ) because Scanner may be need close for good practice .
Gautam0709 why we need static here
tuyendev Read more : What is static keyword ?
prasantabnrg4 what is the use of flag here?
viktor_kirjanov [deleted]
prasantabnrg4 how is this code giving the result of the parallelogram?
tuyendev Do you want the code what can check it's a parallelogram ?
Psycologus I thought static block was mainly used to initialize variables. It gives error if I add declaration in static block. Any Suggestions?
alexgiby good one. help to identify improve on my sloution
kwhwj Thank you for helping.
blandforrestmail Thanks, that helped out a lot. I did mine differently but I didn't know you had to define the variables outside of the static block for them to be static.
lipsachhotray014 what if we don't use static flag variable?
shry39412 Nice solution! Please elaborate the use of 'throw' keyword here..
vasukapil2196 we are declaring flag as static boolean flag=true. As it is static so it shouldn't make changes but when inside the static block we give flag=false,the flag becomes false.Why?
aks130013 but its not running
jissie I don't think the try catch is necessary. Just use println is OK;
//Write your code here static boolean flag = true; static int B, H; static { Scanner scanner = new Scanner(System.in); B = scanner.nextInt(); scanner.nextLine(); H = scanner.nextInt(); scanner.close(); if (B <= 0 || H <=0){ flag = false; System.out.println("java.lang.Exception: Breadth and height must be positive"); } else { //System.out.println(B * H); } }
malathisakamuri1 [deleted]
bira_999_br bro where is the product (area of parallelogram ) shown
jissie The area is printed in the main method. The flag variable is uesd to control the output of the area.
NandagiriAditya can some body explain the role of flag in the above example ?? why do we need a flag , what does it do?i tried the code without the flag but it dosent work!please help
just_1_beginer see the main method There it's written that print B*H (if and only if flag ==true) And since we know that static block gets initialised as soon a s the class is loaded, So, as soon as the class is loaded and the values of B and H are loaded it's checked that is flag still equal to true
I HOPE THAT YOU GOT IT NOW But still if you have any doubt feel free to ask
NandagiriAditya yaa ..thank you very much brother ..
danagul_kz1 [deleted]huyen_nguyen_011 Why do you need to use static for Scanner, boolean, and int?
hiteshkrsahu A shorter approach
private static Scanner scan = new Scanner(System.in); private static boolean flag = false; private static int B = scan.nextInt(); private static int H = scan.nextInt();
static{
flag = (B>0 && H>0); if(!flag) {System.out.print("java.lang.Exception: Breadth and height must be positive");}}
developersanjeev public static Scanner sc = new Scanner(System.in); public static int B = sc.nextInt(); public static int H = sc.nextInt(); public static boolean flag = init2(); private static boolean init2(){ if(B<=0 || H<=0){ System.out.println("java.lang.Exception: Breadth and height must be positive"); return false; } return true; }
Without try-catch block
shivanibairagi10 static int B; static int H; static boolean flag=true; static { Scanner sc=new Scanner(System.in); B=sc.nextInt(); sc.nextLine(); H=sc.nextInt(); sc.close(); while(H<=0 || B<=0) { flag = false;
System.out.println("java.lang.Exception: Breadth and height must be positive"); break; }}
This is more easy and working well.
sureshdevaraman1 why only try and catch blocks are used why we cant use if-else block can u plzz explain ??
developersanjeev I have used if else block ^^
sureshdevaraman1 okay
Gajukollur Why u used flag ?
Gajukollur what is the use of flag? not used any where
petroskosmas88 well defined and easy to understand try - catch, code. Thanks!
kajulnisha_15cs thanks , i am now clear with our logic
devendranath966 can u please explain why we should keep static for all the code?
dchagoya1 yesssss a throw instead of a brainless print statment :9
kannanbaskaran10 What a horrible program is this!!!!!!!
ionakathryn123 Can someone please explain to me why we have to use static initalizer? Why not just do this logic in the main class?
Thanks
mssvai The exercise calls "Java Static Initializer Block" for a reason. =) To teach how to use Static Initializer Block. Sure it's easy to do the same thing in the main class.
jissie I think the way should be used when you wan't make sure the code is called before any other. And the code will not relay on the current coder to invoke.
LoWesT In my opinion there should be ab better indicator in the description, that the main method contains valuable information like the "flag" variable. Didn´t notice it until the first error, which could be very confusing for new programmers
cloudcityinvest1 System.out... is not an exception! That is a string output to console.
sam_som7007 Code should be this only. If any one of you have used better optimization technique then please do share.
static boolean flag=false; static Scanner sc = new Scanner(System.in); static int B = sc.nextInt(); static int H = sc.nextInt(); static{ try{ if(B> 0 && H > 0) flag=true; else throw new Exception("Breadth and height must be positive"); } catch(Exception e){ System.out.print(e); } }
alex_dkk You can keep the end as simple as: if(!(flag = B > 0 && H > 0)) System.out.println("java.lang.Exception: Breadth and height must be positive");
Since you want to keep it simple lol.
Btw I don't recommend caching an object such as Scanner like that. Its a habit habit for big apps. It won't be deleted from memory that way even thought its not used anywhere else.
arorabhavya23 this could also be tried to simplify a bit the above code.
static boolean flag = true;
static Scanner sc = new Scanner(System.in);
static int B = sc.nextInt();
static int H = sc.nextInt();
static{
if(B<=0 || H<=0){ flag = false; System.out.println("java.lang.Exception: Breadth and height must be positive"); }}
avnishdixit48 but the output for negative integer is wrong
gaboritza Nope. I solved it 99% the same and tested with negative values as well. It also passes the Haker Rank tests. What do you mean it's wrong?
obed_mx I liked you actually throw the exception :)
sandro101 Not worth spending the time even looking at this problem, everyone would spend their time better just looking up static initializers and statics in general. This question does nothing but confuse.
austinoyugi static boolean flag = false; static int B; static int H; static { Scanner input = new Scanner(System.in); B = input.nextInt(); H = input.nextInt(); try { if(B <= 0 || H <= 0) { throw new Exception("java.lang.Exception: Breadth and height must be positive"); } else { flag = true; } } catch(Exception e) { System.out.println(e.getMessage()); } }
garciart Hi austinoyugi! Nice. I used
throw new java.lang.Exception("Breadth and height must be positive");and
catch (java.lang.Exception e) { System.out.println(e.toString()); }
abhiapg11 static means scope is through out whole program so we declare flag , B and H as static and get User in put in B and H
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