vubon // For loop public static void main(String[] args) { Scanner scan = new Scanner(System.in); for(int i = 1; scan.hasNext()== true; i++){ System.out.println(i + " " + scan.nextLine()); } } // While loop System public static void main(String[] args) { Scanner scan = new Scanner(System.in); int i = 0; while(scan.hasNext()){ i++; System.out.println(i + " " + scan.nextLine()); } }Both Loop working for me
polmki99 Don't use for loops like that! :-)
Thriskel What do you mean?
thaophuong10498 can you explain why ?
VIPERCODER1 because in this case the time compkexity is more for the for loop that while loop
shareef_hiasat any link to explain/proof would be great.
sorathia_urvik91 Can you explain how complexity is differed between those loops ?? please
gustavocpassos 1 - It's a bad pattern to compare a boolean with true/false like that: "scan.hasNext() == true" Explanation: if the variable is a boolean, doesn't need "== true"
2 - If you have to do a loop that uses the return of a boolean method (like "scan.hasNext()"), prefer to use while. Explanation: Your code is more beautifull and readable. Clean code is a good practice.
Help this helps you
kawaljit005 Thank you
berkemuftuoglu1 Thanks that was really helpful!
thanhdaikhoa3289 ???
amar301189 hi, how do you terminate taking the input from the console.. It kept asking, i print several runs...how do you exactly make the scan.hasNext() condition false through the console
for eg:hello 1 hello hi 2 hi how 3 how how are you
it still taking the input and i want to terminate here.
Ignis_Qui_Vir Ctrl-Z works for me. But there is a bug in Eclipse: you have to switch the view, then go back to the console and then you can enter Ctrl-Z.
saurabhkumar8112 read my answer on the top
esma3eel97 where ??
berkemuftuoglu1 you can use a statment like if(name.equals(" ")) then break in which you can get out of the loop when you input space
gaurav21pendhari [deleted]gaurav21pendhari will you please explain
System.out.println(i + " " + scan.nextLine());I am new to java
polmki99 It's equivalent to this code:
String line = scan.nextLine(); System.out.println(i + " " + line);
vardhangaharwar replace System.out.println(i + " " + line); with System.out.println(i + " \t" + line);
vaibhavpkc it prints the current i value. And the Double inverted commas work as giving space between the value of i and the value of the scan object value that has been read by the compiler entered by the user.
sweta_jain it prints number(i) plus space(" ") plus the next line
eshandangwal9991 In Stdout first of all current value of i will be print than " " will genrate space and scan.nextLine() will print value read by compiler. For example if i=1 and complier reads input such as "Hello Eveyone !" so it will be printed like this: 1 Hello Eveyone !
namila007 how to terminate the loop?
polmki99 When scan.hasNext() == false.
neelay619 how can that condition be met in the given problem set?? please if anyone can explain.
tenzin392 I'm 20 days late, but the reason is: hasNext() checks if there is a string remaining in the Scanner; and returns false only when nothing is left there, which is equivalent to EOF
ace_dexter But the input is being given line by line. How can you say it will search for EOF? It keeps on asking for input!!!
Arafa I had also the same question, I could not find any easy explenation over net.
Freank_224 You can use a while loop to look for it
e.g.
while (scanner.hasNextLine())
If there is no next line it will terminate the loop
leospostbox I've done some research on this.
Basically, hasNext() will block until Standard input (stdin) is closed.
So when you run in an IDE, this loop never ends if you never kill stdin.
It looks like on Hackerrank, at the end of the file, stdin is being closed thus it ends.
You can also have some kind of 'END_OF_FILE' token to indicate end of input.
tsherisherpa if it reached the end of file then
jamaa_cool_afg I had the very same problem. I didn´t had the idea to put scanner.hasNext()==true in the termination condition.
SwordFish91 that's made wrong this is the right :
Scanner sc = new Scanner(System.in);
while (sc.hasNextLine()) { int x=1; System.out.println(x++ + " " +sc.nextLine()); }
CrossJack Scanner sc = new Scanner(System.in); int count=1; while (sc.hasNextLine()) { System.out.println(count++ + " " +sc.nextLine());
}claw10 @ohdoghdffdogdf, you need place int x=1 out of the while loop. Otherwise, it prints 1 everytime for each subsequent lines.
changhaoran144 that is: each time when you do the loop, you re-initialize x to 1
vikas_yadav_6710 int x=1; is use ouside the loop'
Coder_Zuki initialize x outside while loop man.
int x=1; while (sc.hasNextLine()) { System.out.println(x++ + " " +sc.nextLine()); }
urushakhan1997 But it is not passing all the testcases.
vvbhandare Does it mean like returning TRUE only for any keyboard value?
rcgonzalezf Your for loop can be simplified:
// For loop public static void main(String[] args) { Scanner scan = new Scanner(System.in); for(int i = 1; scan.hasNext(); i++){ System.out.println(i + " " + scan.nextLine()); } }
You don't have to use the == operator to compare with true since the hasNext method returns a boolean.
synhack In fact, you can shorten it even further (at a potential loss of readability) by including the print statement within the for loop:
for(int i = 1; scan.hasNext(); System.out.println(i + " " + scan.nextLine()), i++);
JChiquin Even shorter:
for(int i=1; scan.hasNext() ; System.out.println(i++ +" "+scan.nextLine()));
adarsh98 Good idea!
[deleted] That's pretty sweet! Thanks.
stelabr2 That's not very readable code! A
whileloop is more appropriate.tasfiahm Thanks for a different a view.
praseedakannuz11 [deleted]
quannv Oh...thats great idea =))
pandeyvirat12345 what scan.hasNext do?
Surya07081995 //shortest for(int i=1;in.hasNext() ;out.println(i++ +" "+in.nextLine()) );
nimish_gupta what does Scan.hasNext() exactly means??
And while in the for loop we are not comparing Scan.hasNext() ??
JeanDuarte U shall read Scan.hasNext like this.... does "Scan" have next element ? If it has, it returns TRUE, so while something is TRUE, keep it going. That's why you don't need any comparing.
Surya07081995 sorry for being late,scan.hasNext() means after getting one input sacanner will wait for the next input .
tasfiahm Thanks a lot for sharing your knowledge..
Nageswaran Thanq bro
plodder1317 vubon, Your loop is an infinte loop,I don't know how it works in Hackerrank, but it's not a correct way.
abir4u2011 It's not actually.When there is no more line of inputs the hasNext method returns false and the loop terminates.
ReusMandal what is scan.hasnext ? scan is your scanner name ? but what is the meaning of this ?
JeanDuarte That's it, scan is his scanner's name, that's the name he gave to the object created. Scanner x = new Scanner(System.in); ::: x is the scanner's name
vikaskumar_vsk Even if I dont put scan.hasNext == true; it works fine. Can you explain why?
Scanner scan = new Scanner(System.in); //for loop for(int i =1; scan.hasNext(); i++){ System.out.println(i + " " + scan.nextLine()); }
JChiquin Because, the for loop check if the condition is true but
scan.hasNext()is already boolean.- When scan.hasNext() return TRUE:
- TRUE == TRUE -> TRUE
- When scan.hasNext() return FALSE:
- FALSE == TRUE -> FALSE
So you see, it's not necesary compare with other boolean.
Other example: the method isEmpty(Returns true if this list contains no elements.), We will mostly need to know when the collection contains elements, so we use it this way:
if(!myCollection.isEmpty()){ //do something with the elements... } //Or we can compare with false if(myCollection.isEmpty()==false){ //do something with the elements... }
We simply deny it.
(Sorry my english)
vikaskumar_vsk Got it. Thank you.
johnmlhll Worked for me too..
Scanner input = new Scanner(System.in); int counter = 1; String line = ""; while(input.hasNext()) { line = String.format(counter+" "+input.nextLine()); counter++; System.out.println(line); }
prince8809 nice for loop ;-)
daothisi thanks!
esma3eel97 which part of this code reads from user input is it the "scan.hasNext()" or the "scan.nextLine()" ???
girish_yadawad scan.nextLine();
Sushant000786 can you explain this code ???
Sowmeya Read some unknown n lines of input from stdin(System.in) where is this used in your program
choudharymanisha Hey could you please help me with this problem... I am using a while loop just like you but in the output the Hello part is not being printed..
this is my code--> Scanner sc=new Scanner(System.in); String s=sc.next(); int i=1; while(sc.hasNextLine()) { String s1=sc.nextLine(); System.out.println(i+" "+s1); i++; } Expected Output--> 1 Hello world 2 I am a file 3 Read me until end-of-file. My output--> 1 world 2 I am a file 3 Read me until end-of-file.navneetlahiri line 2: String s=sc.next(); //this scans for "Hello" thats why you are not getting it in the output, try removing it.
Ziyad010 what is the alterinitaive?
cs_chanderkants1 Write String s= sc.nextLine(); not sc.next();
wizyash sc.next(); ----> searches for the first space in the input sc.nextLine(); --> looks for the line change or \n in the input
richardm96123 while(scan.hasNext()) should be the line you use. Also, don't use the strings. The only time String should be in the code is the very beginning.
Scanner scan = new Scanner(System.in);
int lineNum = 1; while(scan.hasNext()) { System.out.println(lineNum + " " + scan.nextLine()); lineNum++;This was my code that I used, and it works flawlessly.
meethparikh01 here how does it scan the input because here it first check whether it has input or not and than it prints the output or can you please explain how does it take input?
rithikacariappa How exactly is it accepting input?
lambert_jeanphi1 in the for loop, no need for scan.hasNext() == true, just scan.hasNext, like in your while loop
dimio Why "hasNext()", not "hasNextLine()"?
My solution with using "hasNextLine()" are working correctly or not?
public static void main(String[] args) { Scanner in = new Scanner(System.in); long i = 0; while ( in.hasNextLine() ){ System.out.println(++i + " " + in.nextLine()); } in.close(); }
richardm96123 the i++; has to be after the output println. Otherwise, its good.
chuucksc This particular example is not assuming that the "end-of-file" string will come at the latest line, therefore this solution will fail.
nishatfarheen3 [deleted]derashrichitran1 At first hasNext() is pointing to wempty string so how it is running
melobruk1 That will give a timeout error
alishaksp33 why we learn this? can you give me any example where this thing will be useful?
ratan_singh98 here is my solution
import java.io.*; import java.util.*; public class Solution { public static void main(String[] args) { Scanner in=new Scanner(System.in); String a; int i=1; while(in.hasNext()) { a=in.nextLine(); System.out.println(i+" "+a); i++; } } }
nooooooobie Initially 'in' will be empty right?Then,how will 'in.hasNext()' return true?
ratan_singh98 exactly, but it checks whether the next line is empty or not, not the current line.
sohal_sheetal My test case is failing since the input section is showing test with “{-truncated-}”. On downloading {truncated} does not display.
Code: public static void main(String[] args) {
Scanner in = new Scanner(System.in); for(int i=0; in.hasNext(); i++){ System.out.println(i+ " " +in.nextLine()); } }Input (stdin)Download Hello world
I am a file
Read me until end-of-file. {-truncated-}
Your Output (stdout) 0 Hello world 1 I am a file 2 Read me until end-of-file.
alkhandani I Have this problem too.
rgrochol i should start from 1, instead 0
shravankumar_ta1 import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { int i=1; Scanner sc = new Scanner(System.in); while(sc.hasNext()) { System.out.println(i++ +" "+sc.nextLine()); } } }
aaryanmukherjee1 Thanks bro
sossio_pascaXXX thanks bro
saranyaalluri03 import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;
public class Solution {
public static void main(String[] args) { Scanner sc=new Scanner( System.in); String a=sc.nextLine(); String b=sc.nextLine(); String c=sc.nextLine(); for(int i=1; sc.hasNext(); i++) { System.out.println(i+" " +sc.nextLine()); } }}
Sir, they said we have to take input. But you have written the code only for checking and printing. But where is the input that we should enter.
nooooooobie Initially 'sc' will be empty right?Then,how will 'sc.hasNext()' return true?
jvinamra776 More efficient then scanner and a bit faster although this is a small code so there will be not much big difference
public static void main(String[] args)throws IOException { int i=1; String str = ""; BufferedReader in = new BufferedReader(new InputStreamReader(System.in)); while((str=in.readLine())!=null){ System.out.println(i+" "+str); i++; } }
1ojusticeo1 Scanner Scan = new Scanner(System.in); int Count = 0; while(Scan.hasNext()) System.out.println(++Count + " " + Scan.nextLine() );
Neer1416 My Solution:
BufferedReader input = new BufferedReader(new InputStreamReader(System.in)); String str ; int n =0; while((str=input.readLine())!=null ) { n = n+1; System.out.println(n +" "+str); }
vivek5287445 Here is my solution:
import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */ Scanner sc = new Scanner(System.in); for (int i=1; sc.hasNext(); i++){ System.out.println(i+" "+sc.nextLine()); } } }
mishra_sarthak71 for(int i=1; scan.hasNext()==true; i++){
System.out.println(i + " " + scan.nextLine()); if(scan.hasNext()==false) System.exit(0); } #Generally System.exit(0) is used to terminate the JAVA Program
codebloode_123 public class Solution { public static void main(String[] args) { Scanner sc=new Scanner(System.in); int i=1; while(sc.hasNextLine()){ String s=sc.nextLine(); System.out.print(i+" "); System.out.print(s); System.out.println(); i++; } }}
codebloode_123 This work fine..............
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