error in taking input string after integer in java

Question

I was trying to do this :

int n = myScanner.nextInt();
for(int i=0;i<n;i++){
   String str = myScanner.nextLine();
   .
   .
   .
}

when i compiled it shows some errors java.util.Scanner.nextInt(Scanner.java:2117). initially i thought it is a problem of nextLine() so i used next() . then i found out if i add myScanner.nextLine() after taking input in n i.e

    int n = myScanner.nextInt();
    myScanner.nextLine();

Then it worked fine. I want to know why this happened?

Take a look at stackoverflow.com/questions/13102045/… for the explanation. — Codebender
– Codebender, Commented Aug 25, 2015 at 6:36
@Codebender I visited the above link. In the solution,exception handling is done while using Integer.parseInt(), But when I used parseInt, it does not throw any exception. why is it so? — hermit
– hermit, Commented Aug 25, 2015 at 6:58
@hermit, a NumberFormatException will be thrown only if it cannot be parsed. But a NumberFormatExcetion is an unchecked (extending RuntimeException) exception, so you don't have to explicitely write throws or handle it (though it will be thrown up if you dont). Hope that's clear. — Codebender
– Codebender, Commented Aug 25, 2015 at 7:04
@Codebender I thought trying to parse the newline character into integer value would throw an exception. — hermit
– hermit, Commented Aug 25, 2015 at 7:06

moffeltje · Accepted Answer · 2015-08-25 06:40:14Z

3

You need to consume the newline character you enter when passing the integer:

int n = myScanner.nextInt(); //gets only integers, no newline
myScanner.nextLine(); //reads the newline
String str;
for(int i=0;i<n;i++){
   str = myScanner.nextLine(); //reads the next input with newline
   .
   .
   .
}

answered Aug 25, 2015 at 6:40

moffeltje

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Comments

Shah Rukh Qasim · Accepted Answer · 2015-08-25 06:44:34Z

2

There is a newline character left in the stream. Use the code by @moffeltje or probably try this:

int n = Integer.parseInt(myScanner.nextLine());
for(int i=0;i<n;i++){
   String str = myScanner.nextLine();
   .
   .
   .
}

answered Aug 25, 2015 at 6:44

Shah Rukh Qasim

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3 Comments

hermit Over a year ago

You have to handle the NumberFormatException if you are going to use this method because the parseInt will try to parse the newline character into a integer value

Codebender Over a year ago

@hermit, Scanner.nextLine() will consume the new line character, but it wont be returned to the caller. So the newline character is not going to be parsed.

Codebender Over a year ago

From the Scanner java doc, This method returns the rest of the current line, excluding any line separator at the end. The position is set to the beginning of the next line.

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