java.util.InputMismatchException

[Roopam Samal]

import java.util.Scanner;
public class test12
{
public static void main(String[] args)
{
Scanner c=new Scanner("Sue,5true,3");
c.useDelimiter(",");
int  i=c.nextInt();
System.out.println(i);
}
}

here the output is java.util.InputMismatchException

what I have understood is if useDelimiter(",") were not used here then as default delimiter is space and jvm doesn't find any space in given scanner input and as a result of which output is InputMismatchException
If I'm wrong please correct me.
can you please highlight the reason of this exception in my code??
also explain when and why this exception comes if possible with examples

thanks in advance

[Joel Christophel]

You're correctly setting the delimiter to a comma. The problem is in line 8 where you try to use the nextInt() method. "Sue" is not an int.

[Roopam Samal]

Hello Joel

If I change my code to

import java.util.Scanner;
public class test12
{
public static void main(String[] args)
{
Scanner c=new Scanner("Sue,5,true,3");
c.useDelimiter(",");
int  i=c.nextInt();
System.out.println(i);
}
}

then also I'm getting same InputMismatchException.
what i know is nextxxx() get next token and then move to next token.
so 1st token Sue isn't matched then next token 5 is matched thus output should be 5 from my perspective but it doesn't. can't understand why???
please explain it


thanks

[Campbell Ritchie]

No, not at all.
Read the documentation, which you should always do whenever you use an unfamiliar piece of code. It assume that the next token is an int. You can get rid of tokens with a while loop and the hasNextInt method.

[Roopam Samal]

thanks Ritchi. I understood now.
If next token doesn't match Integer regular expression or out of range then InputMismatchException occurs
and here next token is "sue" which doesn't match Integer regular expression so this exception occured

And if I modify the above code to

import java.util.Scanner;
public class test12
{
public static void main(String[] args)
{
Scanner c=new Scanner("5,sue,true,3");
c.useDelimiter(",");
int  i=c.nextInt();
System.out.println(i);
}
}

then output is 5 since next token matches Integer regular Expression.

[Campbell Ritchie]

Yes, that will work (), but you can use a while loop as I mentioned earlier to discard any tokens before the 5. Try that.

[Winston Gutkowski]

Roopam Samal wrote:If next token doesn't match Integer regular expression or out of range then InputMismatchException occurs
and here next token is "sue" which doesn't match Integer regular expression so this exception occured...

Is there any particular reason you're using Scanner here?
String[] tokens = "5,sue,true,3".split(",");
would work just as well, and then:
1. You have all the tokens in your string, rather than having to work through sequentially to get each one.
2. All those tokens are Strings, and now you can worry about converting them.

Personally, I don't particularly like Scanner - mainly because I don't think it achieves what it set out to do (make getting different types, especially numbers, from Strings and files simpler) - but I know that Campbell and I disagree on this point.

However, that said, don't let me stop you from "discovering" Scanner. That's what programming's all about.

Winston

[Campbell Ritchie]

The problem is that Scanner can find an empty String unexpectedly. You will find details here. The books I have read, and the Java® Tutorials, do not cover that particular pitfall at all well.
Have you made any progress with discarding tokens?

[Roopam Samal]

Hi Ritchi
I have understood how to discard tokens.

import java.util.Scanner;
public class test2
{
public static void main(String[] args)
{
Scanner s=new Scanner("sue,5,rrr,3");
s.useDelimiter(",");
while(s.hasNext())
{
if(s.hasNextInt())
System.out.print(s.nextInt()+" ");
else 
s.next();
}
}
}

If any other better way is available then please tell me.


thanks

[Campbell Ritchie]

Far better way to do it, which Rob Spoor taught me:-

int i;
while (!myScanner.hasNextInt())
{
   myScanner.next();
   System.err.print("That wasn\u2019t an int: please try again: ");
}
i = myScanner.nextInt();