confused about Scanner class

[Faisal Ansari]

this code is from the K & B scjp 6 book(pg 505)

  
import java.util.Scanner;

class ScanNext
{
	public static void main(String[] args)
	{
		boolean b2, b;
		int i;
		String s, hits = " ";

		Scanner s1 = new Scanner(args[0]);
		Scanner s2 = new Scanner(args[0]);

		while (b = s1.hasNext())
		{
			s = s1.next();
			hits += "s";
		}

		while (b = s2.hasNext())
		{	
			if (s2.hasNextInt())
			{
				i = s2.nextInt();
				hits += "i";
			}
			else if (s2.hasNextBoolean())
			{
				b2 = s2.nextBoolean();
				hits += "b";
			}
			else 
			{
				s2.next();
				hits += "s2";
			}
		}
		System.out.println("hits", + hits);
	}
}

  

if this program is invoked with
% java ScanNext "1 true 34 hi"

it produces the output
hits ssssibis2

can anyone explain me the program....how does it work...i am confused

[Henry Wong]

can anyone explain me the program....how does it work...i am confused

It would help us, if you explain what are you confused with, as the example looks straightforward.

Henry

[Faisal Ansari]

i know its simple...but how does it print the output...that is confusing me

[Ninad Kulkarni]

hi Faisal,
Just check your program you are using "s", "i", "b", "s2" all these are strings not reference variable or primitive variable.

[Faisal Ansari]

i knw they are strings....but how does the pattern matching takes place....

[Ninad Kulkarni]

Would you tell me what you want to know from your question ? You can see Sun's Java Tutorial for Pattern matching.

[Jacob Sonia]

Hi i feel you are confused about the output.

So here the pattern you give is 1 true 34 hi
so the first while loop executes for all 4 and it prints s for each, so this gives the output ssss
now next while loop starts
so first the output is i since the first input is 1
then it sees the next input is true a boolean, so the output is b
then it sees the next input is again an int, so the output is 34
lastly the input is hi, which is neither boolean nor int, so the last case is successful and the output is s2

so this gives ibis2
so the total output is ssssibis2

hope it is clear. The best thing to do now is try with different inputs and see if you getting what you predict

[Asmita Khamkar]

Scanner class has the methods hasNext(), hasNextInt(), hasNextBoolean(). So in the first while loop it is only checking s1 i.e s1.hasNext() and concatenating s in hits variable every time this loop is run. In the next while loop it is checking s2 i.e. s2.hasNext() and within that loop it is checking s2.hasNextInt() or s2.hasNextBoolean() in if and is accordingly concatenating "i", "b" & "s2" in hits. Remember there are variables b and b2 declared as boolean but what you are concatenating in hits are strings "s", "b", "i" and not the values in variable. Try to understand the flow and I hope this clears your doubt.

Asmita

[Faisal Ansari]

thanks a lot ranchers for clearing my silly doubt

[Faisal Ansari]

here's another question

  
import java.util.regex.*;

class Regex2
{
	public static void main(String[] args)
	{
		Pattern p = Pattern.compile(args[0]);
		Matcher m = p.matcher(args[1]);
		boolean b = false;
		while (b = m.find())
		{
			System.out.print(m.start() + m.group());
		}
	}
}
  

when it is invoked with the command line
java Regex2 "\d*" ab34ef

the output is
01234456

can anyone help me as to why the 4 is printed twice

[Jesper de Jong]

Faisal Ansari wrote:here's another question

Please use one thread per question - so if you have a new question, just open a new topic. That way you'll have a better chance of getting your question answered, than when you hide it on the back of another topic.

[Faisal Ansari]

il keep that in mind from the next time onwards....

can anyone help me with the answer to the above question

[Parashuram Hallur]

Faisal Ansari wrote:il keep that in mind from the next time onwards....

can anyone help me with the answer to the above question

Well it goes like this

Pattern is = \d* --> That means zero or more occurrences of the integer
input is = ab34ef

so the pattern matches with all characters and the index of the occurence is returned with the method start() and group() will return the matching string. Because of * a,b, e and f are matched. At the 2nd index 34 will be matched and it will be printed, since it has *, it will also match the index beside the matched string which is 4 here, so 34 is the matching string and the 4 is the matching location for nothing, hence 4 printed twice.

So it is like - (0) a (1) b (2) 34 (3) wont be considered as the string is consumed, thats why 3 is not printed in the output (4) for nothing (5) e (6) f...all other indexes matching except 2nd index ( 34 ) are for the "*", means it matches the empty.

HTH

[Faisal Ansari]

why isnt 3 printed twice...it is also an integer and is present at index 3

[Henry Wong]

Faisal Ansari wrote:why isnt 3 printed twice...it is also an integer and is present at index 3

The "3" is actually at index 2.

Henry

[Henry Wong]

And BTW, this question is actually in the JavaRanch FAQ....

http://faq.javaranch.com/java/ScjpFaq#kb-regexp

Henry

[Faisal Ansari]

Tnx Mr. Henry

[Ken Truitt]

Regarding your confusion on the program (your first question), you know that the source, '1 true 34' or whatever, which is
used to construct the Scanner object, is split by Scanner at every SPACE by default--so there's not any matching per se[/]
going on, it's [i]tokenizing. And you can change the delimiter with the Scanner.useDelimiter() method, as you probably have read.

Then you work through the source with your Scanner object methods next() (gets the next token, and the tokens are all the
strings essentially that were separated by spaces in the source), nextInt() (gets the next int in the source and returns it AS AN
INT), etc.

Because you could get an exception if you call next(), nextInt() or other similar methods when there are no more tokens,
programs that use Scanner always make use of the hasNext() methods to make sure there is another token. The main
thing to remember about these is that the hasNext() method and hasNextInt() method and so on do not retrieve tokens
and so they do not advance the parser through the set of source tokens. That is, you could call hasNext() in an infinite
loop and never reach the end of the source--but if you called next(), or nextInt() in an infinite loop, you would definately
run out of tokens and get an exception.

I just thought that might have been what you were confused about.

[Faisal Ansari]

Tnx Mr.Ken