StringTokenizer

[Tyler Jordan]

Hi,
I have been trying to get this simple function to work as it is stated in the reference library. Instead of changing from true to false after going through the characters and breaking the loop, it keeps resolving as true indefinately. It doesn't break out of the while loop. Have no idea what I am doing wrong.

                int linecount = 1;
                int numLine = 0;
String test = "test,test,test";
while (numLine <= linecount)
{
int numToken = 0;

StringTokenizer parser = new StringTokenizer(test, ",");
while (parser.hasMoreTokens())
{
numToken++;
}

numLine++;
}
  

[Jeff Grant]

I'm betting it is only going through the while (numLine <= linecount) loop once, but it's getting hung at the while (parser.hasMoreTokens()) instead.

You should look into why while (parser.hasMoreTokens()) is never going false.

[miguel lisboa]

i wrote this, and runs fine:

import java.util.*;
 

 
class Simbolo {
 
public static void main(String args[]) {
 
int linecount = 1;
 
int numLine = 0;
 
String test = "test,test,test";
 
while (numLine <= linecount) {
 
int numToken = 0;
 
StringTokenizer parser = new StringTokenizer(test, ",");
 
while (parser.hasMoreTokens()) {
 
numToken++;
 
System.out.println(parser.nextToken());
 
}
 
numLine++;
 
}
 
}
 
}  

prints:

C:\javas>javac Simbolo.java

C:\javas>java Simbolo
test
test
test
test
test
test

[Mark Wuest]

Here's your loop:

while (parser.hasMoreTokens())
{
numToken++;
}

Here's your loop after it has had its Wheaties:

while (parser.hasMoreTokens())
{
System.out.println(parser.nextToken());
numToken++;
}

I hope just giving you the "answer" didn't circumvent you from learning how it works. There's a little sample snippet in the javadocs for StringTokenizer along with an advertisement for the much-easier-to-use (IMHO) String.split() method if you're using 1.4 or later.

Mark

[miguel lisboa]

i ignore the meaning of Wheaties but i guess what you mean is that, if we dont iterate:
parser.nextToken();
then the conditional remains true forever

[Mark Wuest]

Yes. Unless there are *no* tokens at all, in which case hasNext() would always be false.

Did you check out the javadoc at

http://java.sun.com/j2se/1.4.2/docs/api/java/util/StringTokenizer.html

Also, did you check out the java.lang.String.split() method? It gives you back a String[] that I find easier to work with: you can either iterate through with a for(; ;) or you can directly access any element. Again, notice that split() only works in 1.4 and later versions.

Wheaties is a breakfast cereal in the U.S. that used to be advertized as "The Breakfast of Champions." I haven't seen a commercial for it in years, though.

Mark
[ May 20, 2005: Message edited by: Mark Wuest ]

[miguel lisboa]

 String newTest = "t1,t2,t3";
 
for (String token: newTest.split(","))
 
System.out.println(token); 

t1
t2
t3

[Tyler Jordan]

Ok, I just needed to do nextToken() that fixed everything. Thanks

[K Riaz]

If I am not mistaken, it seems as though you are counting the number of tokens in your query word. The StringTokenizer class already has a method called countTokens() which does the same thing.

Just so you know.

[miguel lisboa]

good point!