Tokenizer Help

[jamonathin]

Hey all, this Friday is the computer science competition (Hikaru knows what im talkin bout) and my java skills are pretty poop. Unfortionately for me my c++ is much better and I didn't find out untill last week that we could of used c++ but there's a completely different way of compiling it, and changing would juss screw up my teammates.

So, anyways, what I'm asking is this. There were some sample problems for us to try and some of them included Tokenizer.

Now I understand how to get the "tokens" when there is a space inbetween and convert them to an integer but how about when the user inputs . .

code:

04/21/90

All i've understood so far was how to grab those numbers, not how to extract them out of a string. How would i go about taking the '04', '21' and '90' out of that. And the input will always be in that same format
( dd/mm/yy ) pleh

[wtd]

Something this simple doesn't need any kind of special tokenizer class.

Java:

String original = "04/21/90"; String[] parts = original.split("/");

[jamonathin]

Ok, so basically i make a new array variable that will end up being three because the string is being split. Then i can assign the arrays to whatever var i want.

Thanks wtd, looks simple enough .

[jamonathin]

Since you're so quick at responding wtd . . i have 2 quick questions for you. How do you write or in an if statement. "and" is &&, but what is or?. And how would you go about getting a "long" variable from the user.

[wtd]

Java:

||

Java:

BufferedReader keyboard = new BufferedReader(    new InputStreamReader(System.in)); Long inputLong = Long.parseLong(keyboard.readLine());

[Hikaru79]

Although wtd is right, you don't need a StringTokenizer for something like that, here's what it would have been if you'd chosen to do it.

Java:

StringTokenizer st = new StringTokenizer(myString, "/");

The second, optional, argument is a String, of which every individual character is a delimiter.

[jamonathin]

Sounds good, thanks for your help wtd and hikaru

[jamonathin]

wtd wrote:

Java:

||

Another question about that wtd. I tried using random symbols in my attempts at trying to find out the 'or' sign, and i tried using

Java:

^

inbetween my conditions and the program ran it, then I switched over to the " || ", and it produced the same thing. Is that ^ an or sign or what

[wtd]

jamonathin wrote:

wtd wrote:

Java:

||

Another question about that wtd. I tried using random symbols in my attempts at trying to find out the 'or' sign, and i tried using

Java:

^

inbetween my conditions and the program ran it, then I switched over to the " || ", and it produced the same thing. Is that ^ an or sign or what

http://java.sun.com/docs/books/tutorial/java/nutsandbolts/bitwise.html

It's "bitwise exclusive or".

[1of42]

Hikaru79 wrote:

Although wtd is right, you don't need a StringTokenizer for something like that, here's what it would have been if you'd chosen to do it.

Java:

StringTokenizer st = new StringTokenizer(myString, "/");

The second, optional, argument is a String, of which every individual character is a delimiter.

StringTokenizer is deprecated, don't even bother.

[jamonathin]

This goes with the long question I had for wtd. I think the code you gave me isn't for that long of numbers. When i try: 564696543201678654789456 in my program, i get erros, but smaller numbers are fine.
Here's my code:

Java:

import java.io.*; public class game {                 public static void main(String[] args) {                                 BufferedReader in = new BufferedReader (new InputStreamReader (System.in));                                         long one = 1,two = 1;                 while (one !=0 && two !=0)                 {                                try                         {                                 System.out.print("1st Input: ");                                 one = Long.parseLong(in.readLine());                                 System.out.print("2nd Input: ");                                 two = Long.parseLong(in.readLine());                                                         long result = one + two;                                 if (one != 0 && two !=0)                                 {                                         System.out.println("Result: " + result + "\n");                                 }                         }                                catch (IOException inpEX)                         {                                 System.err.println("Exception: " + inpEX);                         }                 }         } }

[wtd]

A "long" is a 64-bit integer The largest number an unsigned 64-bit integer can hold is 18446744073709551615. Your number is larger than that, so you have problems.

code:

>> 564696543201678654789456 > 18446744073709551615 => true

[jamonathin]

Well then ****. . . lol. Because one of the questions we had to do was

Problem 8: Extended Precision Addition wrote:

Develop a program that takes as input two "long" nonnegative integers and outpus their sum. The inputs will be no longer than 35 digits. The program should repeat until the user enters zero for both inputs.

Sample Program Execution:

1st input: 10371
2nd input: 315
Result: 10686

1st input: 54696543201678654789456
2nd input: 82067690004564356875434
Result: 136764233206243011664890

1st input: 0
2nd input: 0
Result: 0

What would I do if long wont take them . .

[wtd]

http://java.sun.com/j2se/1.4.2/docs/api/java/math/BigInteger.html

[Hikaru79]

1of42 wrote:

Hikaru79 wrote:

Although wtd is right, you don't need a StringTokenizer for something like that, here's what it would have been if you'd chosen to do it.

Java:

StringTokenizer st = new StringTokenizer(myString, "/");

The second, optional, argument is a String, of which every individual character is a delimiter.

StringTokenizer is deprecated, don't even bother.

Huh? How do you figure that? My JDK 1.5.0_4 doesn't seem to think it is (it will usually give you a warning if you're using deprecated objects) and the API ref at http://java.sun.com/j2se/1.5.0/docs/api/java/util/StringTokenizer.html doesn't say it is (it usually does). Where did you get the idea that such a useful thing is deprecated?