Want to win a PS4? Go Premium and enter to win our High-Tech Treats giveaway. Enter to Win

x
?
Solved

Character class. Shorten code for checking upper and lower case.

Posted on 2004-08-15
14
Medium Priority
?
286 Views
Last Modified: 2006-11-17
Can anyone help advice on the following?

Is there any way to shorten the condtion for checking upper and lower case?

Given that the loop accept only 'Y', 'y', 'N' and 'n' for yes and no?

do {
      ......
      ......
} while ((toCon!='Y') && (toCon! ='y') && (toCon!='N') && (toCon!= 'n'));

Do any of the these methods come into use for solving the above-mentioned qn?

Character.isLowerCase(chr)
Character.isUpperCase(chr)
Character.toLowerCase(chr)
Character.toUpperCase(chr)
Character.isLetter(chr)

Thanks in advance.
0
Comment
Question by:coffee_bean
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 7
  • 3
  • 2
  • +1
14 Comments
 
LVL 86

Accepted Solution

by:
CEHJ earned 800 total points
ID: 11804809
Yes - use toLowerCase

toCon = Character.toLowerCase(toCon);
0
 
LVL 86

Assisted Solution

by:CEHJ
CEHJ earned 800 total points
ID: 11804818
do {
    //
}while (toCon != 'n' && toCon != 'y');
0
 
LVL 7

Expert Comment

by:JugglerW
ID: 11804883
You may also use something like:

    public static boolean isYNyn( char ch )    
    {
        ch &= ~32;        
        return ch == 'Y' || ch == 'N';
    }

0
Industry Leaders: We Want Your Opinion!

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

 
LVL 86

Expert Comment

by:CEHJ
ID: 11804900
>>You may also use something like ...

Although it's generally not too good to be assumptive about character encodings

0
 
LVL 7

Expert Comment

by:JugglerW
ID: 11805118
Your'e right but the coding for y, n, Y, N is same in every encoding (beside EBCDIC perhaps) :-)
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 11805127
>>but the coding for y, n, Y, N is same in every encoding

It is *at the moment* yes ... ;-)
0
 
LVL 92

Assisted Solution

by:objects
objects earned 400 total points
ID: 11805933
do {
      ......
      ......
} while ("YyNn".indexOf(toCon)>=0);

That way you can easily add exctra options.

or

do {
      ......
      ......
} while ("YN".indexOf(Character.toUpperCase(toCon))>=0);
0
 

Author Comment

by:coffee_bean
ID: 11845812
Thanks CEHJ for the reply.

I tested out the codes on simple stt below and it worked if I enter Y or y or N or n to terminate a loop.

But I couldn't get it to work the other way round if it's for continuing a loop using while (toCon == 'n' && toCon == 'y');

-----------------------------------
char toCon;
char option;

do {
      System.out.println("End loop? ");
      option = in.readLine().charAt(0);
      toCon = Character.toLowerCase(option);
}while (toCon != 'n' && toCon != 'y');

      
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 11845911
I liked objects' one the best myself ;-)

>>while (toCon == 'n' && toCon == 'y');

How could both conditions be simultaneously true? ;-)
0
 

Author Comment

by:coffee_bean
ID: 11846131
objects, Thanks for the answer.

I tested out ur codes and it works well if it's for continuing loop.

char toCon;
do {
     System.out.println("To loop? ");
     toCon = in.readLine().charAt(0);
}while ("YN".indexOf(Character.toUpperCase(toCon))>=0);

Couldn't get it terminated ("End loop?") if Y or y or N or n  is entered when  condition is changed to !=. Similarly case for 'while ("YyNn".indexOf(toCon)>=0);' .

I assume this code read the first char of letter entered. It will only terminate when Y is entered to a 'while ("YyNn".indexOf(toCon) !=0);'.
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 11846212
It seems rather strange that you should wish it to terminate when either 'y' or 'n' is entered. What is the functionality required here?
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 11846267
A normal usage pattern would be:

do {
      // do something, then ask:
      System.out.print("Do you want to stop now? (y\n) ");
      char toCon = in.readLine().charAt(0);
} while("Yy".indexOf(toCon) < 0);
0
 

Author Comment

by:coffee_bean
ID: 11846610
Well, I am new to Java. A beginner learner.

I was given this qn without any stt provided in the do while loop.  I was kind of obscured with trying to shorten the condtion stt while getting it to work (loop). I didn't notice the conflicting condition until much later.  So I treat the lower & upper case Y & N as any other letter to get the stt to loop or otherwise.

Really appreciate all the answers. It has helped me learn something on the Char class method which I won't be able on my own or from my classmates.
0
 
LVL 92

Expert Comment

by:objects
ID: 11846679
0

Featured Post

VIDEO: THE CONCERTO CLOUD FOR HEALTHCARE

Modern healthcare requires a modern cloud. View this brief video to understand how the Concerto Cloud for Healthcare can help your organization.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

An old method to applying the Singleton pattern in your Java code is to check if a static instance, defined in the same class that needs to be instantiated once and only once, is null and then create a new instance; otherwise, the pre-existing insta…
By the end of 1980s, object oriented programming using languages like C++, Simula69 and ObjectPascal gained momentum. It looked like programmers finally found the perfect language. C++ successfully combined the object oriented principles of Simula w…
The viewer will learn how to implement Singleton Design Pattern in Java.
This tutorial covers a practical example of lazy loading technique and early loading technique in a Singleton Design Pattern.
Suggested Courses

609 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question