RE: quick java question

Okay, thanks! I will attach the .java file to an email and send it to you privately. It will save my poor cs professor from answering all my questions and give me a new perspective. Thanks again. 

Have a great day,
Alex


----- Original Message -----
From: "Suzanne Balik" <spbalik@xxxxxxxxxxxxxx
To: programmingblind@xxxxxxxxxxxxx
Date sent: Thu, 29 Nov 2007 21:30:35 -0500 (EST)
Subject: RE: quick java question
You typically use double's rather than float's in Java. How 
about this --

would this work for you?



Scanner sc = new Scanner(System.in);
if (sc.hasNextDouble())
 double value = sc.nextDouble();
I'll be in my office some time after 3:00 PM tomorrow (Friday). 
If you

want to send me your code, I can help you with it then.



Suzanne Balik







I tried it, but have changed the var to a float after I sent the
message.  I tried var.hasNextFloat() but I get an error on that
line that says something like "float cannot be dereferenced".  I
am checking to be sure a number (the var in question) is greater
than 1 but less than another var and so thought I could just put
this simple statement as another condition in the if statement,
but I got the above error.  I am using the latest Java.



Have a great day,
Alex



----- Original Message -----
From: "Suzanne Balik" <spbalik@xxxxxxxxxxxxxx
To: programmingblind@xxxxxxxxxxxxx
Date sent: Thu, 29 Nov 2007 21:00:11 -0500 (EST)
Subject: RE: quick java question



You can use the Scanner class to do what you want to do easily,

if you're

using Java 5.0.  For example,



Scanner sc = new Scanner(System.in);
if (sc.hasNextInt())
 int value = sc.nextInt();





Suzanne Balik



NC State University      EB II 2318    (919)515-5617



We?ll take a cup o? kindness yet, for auld lang syne.
                                      --Robert Burns




Do a search for
Try catch tutorial
Try statements and catch clauses are the way to go if you are

going to

seriously work with java.




-----Original Message-----
From: programmingblind-bounce@xxxxxxxxxxxxx
[mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf Of Alex

Parks

Sent: Thursday, November 29, 2007 7:55 AM
To: programmingblind@xxxxxxxxxxxxx
Subject: RE: quick java question




I am new to this language.  What do I put in place of "aString"
and how do I use the exception in my if statement? Add a
throwsException to the end? Thanks.



Have a great day,
Alex



----- Original Message -----
From: "Sina Bahram" <sbahram@xxxxxxxxx
To: <programmingblind@xxxxxxxxxxxxx
Date sent: Thu, 29 Nov 2007 07:07:27 -0500
Subject: RE: quick java question



If you're expecting an integer, then you can do



Integer.parseInt(aString)



That will throw a NumberFormatException if it isn't a number.



Take care,
Sina




-----Original Message-----
From: programmingblind-bounce@xxxxxxxxxxxxx
[mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf Of Alex

Parks

Sent: Thursday, November 29, 2007 12:51 AM
To: programmingblind@xxxxxxxxxxxxx
Subject: quick java question



Hi all,
Is there a quick way of determining if the user input is a

number? I need it

to be a number; it throws an exception and exits the program if

it isn't.

Maybe something like:
if (input!=\int
I have no idea about the syntax, but something along those lines

is what I

am looking for.  Thanks for any help.



Have a great day,
Alex
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