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'reusing 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 aregoing toseriously work with java.
-----Original Message----- From: programmingblind-bounce@xxxxxxxxxxxxx [mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf Of AlexParksSent: 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 AlexParksSent: 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 anumber? I need itto be a number; it throws an exception and exits the program ifit isn't.Maybe something like: if (input!=\int I have no idea about the syntax, but something along those linesis what Iam looking for. Thanks for any help.
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