Re: quick java question

  • From: Alex Parks <mehgcap@xxxxxxx>
  • To: programmingblind@xxxxxxxxxxxxx
  • Date: Thu, 29 Nov 2007 23:32:09 -0500

I think I found what is going on. In your example you don't get input until after you call hasNext... but in my program I call it after I get input. I, too, am quite tired and am heading to bed, but tomorrow when I can type straight I will play with the arrangement of all those little statements until I get something like your example instead of its opposite. Thanks again. 


Have a great day,
Alex


----- Original Message -----
From: "Jeffrey Fidler" <jfiddler2@xxxxxxxxxxx
To: <programmingblind@xxxxxxxxxxxxx
Date sent: Thu, 29 Nov 2007 23:12:51 -0500
Subject: Re: quick java question
This will test for a double. Not exactly elegant -- I just did a 
"find and
replace" to get you the double. Compiled and tested just in case 
my tired
mind has completely stopped working, but it is all ok. :) The 
neat thing I
noticed is that you'll get an integer number to work here, too as 
it seems
that the Scanner class will do the conversion from int to double 
for you.

In any case, I'm truly off to bed now!



import java.util.Scanner;



public class TestDbl {



public static void main(String[] args) {
 System.out.print("Please enter a double: ");
 Scanner sc = new Scanner(System.in);
 if (sc.hasNextDouble()) {
 System.out.println("Your double is " + sc.nextDouble());
 } else {
 System.out.println("You did not enter an double!");
 }
 }
}




Kind regards,
Jeff



----- Original Message -----
From: "Alex Parks" <mehgcap@xxxxxxx
To: <programmingblind@xxxxxxxxxxxxx
Sent: Thursday, November 29, 2007 11:07 PM
Subject: Re: quick java question
Thanks. My var is a double called bet (the assignment was a 
blackjack
game, and no you are not helping me cheat by helping--we can use 
whatever
to get it done). I am ensuring bet is valid; that is, above 0, 
below the

user's current balance, and a number and not a letter.  I have:
if (bet<pts || bet>0 || bet.hasNextDouble()




I get an error about bet not being able to be dereferenced.



Have a great day,
Alex



----- Original Message -----
From: "Jeffrey Fidler" <jfiddler2@xxxxxxxxxxx
To: <programmingblind@xxxxxxxxxxxxx
Date sent: Thu, 29 Nov 2007 22:53:27 -0500
Subject: Re: quick java question



Here is an example using an integer.  (Sorry, but I had not

remembered that

you were using a double when I began quickly working up this

example, and

I'm off to bed for now.)  Remember that if you enter a char type,

it will

still work as it will output the ascii value!  You can swap the

int for a

double and it should work.  Hope this helps!



import java.util.Scanner;




public class TestInt {




public static void main(String[] args) {



System.out.print("Please enter an integer: ");



Scanner sc = new Scanner(System.in);



if (sc.hasNextInt()) {



System.out.println("Your integer is " + sc.nextInt());



} else {



System.out.println("You did not enter an integer!");



}



}




}





Kind regards,



Jeff





----- Original Message -----
From: "Sina Bahram" <sbahram@xxxxxxxxx
To: <programmingblind@xxxxxxxxxxxxx
Sent: Thursday, November 29, 2007 10:19 PM
Subject: RE: quick java question




If you'd like some help on this tonight, shoot me the file off

list ...

I'm
busy tomorrow, but hopefully  Suzanne can help you then, if we

don't get

it
tonight?



sbahram@xxxxxxxxx



Take care,
Sina




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

Parks

Sent: Thursday, November 29, 2007 9:55 PM
To: programmingblind@xxxxxxxxxxxxx
Subject: 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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