RE: quick java question

  • From: "Suzanne Balik" <spbalik@xxxxxxxxxxxxxx>
  • To: programmingblind@xxxxxxxxxxxxx
  • Date: Thu, 29 Nov 2007 21:30:35 -0500 (EST)

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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