Re: Array Question

From: Tony Morris (dibblego_at_optusnet.com.au)
Date: 12/28/03

  • Next message: Tony Morris: "Re: simple beginners question" 
    Date: Mon, 29 Dec 2003 00:27:40 +1000

    Quick criticism (for constructive purposes and also because it irks me!)

    ArrayList al = new ArrayList(); // yuk !
    List al = new ArrayList(); // much better

    Use the interface type wherever appropriate (and it usually is with the
    Collections API).
    For more info., "Effective Java Programming Guide", Joshua Bloch. 

    --
Tony Morris
(BInfTech, Cert 3 I.T., SCJP[1.4], SCJD)
Software Engineer
IBM Australia - Tivoli Security Software
"Anthony Borla" <ajborla@bigpond.com> wrote in message
news:wDqFb.61002$aT.22966@news-server.bigpond.net.au...
> "Andrew Dixon - Depictions.net" <andrew.dixon@NOREPLY.depictions.net>
wrote
> in message news:%VjFb.2724$8x3.24779027@news-text.cableinet.net...
> > Hi Everyone.
> >
> > Bit new to Java and I have question about arrays. I have wrote
> > some code to look for certain substrings within a larger string
> > which work fine. I would like to store each substring in an array,
> > however I don't know how many there will be. All the example I
> > have found for arrays only show how to predefine the size of the
> > array, however I would like to expand it by one each time I need to.
> > Can anyone let me know how to do this.
> >
>
> Sorry, no can do: arrays are fixed-size objects. In order to have a
> 'growable' or 'resizable' array you:
>
> * Allocate a new, larger array, copy all elements from old array
>    to the new array; this, effectively, disposes of the old array
>
>    Pretty cumbersome approach - not recommended
>
> * Use a Collection-based object for storage, and only create
>    an array from this when needed [assuming you *must*
>    have an array with which to work]. You might use an
>    'ArrayList' or a 'Vector'. Quick example:
>
>         String[] getArrayOfSubstrings(String bigString)
>        {
>            ArrayList list = new ArrayList();
>            ...
>            for (...)
>            {
>                ...
>                list.add(bigString.substring(...));
>                ...
>            }
>            ...
>            return ((String[]) list.toArray(new String[list.size()]));
>        }
>
> I hope this helps.
>
> Anthony Borla
>
>
>
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