Stupid, maybe, still the boss, definitely! The fact that most boss' can be
stupid from time to time (if not all the time) doesn't change the fact that
they are still the boss!!!
----- Original Message -----
From: Geert Van Damme <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Wednesday, December 20, 2000 5:33 PM
Subject: Re: arrays
> If that's true, your boss is stupid.
>
> Java costs memory and time too, why don't you use assembler?
> Most probably, the time it took you (and the $) is much higher than the
> aggregated difference between sing a vector and an array will ever be.
> That's why this is absolutely stupid!
>
> If I program 1 hour to gain a millisecond, it takes 3.600.000 runs of the
> function before there's any gain.
> 1 hour is nothing.
> 3.600.000 is a lot ;-)
>
>
> Geert Van Damme
>
>
> > -----Original Message-----
> > From: A mailing list about Java Server Pages specification and reference
> > [mailto:[EMAIL PROTECTED]]On Behalf Of Shirley Chen
> > Sent: woensdag 20 december 2000 22:11
> > To: [EMAIL PROTECTED]
> > Subject: Re: arrays
> >
> >
> > Thanks for replying. But my boss insists that I use array because
vector
> > costs more memory and time.
> >
> >
> >
> > >From: Shawn Zhu <[EMAIL PROTECTED]>
> > >Reply-To: A mailing list about Java Server Pages specification and
> > > reference <[EMAIL PROTECTED]>
> > >To: [EMAIL PROTECTED]
> > >Subject: Re: arrays
> > >Date: Wed, 20 Dec 2000 13:06:06 -0800
> > >
> > >Why don't you just use Vector?
> > >--------
> > >Vector v_queryResult = New Vector();
> > >...
> > >if(tmpString.indexOf('C')!=-1)
> > >{
> > > v_queryResult.addElement((Object)tmpString);
> > >}
> > >--------
> > >Then there's no need for Array2.
> > >
> > >
> > > > -----Original Message-----
> > > > From: Shirley Chen [mailto:[EMAIL PROTECTED]]
> > > > Sent: Wednesday, December 20, 2000 11:39 AM
> > > > To: [EMAIL PROTECTED]
> > > > Subject: arrays
> > > >
> > > >
> > > > Hi,
> > > >
> > > > I have a question, somebody please help me:
> > > >
> > > > I have a vector which will keep all the names the user typed
> > > > in. Then the
> > > > user supposes can do a query like "give me all the name which
> > > > starts with
> > > > 'C'". In order to do that, I do something like this:
> > > >
> > > > ...
> > > > String[] array1 = new String[myVector.size()];
> > > > String tempString = "";
> > > > int j=0;
> > > >
> > > > for (int i=0; i<myVector.size(); i++)
> > > > {
> > > > tmpString = (String)myVector.get(i);
> > > > if (tmpString.indexOf('C') != -1)
> > > > {
> > > > array1[j] = tmpString;
> > > > j++;
> > > > }
> > > > }
> > > >
> > > > file://since array1 may not be filled up with data, I transfer it
> > > > to another
> > > > String array so I know how many element in the array
> > > >
> > > > String array2 = new String[j];
> > > > for (int i=0; i<j; i++)
> > > > {
> > > > array2[i] = array1[i];
> > > > }
> > > >
> > > > // then I will do display. Supposedly, say, array2.length is
> > > > 20, if the user
> > > > wants to display first 5, I will display only the first 5
> > > > records, if the
> > > > user wants to display 30, and I will display 20 instead.
> > > > ...
> > > >
> > > > I can display all the situations ok except when myVector
> > > > contains the names
> > > > that all start with "C" (which means array2.length should be
> > > > myVector.size()). then the program will crash. I think it
> > > > might be array
> > > > out of bound problem, but I don't know how to fix it. Please help
me.
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