OpenEuphoria

Re: if statement not working

• Posted by Derek Parnell <ddparnell at bigpond.com> Mar 22, 2002
• 948 views

I'm sorry. I've had a decent sleep now. Yesterday was not a good day at the
office.

----- Original Message -----
From: "Graeme" <graemeburke at hotmail.com>
To: "EUforum" <EUforum at topica.com>
Subject: Re: if statement not working


>
> At 07:04 AM 3/22/02 -0800, you wrote:
> >
> >Graeme wrote:
> >
> >> But seriously, having the interpreter return a valid result
> >> from accessing non-existant data is just way stupid.
> >
> >Is there really a need to make the attack so personal?
>
>
> Well that's EXACTLY what i thought.....
>
> Sorry David. There was nothing of a personal nature in my
> post, unless you mean the word 'stupid'. I used it because
> its the same word Derek used to describe his opinion in the
> post I was replying to. I quoted his comment in my post. You
> have removed this from the top of your reply, as I'm sure you
> must be aware.
>

And Graeme, you are absolutely correct. I was angry over some poor decisions
that colleagues of mine had made; committing my time to projects interstate
for months on end, without consulting me, or my bosses. I've calmed down a
lot now.


>
> >So if i were to ask you to return all 100K bytes of this
> >email, you'd land in a mental institution? Eu would,
> >figuratively speaking.
>
> >Kat
>
>
> So if you asked me for $100 and I diddnt have it so i gave
> you 3 $10 notes and the rest in monopoly money that would
> be ok?
>
> AND HERES THE REAL PROBLEM --->
>
> What if you dont notice, then you go to the supermarket and
> and try to pay for your groceries? Will the cashier accept
> the monopoly money? Maybe. Hopefull not, because every time
> somebody accepts it, the bug is one step further down the line,
> and when it enevtaully stops, you've got a hell of a mess trying
> to figure out who's handing out the bad cash...
>

It is good that Euphoria doesn't allow you access to non-existant sequence
elements. Thus :

   s = "abc"
   t = s[1..4]

Eu is proper to reject this.

My problem still remains though, and I don't have a reasonable solution yet.

Simply put, I don't care how many elements the sequences have, all I want to
know is : does one object begin with another object? I suppose the simplest
approach is:

   s = upper(trim(t))
   if match("*AUTHOR ", s) = 1 then
     GameInfo[vAuthor] = t[9.. length(t)]
     ...
   elsif match("*COPYRIGHT ", s) = 1 then
     GameInfo[vCopyright] = t[12.. length(t)]
     ...
   elsif match("*TITLE ", s = 1) then
     GameInfo[vTitle] = t[8.. length(t)]
     ...

However, in my never-ending quest for readable code, the phrase "match(a,b)
= 1" is not obviously telling the reader that I'm trying to see if 'a'
begins with 'b'. That's part of the ambiguitity of using literal numbers in
code; you don't know if they are special (magic) numbers or not. So I could
make it a bit better by doing:

   constant beginning = 1
   . . .
   if match("*AUTHOR ", s) = beginning then

but that looks a bit ambiguous too. Or :

   if begins(s, "*AUTHOR ") then

but that sounds unusual if you say it out loud.

What I'd really, really like to write is something like:

   structure GameInfo
      sequence Title
      sequence Author
      sequence Copyright
   end structure
   GameInfo theGame
   . . .
   case upper(trim(t)) do
     when begins "*AUTHOR " then
       theGame.Author = t[9..$]
     ...
     when begins "*COPYRIGHT " then
       theGame.Copyright = t[12..$]
     ...
     when begins "*TITLE " then
       theGame.Title = t[8..$]
     ...
   end case

but I know that ain't going to happen 

-----------
cheers,
Derek