OpenEuphoria

Re: A question about certain language features

• Posted by Derek Parnell <ddparnell at bigpond.com> Mar 21, 2002
• 591 views

Karl,
I'm thinking that a distinct token be placed in the slice notation that
indicates end-of-sequence rather than omitting the end slice value. Maybe a
"$" symbol.

    foo = "abcdefghij"
    a = foo[2..$]    -- everything except the first element.
         a==> "bcdefgihj"
    b = foo[1..$-6] -- everything except the last 6 elements.
         b==>"abcd"
    c = foo[$-4..$-1] -- only the elements 6 thru 9.
         c==>"fghi"

The problem with omission is that it you are never certain that the omission
was deliberate or not. A specific symbol inserted is more rarely a mistake.
Of course we don't need a special symbol to represent the beginning of a
sequence because the value 1 suffices.

The other nice slice extention is to allow a list of subslices in the
notation:

   d = foo[1..3, $-2..$]
      d==> "abchij"

but that might be stretching the friendship 

------------
Derek.

----- Original Message -----
From: <kbochert at ix.netcom.com>
To: "EUforum" <EUforum at topica.com>
Sent: Friday, March 22, 2002 6:36 AM
Subject: RE: A question about certain language features



You wrote on 3/21/02 10:54:37 AM:

>Karl,
>
>   Consider this syntax for shorthand slicing. Your current syntax sort
>of defeats the sequence bounds checking.
>
>seq[1..0] -- seq[1..length(seq)]
>seq[1..0-1] -- seq[1..length(seq)-1]
>
>seq[1..1-1] -- {} (reverse slice)
>seq[1..-1] -- index [-1] out of bounds
>
>integer index   index=0
>seq[1..index] -- {} (reverse slice)
>
>'O' is explicit, it cannot be implied. If it is, the original EU rules
>apply.
>
>I haven't thouroughly looked. Does your shorthand work if it's implied
>with a variable?
>
>
>Chris

I handle 'foo[2..]' by expanding it textually. I think of it
as a macro.

when I see the '..' followed by a ']' or '-', I insert the text
'length(foo)' directly into the input stream. (Having previously
saved the 'foo').
'foo[2..]' causes the interpreter to actually see 'foo[2..length(foo)]'
and 'foo[2..-(a*b)]' is seen as 'foo[2..length(foo)-(a*b)]'


As a result, all the normal Euphoria processing is left intact.

Thanks for the link
Karl Bochert