Re: How does Ada.Text_IO.Enumeration_IO work?

[Jerry <lanceboyle@qwest.net>]

On Nov 4, 4:31 pm, Adam Beneschan <a...@irvine.com> wrote:
> On Nov 4, 3:46 pm, Jerry <lancebo...@qwest.net> wrote:
>
>
>
>
>
> > > 'Cuz the rules say so.  A.10.6(5) is the important one here: "Next,
> > > characters are input only so long as the sequence input is an initial
> > > sequence of an identifier or of a character literal (in particular,
> > > input ceases when a line terminator is encountered). The character or
> > > line terminator that causes input to cease remains available for
> > > subsequent input."  A.10.6(10): "The exception Data_Error is
> > > propagated by a Get procedure if the sequence finally input is not a
> > > lexical element corresponding to the type, in particular if no
> > > characters were input ...".  That's the case when your input (skipping
> > > leading blanks) starts with an invalid character like a comma.  Since
> > > the comma can't be the first character of an enumeration literal, the
> > > comma "remains available for subsequent input", and thus "no
> > > characters are input" and Data_Error is raised.
>
> > Thanks, Adam. I believe that the clue for me is that "remains
> > available for subsequent input" means that it is available only to non-
> > enumeration input, characters excepted.
>
> I'm not sure what you mean by this.  If the first character in the
> input (other than space) cannot be the start of an enumeration
> literal, the input will stop right there; and that character remains
> available in the input.  If you then try to perform input using
> Enumeration_IO (without any other input routines in between), that
> same character is "available" in the input but causes Enumeration_IO
> to fail again for the same reason.  Perhaps there's some confusion
> about what the RM means by "available".  "The character remains
> available for subsequent input" means that *any* input routine will
> see that character first (and could raise an exception if that
> character is not legal).  It doesn't mean that the input routine will
> succeed.  But you could suck the character up with Text_IO.Get (to get
> one character), and then the character isn't available for input any
> more.  I hope this doesn't confuse you more, but I couldn't tell from
> your comment whether things were clear to you; and if not, I was
> hoping to try to make it clearer.
>
>                             -- Adam

The source of confusion here is my poorly worded reply. What I meant
to say (he said, sounding like a politician) is that the character
will be "gotten" only by non-enumeration input attempts, character
types excepted. Both RM and you are clear.

Jerry