Re: How does Ada.Text_IO.Enumeration_IO work?

["Dmitry A. Kazakov" <mailbox@dmitry-kazakov.de>]

On Fri, 4 Nov 2011 16:31:33 -0700 (PDT), Adam Beneschan 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.

For what it is worth. Any scanner when detects an error has only two
options:

1. Consume input and then raise an exception
2. Leave input as is and raise

As a side note, it is unfortunate that Enumeration_IO uses the same
exception (Data_Error) in both cases. The behavior 1 means: input
recognized but illegal. The behavior 2 means: nothing there, repeat.

-- 
Regards,
Dmitry A. Kazakov
http://www.dmitry-kazakov.de