Re: [Scheme-reports] Is bytevector a token? Takashi Kato 29 Oct 2013 08:07 UTC

I think it's on purpose.

As my understanding (correct me if I'm wrong), <token> meant be the smallest unit that a lexer reads and <bytevector> is a compound datum that a parser needs to construct. Suppose your reader gets #u8(1 2 3) as its input, then first your lexer needs to return a token to let your parser know what type of datum needs to be constructed. Now, first the lexer returns #u8( then the parser understands this is a bytevector. However if <bytevector> is in <token> then your lexer needs to read the input as a bytevector. Then inside of the bytevector it has a delimiter and token so lexer needs to read your input recursively. I'm not good with theory but seems something is wrong.

Hope it can help you.

_/_/
Takashi Kato
E-mail: ktakashi@ymail.com

On Tuesday, 29 October 2013, 5:48, Yuichi Nishiwaki <yuichi.nishiwaki@gmail.com> wrote:
Hi, all. I'm very excited to see the final R7RS draft published. Thank
you all for the great work.
Reading the final draft, I have one question about the formal syntax
definition (7.1.1). <bytevector> is not listed in <token> line. Is it
by purpose? Or just a missing?

-- Yuichi Nishiwaki

_______________________________________________
Scheme-reports mailing list
Scheme-reports@scheme-reports.org
http://lists.scheme-reports.org/cgi-bin/mailman/listinfo/scheme-reports

_______________________________________________
Scheme-reports mailing list
Scheme-reports@scheme-reports.org
http://lists.scheme-reports.org/cgi-bin/mailman/listinfo/scheme-reports