[r6rs-discuss] Linefeed and Line Endings

[Alan Watson]

Michael Sperber wrote:
> Alan Watson <alan at alan-watson.org> writes:
> 
>> The response to Formal Comment #46 says:
>>
>>  > The R6RS grammar is meant to apply to a decoded stream. That is,
>>  > specific bytes that represent a newline some stream are meant to be
>>  > converted to the LF character in the character stream that is parsed
>>  > as a program.
>>
>> I cannot find this in section 4 of the latest draft. Is it somewhere else?
>>
>> Related to this, shouldn't "\<line ending>" in strings be interpreted in 
>> the same way as "\<linefeed>"?
> 
> I'm confused: The grammar says something about \<line ending>, not about
> \<linefeed>.  The section on "Line endings" has:
> 
>>> In a string literal, a line ending not preceded by a \ stands for a
>>> linefeed character, which is the standard line-ending character of
>>> Scheme.

There are three places this is mentioned:

(b) 4.2.1 has "\<line ending>" as a <string element>.

(b) 4.2.2 says "In a string literal, a line ending not preceded by a \ 
stands for a linefeed character, which is the standard line-ending 
character of Scheme." and also talks about merging crlf and crnl.

(c) 4.2.7 says that "\<linefeed>" is nothing.

So, the grammar in 4.2.1 allows "\<line ending>", but I can see nothing 
in 4.2.2 or 4.2.7 that say how this should be interpreted if the line 
ending is not a line feed.

Also, I asked if the draft R6RS requires that source be "decoded". If it 
is, then for source (but not necessarily data), all line endings will be 
converted to linefeeds.

Regards,

Alan