Catcode of \endlinechar when its charcode is less than 1

Question

When \endlinechar is less than 1, how do we access its catcode? The following gives the error shown below:

! Bad character code (-1).
l.34 ...ndlinechar\string=\the\catcode\endlinechar
                                                  \relax%
? x

\endlinechar=\m@ne
\edef\restorecatcodes{%
  \let\noexpand\restorecatcodes\relax%
  \endlinechar\string=\the\endlinechar\relax%
  \catcode\the\endlinechar\string=\the\catcode\the\endlinechar\relax%
  \catcode\string`\noexpand`\string=\the\catcode\string``\relax%
  \catcode\string`\noexpand=\string=\the\catcode\string`=\relax%
}%

EDIT

I have seen an approach by Michael Downes (in the pcatcode package). It looks like the following:

\begingroup%
\let\e\endlinechar\edef\p{\the\e}\e13\edef\n{\the\catcode\e}%
\xdef\restorecatcodes{%
  \let\noexpand\restorecatcodes\relax%
  \endlinechar=\p\relax\catcode\number\e=\n\relax%
  \catcode\string`\noexpand`\string=\the\catcode\string``\relax%
  \catcode\string`\noexpand=\string=\the\catcode\string`=\relax%
}%
\endgroup%

I think his assumption is that changing the charcode will not affect the catcode.

Answer 1

In TeX a character code has to be between 0 and 255 to be a valid character (as TeX deals with 8 bit encodings). Setting \endlinechar to -1 means that no endline character is inserted (bit of a supported hack, as Don thought this would be useful I guess). As -1 is not a valid character code there is no \catcode associated with it, which is why you get this error if you try to access it.

Update to match edit in question

In TeX there are 256 different characters tokens denoted by character codes 0 to 255. Each character has an associated \catcode which is bound to the character code. The three lines

\catcode 98=13
\catcode `b=13     % only possible if the char (here "b") has not a special catcode like comment etc
\catcode `\b=13    % always safe

are all doing the same, i.e. they set the catcode of character 98 which is ascii "b" to 13 (active).

The \endlinechar holds a number which denotes the character code to be appended at the end of the line (if its between 0 and 255). The catcode of this endline character is the catcode of the character token it points to. So if you do

\endlinechar=98

then at end end of each line a "b" is appended with whatever catcode "b" currently has. In other words, if you temporarily change the \endlinechar, the catcode of the original character is not modified. The only thing that happens is that a different character is appended at the end of the line (or none in case of -1). Thus all you have to do to get back to the old situation is to reset \endlinechar to its previous number.

Answer 2

Just check whether the \endlinechar can have a catcode:

\edef\restorecatcodes{%
  \let\noexpand\restorecatcodes\relax%
  \endlinechar\string=\the\endlinechar\relax%
  \ifnum\endlinechar>-1 \ifnum\endlinechar<256%
    \catcode\the\endlinechar\string=\the\catcode\the\endlinechar\relax%
  \fi\fi
  \catcode\string`\noexpand`\string=\the\catcode\string``\relax%
  \catcode\string`\noexpand=\string=\the\catcode\string`=\relax%
}%

\endlinechar=-1

\restorecatcodes
\showthe\catcode\endlinechar

The answer is

> 5.
l.16 \showthe\catcode\endlinechar

This shows that the endlinechar is restored to what it was before. If we say

\catcode`\^^M=\active

\edef\restorecatcodes{%
  \let\noexpand\restorecatcodes\relax%
  \endlinechar\string=\the\endlinechar\relax%
  \ifnum\endlinechar>-1 \ifnum\endlinechar<256%
    \catcode\the\endlinechar\string=\the\catcode\the\endlinechar\relax%
  \fi\fi
  \catcode\string`\noexpand`\string=\the\catcode\string``\relax%
  \catcode\string`\noexpand=\string=\the\catcode\string`=\relax%
}%

\endlinechar=-1

\restorecatcodes
\showthe\catcode\endlinechar

the answer will be

> 13.
l.16     \showthe\catcode\endlinechar

so we see that the endlinechar is restored to what it was before (character code and category code).

The same if we say \catcode\^^M=5instead of\endlinechar=-1`.