[I encourage you to check out Phelype's impressive approach]
I hear about macros that do things like "expand until they reach an unexpandable token" and this question may (or may not) be related to that kind of thing and the \romannumeral trick. In particular, I'm thinking of the behavior of things like \numexpr which terminate upon encountering a \relax token. However, they also terminate without a \relax token, if an end-of-group is reached (and apparently also when a begin-group is reached). Thus
\the\numexpr 1+1+1\relax
and
{\the\numexpr 1+1+1}
both work.
I have an interest in a recursive version of \numexpr, call it \rnumexpr, that will expand groups in its argument, continuing the calculation using the previously grouped data.
Here it is and it seems to work great. It relies on a feature of tokenization that if a group is passed as an argument, the grouping is stripped and the contents of the group become the actual argument.
However, with my coding, it requires an explicit terminator (in this case, \rrelax).
EDITED to handle up to 8 nesting levels (i.e., 8 successive left braces), but it still can't handle an implicit delimiter
\documentclass{article}
\makeatletter
\let\@relax\relax
% CAN HANDLE 8 SUCCESSIVE LEFT BRACES
\def\rnumexpr#1\rrelax{\numexpr\@rnumexpr
\@empty\@empty\@empty\@empty\@empty\@empty\@empty\@empty\@empty
#1\relax \@empty\@empty\@empty\@empty\@empty\@empty\@relax}
\def\@rnumexpr#1#2#3#4#5#6#7#8#9\@relax{%
#1\ifx\relax#2\relax\else\@rnumexpr#2#3#4#5#6#7#8#9\@relax\fi}
\makeatother
\begin{document}
\the\numexpr+1+1+1+1+1\relax,
\the\numexpr+1+1{+1+1+1}\relax,
\the\numexpr+1+1{+1{+1+1}}\relax
\the\rnumexpr+1+1+1+1+1\rrelax,
\the\rnumexpr+1+1{+1+1+1}\rrelax,
\the\rnumexpr+1+1{+1{+1+1}}\rrelax,
Expandable! \edef\z{\the\rnumexpr+1+1{+1{+1+1}}\rrelax}\z
\the\rnumexpr+1+1+1+1+1\rrelax,
\the\rnumexpr+1+1{+1+1+1}\rrelax,
\the\rnumexpr+1+1{+1{+1+1}}\rrelax,
\the\rnumexpr{+1{+1{+1{+1{+1{+1{+1{+1{+1{+1}}}}}}}}}}+1\rrelax,
Can handle up to 8 successive left braces:
\the\rnumexpr{+1{{{{{{{{+1}+1}+1}+1}+1}+1}+1}+1}+1}+1\rrelax{},
\the\rnumexpr{+1{{{{{{{{+1}}}}}}}}}+1\rrelax{},
\the\rnumexpr{{{{{{{{+1}}}}}}}}\rrelax{}
{\the\numexpr1+1+1} numexpr uses implicit delimiter
%{\the\rnumexpr1+1+1}
but rnumexpr won't work...EXPLICIT DELIMITER EXPECTED
\end{document}
The first two lines compare the results of \numexpr and \rnumexpr, showing how \numexpr appears to stop when it reaches the begin-group, whereas \rnumexpr extracts it and continues the calculation. It is even shown to be expandable!
The 3rd and 4th lines show put \rnumexpr to a tougher test. Phelype pointed out that my original request was quite limited as to how many levels of nesting it could handle. This edited approach can handle more nesting levels (up to 8 successive left braces), but still has a finite limit.
The 5th line of output shows how \numexpr can terminate without an explicit \relax. Attempting such a syntax with \rnumexpr does not work because I've coded it to expect an explicit delimiter.
Is there a way to redefine \rnumexpr to also end when reaching an end-of-group rather than an explicit terminator (while at the same time not ending when reaching a start-of-group)
Note: The purpose here is not to develop a logical approach to nested calculations. While that may be a desirable thing in certain applications, that is not what is being attempted here. Thus, approaches that suggest using parens rather than braced subunits do not address my concern.
As I replied to David, the process I am really interested in is counting certain "qualified" tokens across an arbitrary argument. Using the approach I am taking to this larger question, for example, I ignore "unqualified" tokens, but when I come across "qualified" tokens, I place a +1 in the output macro. However, the process I have developed also retains the grouping of the original argument in the output macro.
So when I am done examining the argument token-by-token (with grouping retained), the output contains an arbitrary number of +1 tokens within the argument's original grouping structure. It is this output macro that I hope to operate on with \rnumexpr. Since I am writing the code, I can always be sure that I add the \rrelax at the end, but this question has more to do with me wondering if it was possible to rewrite \rnumexpr without the closing delimiter.
{}groups rather than()? especially\rnumexpr+1+1{+1+1+1}\rrelaxwhich looks like it should be an error rather than\rnumexpr+1+1+{1+1+1}\rrelax? – David Carlisle May 24 '19 at 15:09{}work and it is not at all clear how they are supposed to work, If they work like(..)then the examples would be in error, or are they ignored, what answer would you want for{1}{2}or{1+2}*{3+4}? Are they the same as12and1+2*3+4? – David Carlisle May 24 '19 at 15:21\relaxor brace-delimited, or anything else). With an optional delimiter like for\numexprit could perhaps be done un-expandably (using\futureletand a huge processing time). Both I think it's not possible without making the code extremely (and I mean it) fragile. Would you be okay with a mandatory delimiter? – Phelype Oleinik May 24 '19 at 15:31\numexprwas easily do-able, or not. See my "Note" in my edited question, which explains better the motivation. – Steven B. Segletes May 24 '19 at 15:36\numexprexpression with ignorable relax isn't feasible with anything like a reasonable amount of code. – David Carlisle May 24 '19 at 15:59\numexpr, but I can't think of an expandable way to test whether that token is}without running intoextra }errors. – siracusa May 24 '19 at 17:48\numexpris a built-in command, rather than a macro, it operates under a separate set of rules in accomplishing its argument digestion. – Steven B. Segletes May 24 '19 at 17:53