perlop: clarify \U, \L, \F behaviour

[mfontani]

... as they do not stop "at \E or end of string" but are also
stopped by another \U, \L or \F.

See also:
https://www.nntp.perl.org/group/perl.perl5.porters/2022/07/msg264490.html

[khwilliamson]

People may find #11145 instructive

[mfontani]

@khwilliamson my goal here is to "at least" document (and have tests for) the current non-stacking behaviour, so that it's clearer that's what's happening / isn't as nebulous as before.

It's hopefully a simple change to merge.

[bram-perl]

Looking at some tickets; this change in docs was also suggested in #19670
but a reply (from @khwilliamson) was "It's worse then that";
Unfortunately the reply didn't specify in what way it worse/in what way documenting it like that would be incorrect :(

[khwilliamson]

But, I added a follow-up immediately after to that reply that did give concrete examples:

Here are some tickets #8848 #11145 #18981

[iabyn]

On Wed, Jul 27, 2022 at 04:46:51AM -0700, Marco Fontani wrote: ... as they do not stop "at \E or end of string" but are also stopped by another \U, \L or \F.

Given that we're currently unclear as to what the correct behaviour should be in all circumstances, and we may well change things once we are clear, I don't think we want to be fixing the current behaviour in the documentation.

…

-- A major Starfleet emergency breaks out near the Enterprise, but fortunately some other ships in the area are able to deal with it to everyone's satisfaction. -- Things That Never Happen in "Star Trek" #13

[mfontani]

Sounds good, closing this then.

[bram-perl]

Having documented some(/all?) the quirks/caveats/... for case modifiers¹ it puts me in a better position to comment.

The text is mostly correct but based on the current behavior it's missing the caveats:

1. \U\lfoo is transformed into \l\Ufoo;
2. \L\ufoo is transformed into \u\Lfoo;
3. A \E which follows immediately after a \U, \L, \F, \Q, \u, \l causes both symbols to be completely ignored. Examples:
   • \Ufoo\L\Ebar is parsed as \Ufoobar -> There no longer is a \L or \E!
   • applying 1. and 2.: \Ufoo\L\u\Ebar is first transformed into \Ufoo\u\L\Ebar and then into \Ufoo\ubar -> Again there is no longer a \L or \E

So if after 1., 2., 3. there are still two occurrences of \U, \L, \F in the string then the second occurrence will terminate the first \U, \L, \F.

Also note: a \L also ends a previous \L, i.e. \LFOO\LBAR\EBAZ is equal to foobarBAZ (same applies for \U and \F)

Footnotes

1. see Interaction of case-modifiers (\U, \L, \u, \l, \F, \Q, \E) in double quoted strings #20042 ↩