in reply to I thought I found a bug of Perl

See the documentation of s///:

If the pattern evaluates to the empty string, the last successfully executed regular expression is used instead. See perlre for further explanation on these.

In your code, s///r has the empty pattern in it.

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Re^2: no way it's not a bug of Perl
by vincentaxhe (Scribe) on Jul 14, 2024 at 10:48 UTC
    unexpected quantum influence
      Perhaps, but what did you expect it to do? A no-op?

      You'll find a lot of places in Perl where they decided that a code expression that doesn't do anything useful should get a special case so that it does something useful.

        > A no-op?

        Please note that in mathematics an empty set is always a sub set of all other sets.

        And so does the empty pattern in Perl always match.

        If no match has previously succeeded, this will (silently) act instead as a genuine empty pattern (which will always match).

        Hence not a no-op!°

        > that a code expression that doesn't do anything useful should get a special case so that it does something useful.

        I disagree. The "repeat the last match" feature should be dependent on an explicit modifier like eg s///b ²

        I'd welcome a feature to switch off the current default behavior in favor of an explicit modifier.

        Reasoning: The ratio of useful vs confusing is too bad.

        Update

        °) Demo:

        ~ $ perl -de0 ... DB<1> $_='XXXX' DB<2> s/(?:)/Y/ # genuine empty pattern DB<3> p YXXXX DB<3> s/(?:)/Z/g DB<4> p ZYZXZXZXZXZ DB<4>

        ²) in an earlier version I suggested s///l but this modifier is already taken. "b" (for before) seems to be free.

        Cheers Rolf
        (addicted to the Perl Programming Language :)
        see Wikisyntax for the Monastery

        I expected it just do nothing, But it's more useful to take last matched pattern.