Re: no way it's not a bug of Perl

[Corion]

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.

[vincentaxhe]

unexpected quantum influence

[NERDVANA]

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.

[LanX]

> 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.

[vincentaxhe]

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