my $s = 'x';
print("\$s is $s,\$s =~ s///r is ",$s =~ s///r, ".\n");
$s =~ /x/;
print("\$s is $s,\$s =~ s///r is ",$s =~ s///r, ".\n");

prints

$s is x,$s =~ s///r is x.
$s is x,$s =~ s///r is .

So what is changed after the third line?

CodePudding user response：

Apparently, under /r modifier the substitution operator actually returns the rest of the original string from the position in the string after the last previous successful regex -- if the pattern was an empty string (and the string has not been changed)

perl -wE'$s = "x arest"; $s =~ /x a/; $r = ($s =~ s///r); say $s; say $r'

Prints the same $s, and then rest

It is the particular behavior with the empty-string pattern that affects this, per perlop

If the pattern evaluates to the empty string, the last successfully executed regular expression is used instead.

The way I see this to explain the behavior is that the // (empty string) pattern is literally replaced by the last match and that is removed from the string, so the rest returned

perl -wE'$s = "x arest"; $s =~ /x a/; $r = ($s =~ s//X/r); say $r'

Prints Xrest.

Notes from hunting for a better understanding of this behavior, in case there is more to it than the above lone comment from perlop

• It is not affected by pos, as this works the same way

  perl -wE'$s = "x arest"; $s =~ /x a/g; pos $s = 2; $r = ($s =~ s///r); say $r'
  

  I've added /g so to be able to use pos. Just uses $&?

• The section on Repeated matches with zeo-length substring is clearly interesting in this regard, perhaps to explain the rationale for the behavior