Just to complete the above: (?{}) sets $_ to the LHS side also in the substitution part of s///e, even if the code is a no-op:

qwurx [shmem] ~> perl -le '$s = "foo"; $_="bar"; $s =~ s/./print/e' 
bar
qwurx [shmem] ~> perl -le '$s = "foo"; $_="bar"; $s =~ s/.(?{print})/p
+rint/e' 
foo
foo
qwurx [shmem] ~> perl -le '$s = "foo"; $_="bar"; $s =~ s/.(?{})/print/
+e' 
foo
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