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Postings from February 2009
Bram wrote:
>Why use (?>[a-z])* ? Is it any different than ([a-z])* ?
/\A([a-z])*\z/ or /\A(?:[a-z])*\z/ doesn't tickle the bug. Apparently the
regexp engine is clever enough to recognise that a subpattern has a fixed
length of 1, and quantifies it in a different way. /\A(?>[a-z])*\z/
makes the subexpression sufficiently complex to go the heavyweight path
and encounter the bug.
Another way to make the expression sufficiently complicated is
/\A(?:[a-z][a-z])*\z/. You have to double the length of the string,
of course, but then it shows the bug in the same way as the (?>) form:
$ for pver in 5.{6.2,8.{0,8,9},9.4,10.0}; do echo $pver $(~/usr/perl/util/perlver $pver-i32-f52 perl -lwe '$SIG{__WARN__}=sub{print"(warn) "}; $a="xyzt"x20000; print $a =~ /\A(?:[a-z][a-z])*\z/ ? "ok" : "bug"'); done
5.6.2 ok
5.8.0 ok
5.8.8 ok
5.8.9 ok
5.9.4 bug
5.10.0 bug
$ for pver in 5.{8.{0,8,9},9.4,10.0}; do echo $pver $(~/usr/perl/util/perlver $pver-i32-f52 perl -lwe '$SIG{__WARN__}=sub{print"(warn) "}; $a="xyzt"x20000; utf8::upgrade($a); print $a =~ /\A(?:[a-z][a-z])*\z/ ? "ok" : "bug"'); done
5.8.0 bug
5.8.8 bug
5.8.9 ok
5.9.4 bug
5.10.0 bug
$
You can also make the subexpression sufficiently complicated by adding
something that can match multiple ways. /\A(?:X?[a-z])*\z/ triggers
the bug in a different manner, encountering the "recursion limit" warning.
-zefram
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