Interestingly enough, if you use a named parameter class instead of a the \s assertion, it finds it correctly in both strings.
$ascii = "\xa0\x{a0}";
$unicode = "\x{100}\xa0\x{a0}";
print "Latin-1 \\s : ", $ascii =~ /\s/ ? "yes":"no","\n";
print "Latin-1 \\p{Space}: ", $ascii =~ /\p{Space}/ ? "yes":"no","\n
+\n";
print "Unicode \\s: ",$unicode =~ /\s/ ? "yes":"no","\n";
print "Unicode \\p{Space}: ",$unicode =~ /\s/ ? "yes":"no","\n";
I certainly wouldn't expect non-breaking space to be recognized or not as a space depending on what else was in the string.
(I even experimented with whether the enclosing brackets were significant... apparantly not.)
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