give modern advice on \p and \P

perl6 · Mar 5, 2012 · 1ad3292 · 1ad3292
1 parent 292f669
commit 1ad3292
Showing 1 changed file with 4 additions and 4 deletions.
diff --git a/STD.pm6 b/STD.pm6
@@ -5054,11 +5054,11 @@ grammar Regex is STD {
         :my $m;
         :my $p;
         <sym=[pP]>
-        { $s = $<sym>.Str; $m = $s lt 'a' ?? '-' !! '+'; }
+        { $s = $<sym>.Str; $m = $s lt 'a' ?? '-' !! ''; }
         [
-        || (\w) { $p = $0.Str; $¢.obs("\\$s$p", '<' ~ $m ~ "is$p>"); }
-        || '{' $<param>=[\w+] '}' { $p = $<param>.Str; $¢.obs("\\$s\{$p\}", '<' ~ $m ~ "is$p>"); }
-        || '{' $<param>=[\w+] \= $<val>=[<-[\}]>*] '}' { $p = $<param>.Str; my $v = $<val>.Str; $¢.obs("\\$s\{$p=$v\}", '<' ~ $m ~ "is$p\('$v')>"); }
+        || (\w) { $p = $0.Str; $¢.obs("\\$s$p", '<' ~ $m ~ ":$p>"); }
+        || '{' $<param>=[\w+] '}' { $p = $<param>.Str; $¢.obs("\\$s\{$p\}", '<' ~ $m ~ ":$p>"); }
+        || '{' $<param>=[\w+] \= $<val>=[<-[\}]>*] '}' { $p = $<param>.Str; my $v = $<val>.Str; $¢.obs("\\$s\{$p=$v\}", '<' ~ $m ~ ":$p\('$v')>"); }
         ]
     }
     token backslash:Q { <sym> <.obs('\\Q as quotemeta', 'quotes or literal variable match')> }