Omit needless words: don't produce parameters with the argument label "with".

When dropping the redundant type information from a parameter name
would leave us with "with", instead drop the "with" and lowercase the
rest of the parameter name. It's still redundant information, but it's
less bad than simply "with".

Swift SVN r31835

lib/Basic/StringExtras.cpp

@@ -559,7 +559,20 @@ StringRef swift::omitNeedlessWords(StringRef name, OmissionTypeName typeName,
           break;
         }
 
-      SWIFT_FALLTHROUGH;
+        // If removing the type information would leave us with a
+        // parameter named "with", lowercase the type information and
+        // keep that instead.
+        if ((role == NameRole::FirstParameter ||
+             role == NameRole::SubsequentParameter) &&
+            *nameWordRevIter == "with" &&
+            std::next(nameWordRevIter) == nameWordRevIterEnd) {
+          name = toLowercaseWord(
+                   name.substr(nameWordRevIter.base().getPosition()),
+                   scratch);
+          break;
+        }
+
+        SWIFT_FALLTHROUGH;
 
       case PartOfSpeech::Verb:
       case PartOfSpeech::Gerund:
@@ -574,6 +587,7 @@ StringRef swift::omitNeedlessWords(StringRef name, OmissionTypeName typeName,
       }
       break;
     }
+
   } else if (role == NameRole::Property) {
     // For a property, check whether type information is at the beginning.
     name = omitNeedlessWordsForResultType(name, typeName, true, scratch);
@@ -583,11 +597,11 @@ StringRef swift::omitNeedlessWords(StringRef name, OmissionTypeName typeName,
   if (name == "get" || name == "set")
     return origName;
 
-  // If we ended up with a keyword for a property name or base name,
-  // do nothing.
   switch (role) {
   case NameRole::BaseName:
   case NameRole::Property:
+    // If we ended up with a keyword for a property name or base name,
+    // do nothing.
     if (isKeyword(name))
       return origName;
     break;