Lazily require types when validating Extensions API calls

chrome/renderer/resources/extensions/utils.js

 // Copyright (c) 2012 The Chromium Authors. All rights reserved.
 // Use of this source code is governed by a BSD-style license that can be
 // found in the LICENSE file.
 
 var chrome = requireNative('chrome').GetChrome();
+var schemaRegistry = requireNative('schema_registry');
 
 function forEach(obj, f, self) {
   // For arrays, make sure the indices are numbers not strings - and in that
   // case this method can be more efficient by assuming the array isn't sparse.
   //
   // Note: not (instanceof Array) because the array might come from a different
   // context than our own.
   if (obj.constructor && obj.constructor.name == 'Array') {
     for (var i = 0; i < obj.length; i++)
       f.call(self, i, obj[i]);
@@ skipping 14 matching lines @@
   if (matches.length == 0) {
     return undefined;
   } else if (matches.length == 1) {
     return matches[0]
   } else {
     throw new Error("Failed lookup of field '" + field + "' with value '" +
                     value + "'");
   }
 }
 
-// Specify |currentApi| if this should return an API for $refs in the current
-// namespace.
-function loadRefDependency(ref, currentApi) {
-  var parts = ref.split(".");
-  if (parts.length > 1)
-    return chrome[parts.slice(0, parts.length - 1).join(".")];
-  else
-    return currentApi;
+function loadRefType(ref, defaultSchema) {

[not at google - send to devlin, 2013/03/21 00:01:13]

loadTypeSchema?

ref -> typeName?

(the fact that these are used for looking up $ref isn't interesting to this
method)

[cduvall, 2013/03/21 00:59:51]

Done.

+  var parts = ref.split('.');
+  if (parts.length == 1) {
+    if (!defaultSchema)
+      return undefined;

[not at google - send to devlin, 2013/03/21 00:01:13]

CHECK

[cduvall, 2013/03/21 00:59:51]

Done.

+    var types = defaultSchema.types;
+  } else {
+    var schemaName = parts.slice(0, parts.length - 1).join('.')
+    var types = schemaRegistry.GetSchema(schemaName).types;
+  }
+  for (var i = 0; i < types.length; ++i) {
+    if (types[i]['id'] == ref)

[not at google - send to devlin, 2013/03/21 00:01:13]

.id

[cduvall, 2013/03/21 00:59:51]

Done.

+      return types[i];
+  }

[not at google - send to devlin, 2013/03/21 00:01:13]

return null (nice to be explicit)

[cduvall, 2013/03/21 00:59:51]

returned undefined to be consistent. Is null better?

[not at google - send to devlin, 2013/03/21 01:36:55]

that is always a question with JS.

my rule at least: if it's defined (i.e. if we know what it is), then don't
return undefined. Here we know definitely that it's not found, so it's defined,
therefore it should be null.

OTOH void return functions have an undefined result, so should be undefined.

[cduvall, 2013/03/21 17:16:07]

Done.

 }
 
 exports.forEach = forEach;
-exports.loadRefDependency = loadRefDependency;
+exports.loadRefType = loadRefType;
 exports.lookup = lookup;