If we know a call will actually not reach an element anymore, because we have more information on t…

tests/compiler/dart2js/simple_inferrer_unregister_call_test.dart

+// Copyright (c) 2013, the Dart project authors.  Please see the AUTHORS file
+// for details. All rights reserved. Use of this source code is governed by a
+// BSD-style license that can be found in the LICENSE file.
+
+import 'package:expect/expect.dart';
+
+import 'compiler_helper.dart';
+import 'parser_helper.dart';
+
+const String TEST = """
+var a = '';
+class A {
+  operator+(other) => other;
+}
+
+foo() {
+  // The following '+' call will first say that it may call A::+,
+  // String::+, or int+. After all methods have been analyzed, we know

[kasperl, 2013/04/23 14:39:03]

int::+

[ngeoffray, 2013/04/23 14:42:00]

Done.

+  // that a is of type String, and therefore, this method cannot call
+  // A::+. Therefore, the type of the parameter of A::+ will be the
+  // one given by the other calls.
+  return a + 'foo';
+}
+
+main() {
+  new A() + 42;
+  foo();
+}
+""";
+
+
+void main() {
+  Uri uri = new Uri.fromComponents(scheme: 'source');
+  var compiler = compilerFor(TEST, uri);
+  compiler.runCompiler(uri);
+  var typesInferrer = compiler.typesTask.typesInferrer;
+
+  checkReturnInClass(String className, String methodName, type) {
+    var cls = findElement(compiler, className);
+    var element = cls.lookupLocalMember(buildSourceString(methodName));
+    Expect.equals(type, typesInferrer.internal.returnTypeOf[element]);
+  }
+
+  checkReturnInClass('A', '+', typesInferrer.intType);
+}