Allow type aliases to take (ignored) type arguments

type Foo<T> = Bar<T,T> is treated as Foo = Bar; Foo<T> = @?T isn't supported.

hphp/test/quick/typedef_typearg.php

@@ -0,0 +1,12 @@
+<?hh
+
+type Foo<X> = string;
+
+interface Dummy {}
+
+function main(Foo<Dummy> $x) {
+  echo "Hi\n";
+}
+
+main('42');
+main(42);

hphp/test/quick/typedef_typearg.php.expectf

@@ -0,0 +1,2 @@
+Hi
+HipHop Fatal error: Argument 1 passed to main() must be an instance of Foo, int given in %s on line 9

hphp/test/quick/typedef_typearg.php.hphp_opts

@@ -0,0 +1 @@
+hphp_opts.hip_hop_syntax

hphp/util/parser/hphp.y

@@ -2338,8 +2338,10 @@ class_constant:
  */
 
 sm_typedef_statement:
-    T_TYPE ident '=' sm_type ';'       { only_in_strict_mode(_p);
-                                         _p->onTypedef($$, $2, $4); }
+    T_TYPE sm_name_with_typevar
+    '=' sm_type ';'                    { only_in_strict_mode(_p);
+                                         _p->onTypedef($$, $2, $4);
+                                         _p->popTypeScope(); }
 ;
 
 sm_name_with_type:  /* foo -> int foo */