Fix potential null ref with unknown variable types

Author: joemcbride

src/GraphQL.Tests/Validation/KnownTypeNamesTests.cs

@@ -23,6 +23,28 @@ fragment PetFields on Pet {
             ");
         }
 
+        [Fact]
+        public void unknown_nonnull_type_name_is_invalid()
+        {
+            ShouldFailRule(_ =>{
+                _.Query = @"
+                    query Foo($var: Abcd!) {
+                        user(id: 4) {
+                            pets {
+                                ... on Pet { name },
+                                ...PetFields
+                            }
+                        }
+                    }
+                    fragment PetFields on Pet {
+                        name
+                    }
+                    ";
+
+                _.Error(Rule.UnknownTypeMessage("Abcd", null), 2, 37);
+            });
+        }
+
         [Fact]
         public void unknown_type_names_are_invalid()
         {

src/GraphQL/Types/TypeExtensions.cs

@@ -7,10 +7,16 @@ public static class TypeExtensions
     {
         public static IGraphType GraphTypeFromType(this IType type, ISchema schema)
         {
+            if (type == null) return null;
+
             if (type is NonNullType)
             {
                 var nonnull = (NonNullType)type;
                 var ofType = GraphTypeFromType(nonnull.Type, schema);
+                if(ofType == null)
+                {
+                    return null;
+                }
                 var nonnullGraphType = typeof(NonNullGraphType<>).MakeGenericType(ofType.GetType());
                 var instance = (NonNullGraphType)Activator.CreateInstance(nonnullGraphType);
                 instance.ResolvedType = ofType;
@@ -21,6 +27,10 @@ public static IGraphType GraphTypeFromType(this IType type, ISchema schema)
             {
                 var list = (ListType)type;
                 var ofType = GraphTypeFromType(list.Type, schema);
+                if(ofType == null)
+                {
+                    return null;
+                }
                 var listGraphType = typeof(ListGraphType<>).MakeGenericType(ofType.GetType());
                 var instance = (ListGraphType)Activator.CreateInstance(listGraphType);
                 instance.ResolvedType = ofType;