From 569cf96d06f612e1464704e85b16eef3add37be2 Mon Sep 17 00:00:00 2001
From: Nathan Shively-Sanders <nathansa@microsoft.com>
Date: Mon, 9 Nov 2015 10:48:22 -0800
Subject: [PATCH] Fix new and improved linter errors

---
 src/compiler/checker.ts | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/src/compiler/checker.ts b/src/compiler/checker.ts
index ff30b67d9c6..2494009333b 100644
--- a/src/compiler/checker.ts
+++ b/src/compiler/checker.ts
@@ -2876,7 +2876,7 @@ namespace ts {
             }
             const baseTypeNode = getBaseTypeNodeOfClass(type);
             let baseType: Type;
-            let originalBaseType = baseConstructorType && baseConstructorType.symbol ? getDeclaredTypeOfSymbol(baseConstructorType.symbol) : undefined;
+            const originalBaseType = baseConstructorType && baseConstructorType.symbol ? getDeclaredTypeOfSymbol(baseConstructorType.symbol) : undefined;
             if (baseConstructorType.symbol && baseConstructorType.symbol.flags & SymbolFlags.Class &&
                 !baseTypeHasUnappliedOuterTypeParameters(originalBaseType)) {
                 // When base constructor type is a class with no captured type arguments we know that the constructors all have the same type parameters as the
@@ -2916,8 +2916,8 @@ namespace ts {
         }
 
         function baseTypeHasUnappliedOuterTypeParameters(type: Type): boolean {
-            let originalBaseType = <InterfaceType>type;
-            let originalTypeReference = <TypeReference>type;
+            const originalBaseType = <InterfaceType>type;
+            const originalTypeReference = <TypeReference>type;
             if (originalBaseType.outerTypeParameters) {
                 // an unapplied type type parameter is one 
                 // whose argument symbol is still the same as the parameter symbol