Remove explicit return type of the ranget::map function

Because infering the return type from the template parameters is a
sizeable portion of the function implementation. So using an `auto` type
tells the compiler that the type is in the return statement and saves us
duplicating the same work in two places.

Author: thomasspriggs

src/util/range.h

@@ -418,9 +418,7 @@ struct ranget final
   /// a value through `f`. `f` may take a move-only typed parameter by const
   /// reference. 'f' may also construct and return a move-only typed value.
   template <typename functiont>
-  auto map(functiont &&f) -> ranget<map_iteratort<
-    iteratort,
-    typename std::result_of<functiont(value_type)>::type>>
+  auto map(functiont &&f)
   {
     using outputt = typename std::result_of<functiont(value_type)>::type;
     auto shared_f = std::make_shared<