Change result_of to invoke_result in gloo/transport/uv/libuv.h

Summary: C++20 has [eliminated](https://en.cppreference.com/w/cpp/types/result_of) `result_of` in favour of `invoke_result`. It's mysterious that this code even still works, but, nevertheless, I'm fixing it.

Differential Revision: D56987407

fbshipit-source-id: 3ebe7bb70af42590fa88e49641c5fcd559aa2ad9

Author: r-barnes

gloo/transport/uv/libuv.h

@@ -346,7 +346,7 @@ class Handle : public Resource<T, U>, public BaseHandle {
   }
 
   template <typename F, typename... Args>
-  typename std::result_of<F(Args...)>::type invoke(F&& f, Args&&... args) {
+  typename std::invoke_result<F, Args...>::type invoke(F&& f, Args&&... args) {
     return std::forward<F>(f)(std::forward<Args>(args)...);
   }
 
@@ -383,8 +383,8 @@ class Request : public Resource<T, U>, public BaseRequest {
   // assumption that it is unleaked when the callback gets called.
   template <typename F, typename... Args>
   typename std::enable_if<
-      !std::is_void<typename std::result_of<F(Args...)>::type>::value,
-      typename std::result_of<F(Args...)>::type>::type
+      !std::is_void<typename std::invoke_result<F, Args...>::type>::value,
+      typename std::invoke_result<F, Args...>::type>::type
   invoke(F&& f, Args&&... args) {
     auto err = std::forward<F>(f)(std::forward<Args>(args)...);
     if (err) {