Restoring authorship annotation for <thegeorg@yandex-team.ru>. Commit 1 of 2.

1 files changed, 22 insertions, 22 deletions

diff --git a/contrib/restricted/abseil-cpp/absl/hash/hash.h b/contrib/restricted/abseil-cpp/absl/hash/hash.h
index 8282ea53c6..d26f1d0a9a 100644
--- a/contrib/restricted/abseil-cpp/absl/hash/hash.h
+++ b/contrib/restricted/abseil-cpp/absl/hash/hash.h
@@ -73,8 +73,8 @@
 #ifndef ABSL_HASH_HASH_H_
 #define ABSL_HASH_HASH_H_
 
-#include <tuple>
-
+#include <tuple> 
+ 
 #include "absl/hash/internal/hash.h"
 
 namespace absl {
@@ -216,26 +216,26 @@ ABSL_NAMESPACE_BEGIN
 template <typename T>
 using Hash = absl::hash_internal::Hash<T>;
 
-// HashOf
-//
-// absl::HashOf() is a helper that generates a hash from the values of its
-// arguments.  It dispatches to absl::Hash directly, as follows:
-//  * HashOf(t) == absl::Hash<T>{}(t)
-//  * HashOf(a, b, c) == HashOf(std::make_tuple(a, b, c))
-//
-// HashOf(a1, a2, ...) == HashOf(b1, b2, ...) is guaranteed when
-//  * The argument lists have pairwise identical C++ types
-//  * a1 == b1 && a2 == b2 && ...
-//
-// The requirement that the arguments match in both type and value is critical.
-// It means that `a == b` does not necessarily imply `HashOf(a) == HashOf(b)` if
-// `a` and `b` have different types. For example, `HashOf(2) != HashOf(2.0)`.
-template <int&... ExplicitArgumentBarrier, typename... Types>
-size_t HashOf(const Types&... values) {
-  auto tuple = std::tie(values...);
-  return absl::Hash<decltype(tuple)>{}(tuple);
-}
-
+// HashOf 
+// 
+// absl::HashOf() is a helper that generates a hash from the values of its 
+// arguments.  It dispatches to absl::Hash directly, as follows: 
+//  * HashOf(t) == absl::Hash<T>{}(t) 
+//  * HashOf(a, b, c) == HashOf(std::make_tuple(a, b, c)) 
+// 
+// HashOf(a1, a2, ...) == HashOf(b1, b2, ...) is guaranteed when 
+//  * The argument lists have pairwise identical C++ types 
+//  * a1 == b1 && a2 == b2 && ... 
+// 
+// The requirement that the arguments match in both type and value is critical. 
+// It means that `a == b` does not necessarily imply `HashOf(a) == HashOf(b)` if 
+// `a` and `b` have different types. For example, `HashOf(2) != HashOf(2.0)`. 
+template <int&... ExplicitArgumentBarrier, typename... Types> 
+size_t HashOf(const Types&... values) { 
+  auto tuple = std::tie(values...); 
+  return absl::Hash<decltype(tuple)>{}(tuple); 
+} 
+ 
 // HashState
 //
 // A type erased version of the hash state concept, for use in user-defined