1 write to ExceptionDispatchInfo
System.Text.Json (1)
System\Text\Json\Serialization\JsonSerializerOptions.Caching.cs (1)
382ExceptionDispatchInfo = exception;
1 reference to ExceptionDispatchInfo
System.Text.Json (1)
System\Text\Json\Serialization\JsonSerializerOptions.Caching.cs (1)
388ExceptionDispatchInfo?.Throw();