ATMLModelLibrary.model.equipment.Resources.Deserialize C# (CSharp) Метод

Deserialize() публичный статический Метод

public static Deserialize ( Stream s ) : Resources
s Stream
Результат Resources
        public static Resources Deserialize(Stream s)
        {
            return ((Resources)(Serializer.Deserialize(s)));
        }
        #endregion

Same methods

Resources::Deserialize ( string input ) : Resources
Resources::Deserialize ( string input, Resources &obj ) : bool
Resources::Deserialize ( string input, Resources &obj, Exception &exception ) : bool