ATMLModelLibrary.model.signal.basic.Control.Deserialize C# (CSharp) Метод

Deserialize() публичный статический Метод

Deserializes workflow markup into an Control object
public static Deserialize ( string input, Control &obj, System &exception ) : bool
input string string workflow markup to deserialize
obj Control Output Control object
exception System output Exception value if deserialize failed
Результат bool
        public static bool Deserialize(string input, out Control obj, out System.Exception exception)
        {
            exception = null;
            obj = default(Control);
            try
            {
                obj = Deserialize(input);
                return true;
            }
            catch (System.Exception ex)
            {
                exception = ex;
                return false;
            }
        }

Same methods

Control::Deserialize ( System s ) : Control
Control::Deserialize ( string input ) : Control
Control::Deserialize ( string input, Control &obj ) : bool