ATML1671Translator.model.numbers.Deserialize C# (CSharp) Метод

Deserialize() публичный статический Метод

public static Deserialize ( string input, numbers &obj ) : bool
input string
obj numbers
Результат bool
        public static bool Deserialize(string input, out numbers obj)
        {
            Exception exception;
            return Deserialize(input, out obj, out exception);
        }

Same methods

numbers::Deserialize ( string input, numbers &obj, Exception &exception ) : bool
numbers::Deserialize ( Stream s ) : numbers
numbers::Deserialize ( string input ) : numbers