我想从类型A的现有对象创建类型B的新对象.B继承自A.我想确保将类型A的对象中的所有属性值复制到对象B型是什么?实现这一目标的最佳方法是什么?
class A
{
public int Foo{get; set;}
public int Bar{get; set;}
}
class B : A
{
public int Hello{get; set;}
}
class MyApp{
public A getA(){
return new A(){ Foo = 1, Bar = 3 };
}
public B getB(){
A myA = getA();
B myB = myA as B; //invalid, but this would be a very easy way to copy over the property values!
myB.Hello = 5;
return myB;
}
public B getBAlternative(){
A myA = getA();
B myB = new B();
//copy over myA's property values to myB
//is there a better way of doing the below, as it could get very tiresome for large numbers of properties
myB.Foo = myA.Foo;
myB.Bar = myA.Bar;
myB.Hello = 5;
return myB;
}
}
答案 0 :(得分:4)
class A {
public int Foo { get; set; }
public int Bar { get; set; }
public A() { }
public A(A a) {
Foo = a.Foo;
Bar = a.Bar;
}
}
class B : A {
public int Hello { get; set; }
public B()
: base() {
}
public B(A a)
: base(a) {
}
}
修改强>
如果您不想复制每个属性,可以使A和B可序列化并序列化您的A实例(比如说流到Stream,而不是文件)并使用它初始化您的B实例。但我警告你,这很恶心,并且有很多开销:
A a = new A() {
Bar = 1,
Foo = 3
};
System.Xml.Serialization.XmlSerializer xs = new System.Xml.Serialization.XmlSerializer(typeof(A));
System.IO.Stream s = new System.IO.MemoryStream();
xs.Serialize(s, a);
string serial = System.Text.ASCIIEncoding.ASCII.GetString(ReadToEnd(s));
serial = serial.Replace("<A xmlns", "<B xmlns");
serial = serial.Replace("</A>", "</B>");
s.SetLength(0);
byte[] bytes = System.Text.ASCIIEncoding.ASCII.GetBytes(serial);
s.Write(bytes, 0, bytes.Length);
s.Position = 0;
xs = new System.Xml.Serialization.XmlSerializer(typeof(B));
B b = xs.Deserialize(s) as B;
您可以在ReadToEnd
here上获得更多信息。
答案 1 :(得分:3)
您可以为B:
定义显式的强制转换运算符 public static explicit operator B(A a)
{
B b = new B();
b.Foo = a.Foo;
b.Bar = a.Bar;
b.Hello = 5;
return b;
}
所以你可以这样做:
B myB = (B)myA;
答案 2 :(得分:0)
这在OO机制中是不可能的,但是为此用途创建了AutoMapper。