Jon Jagger
jon@jaggersoft.com
Table of Contents 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Notes DownloadECMA-334 C# Language Specificationpreviousnextprevious at this levelnext at this level 14 Expressionsprevious at this levelnext at this level 14.13 Assignment operatorsprevious at this levelnext at this level 14.13.1 Simple assignment Paragraph 11 The = operator is called the simple assignment operator. 2 In a simple assignment, the right operand must be an expression of a type that is implicitly convertible to the type of the left operand. 3 The operation assigns the value of the right operand to the variable, property, or indexer element given by the left operand. Paragraph 21 The result of a simple assignment expression is the value assigned to the left operand. 2 The result has the same type as the left operand and is always classified as a value. Paragraph 31 If the left operand is a property or indexer access, the property or indexer must have a set accessor. 2 If this is not the case, a compile-time error occurs. Paragraph 41 The run-time processing of a simple assignment of the form x = y consists of the following steps: [Note: The array covariance rules (§19.5) permit a value of an array type A[] to be a reference to an instance of an array type B[], provided an implicit reference conversion exists from B to A. Because of these rules, assignment to an array element of a reference-type requires a run-time check to ensure that the value being assigned is compatible with the array instance. In the example
string[] sa = new string[10];  
object[] oa = sa;  
oa[0] = null;      // Ok  
oa[1] = "Hello";     // Ok  
oa[2] = new ArrayList();  // ArrayTypeMismatchException  
the last assignment causes a System.ArrayTypeMismatchException to be thrown because an instance of ArrayList cannot be stored in an element of a string[]. end note]
Paragraph 51 When a property or indexer declared in a struct-type is the target of an assignment, the instance expression associated with the property or indexer access must be classified as a variable. 2 If the instance expression is classified as a value, a compile-time error occurs. [Note: Because of §14.5.4, the same rule also applies to fields. end note] [Example: Given the declarations:
struct Point  
{  
   int x, y;  
   public Point(int x, int y) {  
      this.x = x;  
      this.y = y;  
   }  
   public int X {  
      get { return x; }  
      set { x = value; }  
   }  
   public int Y {  
      get { return y; }  
      set { y = value; }  
   }  
}  
struct Rectangle  
{  
   Point a, b;  
   public Rectangle(Point a, Point b) {  
      this.a = a;  
      this.b = b;  
   }  
   public Point A {  
      get { return a; }  
      set { a = value; }  
   }  
   public Point B {  
      get { return b; }  
      set { b = value; }  
   }  
}  
in the example
Point p = new Point();  
p.X = 100;  
p.Y = 100;  
Rectangle r = new Rectangle();  
r.A = new Point(10, 10);  
r.B = p;  
the assignments to p.X, p.Y, r.A, and r.B are permitted because p and r are variables. However, in the example
Rectangle r = new Rectangle();  
r.A.X = 10;  
r.A.Y = 10;  
r.B.X = 100;  
r.B.Y = 100;  
the assignments are all invalid, since r.A and r.B are not variables. end example]
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