C#的+=运算符两例

核心提示： Otherwise, the compound assignment is invalid, and a compile-time error occurs.The term “evaluated only once” means that in the evalu

Otherwise, the compound assignment is invalid, and a compile-time error occurs.

The term “evaluated only once” means that in the evaluation of x op y, the results of any constituent expressions of x are temporarily saved and then reused when performing the assignment to x. [Example: In the assignment A()[B()] += C(), where A is a method returning int[], and B and C are methods returning int, the methods are invoked only once, in the order A, B, C. end example]

When the left operand of a compound assignment is a property access or indexer access, the property or indexer shall have both a get accessor and a set accessor. If this is not the case, a compile-time error occurs.

The second rule above permits x op= y to be evaluated as x = (T)(x op y) in certain contexts. The rule exists such that the predefined operators can be used as compound operators when the left operand is of type sbyte, byte, short, ushort, or char. Even when both arguments are of one of those types, the predefined operators produce a result of type int, as described in §14.2.6.2. Thus, without a cast it would not be possible to assign the result to the left operand.

The intuitive effect of the rule for predefined operators is simply that x op= y is permitted if both of x op y and x = y are permitted. [Example: In the following code

C#代码　　　

byte b = 0;
char ch = '';
int i = 0;
b += 1;    // Ok
b += 1000;      // Error, b = 1000 not permitted
b += i;        // Error, b = i not permitted
b += (byte)i;    // Ok
ch += 1;       // Error, ch = 1 not permitted
ch += (char)1;   // Ok

	 
	 
	
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