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7.13.2 Compound assignment

Visual Studio .NET 2003

An operation of the form x op= y is processed by applying binary operator overload resolution (Section 7.2.4) as if the operation was written x op y. Then,

  • If the return type of the selected operator is implicitly convertible to the type of x, the operation is evaluated as x = x op y, except that x is evaluated only once.
  • Otherwise, if the selected operator is a predefined operator, if the return type of the selected operator is explicitly convertible to the type of x, and if y is implicitly convertible to the type of x, then the operation is evaluated as x = (T)(x op y), where T is the type of x, except that x is evaluated only once.
  • 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. For 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.

When the left operand of a compound assignment is a property access or indexer access, the property or indexer must 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 Section 7.2.6.2. 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. In the example

byte b = 0;
char ch = '\0';
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

the intuitive reason for each error is that a corresponding simple assignment would also have been an error.

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