increasing var twice in a statement

[Chris Wolstenholme]

Christian Böhme wrote:
> chris@blueband.demon.co.uk wrote:
> 
>> I feel 10 is the correct answer due to the pre-increment operator 
>> having a
>> higher precedence than the addition operator. This would mean both
>> increments will be done before the addition (creating 5 + 5).
> 
> That's interesting.  With b having been initialized to 3 but ending
> up at 5 on _both_ sides of "+" would imply that b was then incremented
> _four_ times (which is clarly not what the semantics of the statement
> are).
> 
> I suspect a 10 to be the result of overly "smart" playing with the
> operands of the addition on the compiler's part: b is expected to
> be incremented twice as it must be evaluated twice according to
> the definition of the addition operation.  However, the compiler
> also recognizes that the expression "++b" as operand to the "+"
> operator appears twice which it transforms into a multiplication
> of the addition's operand and 2.  Remember also that addition is
> a commutative operation which means it is irrelevant in which order
> its operands are evaluated to produce the result.
> 

My reasoning was as follows:

1) The statement was a = ++b + ++b;
2) b=3 before this statement
3) The operator ++ on the right hand side of the + operator is executed 
first by rules of precedence. As it is pre-increment, b becomes 4.
4)The operator ++ on the left hand side of the + operator is then 
executed. As it is also pre-increment, b is again altered to become 5.
5) Then the operator + is executing adding b (now 5) to b (still 5).

That's only two increments of b, but both before the addition operator.

Chris