Christian Böhme wrote:
chris@xxxxxxxxxxxxxxxxxxxx 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
