--- Chris Wolstenholme <chris@xxxxxxxxxxxxxxxxxxxx> wrote: > 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. This has been answered by others, but for emphaphis I want too reiterate that according to the C and C++ language rules any two modifications of the same variable between sequence points produces undefined behavior. a = ++b + ++b; This is undefined behavior in both C and C++. Undefined behavior in the C++ standard is defined as "behavior, such as might arise upon use of an erroneous program construct or erroneous data, for which this International Standard imposes no requirements." Hope that helps, --------------------- Ivan Novick http://www.0x4849.net
