Why does a=(b++) have the same behavior as a=b++?

Question

I am writing a small test app in C with GCC 4.8.4 pre-installed on my Ubuntu 14.04. And I got confused for the fact that the expression a=(b++); behaves in the

Answer 1

Your expectations are completely unfounded.

Parentheses have no direct effect on the order of execution. They don't introduce sequence points into the expression and thus they don't force any side-effects to materialize earlier than they would've materialized without parentheses.

Moreover, by definition, post-increment expression b++ evaluates to the original value of b. This requirement will remain in place regardless of how many pair of parentheses you add around b++. Even if parentheses somehow "forced" an instant increment, the language would still require (((b++))) to evaluate to the old value of b, meaning that a would still be guaranteed to receive the non-incremented value of b.

Parentheses only affects the syntactic grouping between operators and their operands. For example, in your original expression a = b++ one might immediately ask whether the ++ apples to b alone or to the result of a = b. In your case, by adding the parentheses you simply explicitly forced the ++ operator to apply to (to group with) b operand. However, according to the language syntax (and the operator precedence and associativity derived from it), ++ already applies to b, i.e. unary ++ has higher precedence than binary =. Your parentheses did not change anything, it only reiterated the grouping that was already there implicitly. Hence no change in the behavior.