Why are tokens wrapped by parentheses not r-value expressions?

Question

Consider the following code:

#include <iostream>

struct Foo
{
  Foo() : bar( 0 ) {}

  int bar;
};


int main()
{
  Foo foo;

  ++(foo.bar);

  std::cout<< foo.bar << std::endl;

  system("pause");
  return 0;
};

Why does foo.bar evaluate to 1?

Doesn't the parentheses in (foo.bar) create an unnamed (r-value) expression which is then incremented?

Answer 1

Because the standard explicitly states that in 3.4.2 para 6:

A parenthesized expression is a primary expression whose type and value are identical to those of the enclosed expression. The presence of parentheses does not affect whether the expression is an lvalue.

emphasis mine.

Answer 2

No, parenthesis do not have any meaning besides changing the order of operations.

To create an rvalue you need to use special function std::move(x).