Why are tokens wrapped by parentheses not r-value expressions?
c++, parentheses, rvalue
Solution
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.
Problem
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?