In C99 however, you are allowed to do &٭a because when the operand of & is the result of the unary ٭ operator > neither that operator nor the & operator is eva

In C99 however, you are allowed to do &٭a because when the operand of & is the result of the unary ٭ operator > neither that operator nor the & operator is evaluated and the result is as if both were omitted, except that the constraints on the operators still apply and the result is not an lvalue