is a temporary l-value or not?

Question

Why does the first line in main compiles but the second doesn't? Both are temporaries I think but one is treated as l-value and the other not..

class complex
{
   public:
     complex() : r(0),i(0) {}
     complex(double r_, double i_) : r(r_), i(i_)
    {

    }

  private:
    double r;
    double i;
 };

int main()
{
   complex(2,2) = complex(1,2);
   char() = char(2);
}

Answer 1

On class types, the assignment operator is a member function. That is, a = b is just syntatic sugar for a.operator=(b). And it is perfectly fine to call member functions on temporaries.

Please note that in C++, the term lvalue has nothing to do with the left-hand side of an assignment. As your example demonstrates, it is perfectly fine to assign to rvalues of class type. Also, there are lvalues which you cannot assign to, for example arrays and/or constants, especially string literals.