Method call acting unexpectedly like an l-value

Question

Can anybody explain why this code compiles:

typedef struct longlong
{
  unsigned long low;
  long high;
}
longlong;

typedef longlong Foo;    

struct FooStr         


        

Answer 1

This works because longlong is a class, and as such = is longlong::operator =, the implicitly-defined assignment operator. And you can call member functions on temporaries as long as they're not qualified with & (the default operator = is unqualified).

To restrict the assignment operator to lvalues, you can explicitly default it with an additional ref-qualifier:

struct longlong
{
    longlong &operator = (longlong const &) & = default;
    //                                      ^

    // ...
};