assignment operator - C++ function returns a rvalue, but that can be assigned a new value? -

the code follows:

 #include <iostream>  using namespace std;   class {   };   rtbyvalue() { return a();  }   void passbyref(a &aref) {     // nothing  }   int main() {     aa;     rtbyvalue() = aa;            // compile without errors     passbyref(rtbyvalue());      // compile error       return 0;  } 

the g++ compiler gives following error:

d.cpp: in function ‘int main()’: d.cpp:19:23: error: invalid initialization of non-const reference of type ‘a&’ rvalue of type ‘a’ d.cpp:12:6: error: in passing argument 1 of ‘void passbyref(a&)’ 

it says can't pass rvalue argument of non-const reference, i'm confused why can assign rvalue, code shows.

passing rvalue rtbyvalue() function expects lvalue reference doesn't work because require lvalue reference argument initialized rvalue. §8.5.3/5 describes how lvalue references can initialized – won't quote in full, says lvalue reference can initialized

• either lvalue reference
• or can converted lvalue reference of intermediary type
• or rvalue, if lvalue reference initialize const-reference

since argument need initialize not const-reference, none of applies.

on other hand,

rtbyvalue() = aa;  

i.e., assigning temporary object, possible because of:

(§3.10/5) lvalue object necessary in order modify object except rvalue of class type can used modify referent under circumstances. [ example: member function called object (9.3) can modify object. — end example ]

so works because a of class-type, , (implicitly defined) assignment operator member function. (see this related question further details.)

(so, if rtbyvalue() return, example, int, assignment wouldn't work.)