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

Question

The code is as follows:

 #include <iostream>
 using namespace std;

 class A {

 };

 A rtByValue() {
return A();
 }

 void passByRef(A &aRef) {
    // do nothing
 }

 int main() {
    A aa;
    rtByValue() = aa;            // compile without errors
    passByRef(rtByValue());      // compile with error 

    return 0;
 }

The g++ compiler gives the following error:

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

It says that I can't pass an rvalue as an argument of a non-const reference, but what I'm confused about is why I can assign to this rvalue, just as the code shows.

Answer 1

Passing the rvalue rtByValue() to a function that expects an lvalue reference doesn't work because this would require the lvalue reference argument to be initialized from an rvalue. §8.5.3/5 describes how lvalue references can be initialized – I won't quote it in full, but it basically says that an lvalue reference can be initialized

• either from another lvalue reference
• or something that can be converted to an lvalue reference of an intermediary type
• or from an rvalue, but only if the lvalue reference we initialize is a const-reference

Since the argument we need to initialize is not a const-reference, none of this applies.

On the other hand,

rtByValue() = aa; 

i.e., assigning to a temporary object, is possible because of:

(§3.10/5) An lvalue for an object is necessary in order to modify the object except that an rvalue of class type can also be used to modify its referent under certain circumstances. [ Example: a member function called for an object (9.3) can modify the object. — end example ]

So this works only because A is of class-type, and the (implicitly defined) assignment operator is a member function. (See this related question for further details.)

(So, if rtByValue() were to return, for example, an int, then the assignment wouldn't work.)