关于为函数赋值

Question

I have the following code snippet:

#include <iostream>
using namespace std;


class X {
int i;
public:
    X(int ii = 0);
    void modify();
};

X::X(int ii) { i = ii; }
void X::modify() { i++; }

X f5() {   return X(); }
const X f6() {   return X(); }

void f7(X& x) {
  x.modify();
}

int f()
{
    return 18;
}
int main() {
  f5() = X(1);          /// Why does this work??? Isn't f5() an rvalue ??? (*)
  f5().modify();
///  f7(f5());          /// cannot bind non-const lvalue reference of type 'X&' to an rvalue of type X; not contradictory with (*)?
//!  f6() = X(1);
//!  f6().modify();
//!  f7(f6());
//! f() = 12; this also doesn't work
  return 0;
}

How can f5() = X() work? Isn't f5() a rvalue? Then why doesn't f() = 12 work? What is the difference? Also, the error that f7(f5()) generates doesn't say that f5() is a rvalue? What am I missing?

Answer 1

f5() indeed yields an rvalue and f5() = X(1); invokes implicitly generated move operator = . It will stop working if this operator is deleted: void operator =(X &&) = delete; .

Also the error that f7(f5()) generates does say that f5() is an rvalue:

to an rvalue of type X