Can I assign to the return value of member access operator?

I've read that the Member Access Operators dot operator . and arrow operator -> return values:

The arrow operator requires a pointer operand and yields an lvalue. The dot operator yields an lvalue if the object from which the member is fetched is an lvalue; otherwise the result is an rvalue.

This is from C++ Primer 5 edition.

So I imagine I can assign a value whenever a non-const lvalue is the return of their expression e.g:

struct foo {
    int x_;
    const int y_ = 17; ;
    void bar() { cout << "bar()" << endl;}
}f, *pf;

pf = &f;

(pf->bar()) = 75; // error
cout << f.x_ << endl;
(f.bar()) = 12;// error
(f.x_) = 23;
cout << "f_x: " << f.x_ << endl;
(pf->y_) = 34;// error

I am confused about assigning to the return value of arrow operator. Above it is said that -> always returns an lvalue but it fails if I try to assign to that value.

Can anyone explain to me the paragraph above from the C++ book. Thank you.

Solution

Above it is said that -> always returns an lvalue but it fails if I try to assign to that value.

This is about member variables, not functions. bar returns void so you can never assign to it. x_ works since it gives you an integer lvalue expression. y_ fails because it is const, and you can't assign to a const variable.

So, in

foo bar;
foo* f = &bar;
f->x_;
f->y_;

both member accesses yield an lvalue expression. In

foo f;
f.x_;
f.y_;

you again have lvalues since f is an lvalue. However, in

foo{}.x_;
foo{}.y_;

both member accesses are rvalues since foo{} is an rvalue.