What exactly happens when returning a const reference to a local object?

struct A {
    A(int) : i(new int(783)) {
        std::cout << "a ctor" << std::endl;
    }

    A(const A& other) : i(new int(*(other.i))) {
        std::cout << "a copy ctor" << std::endl;
    }

    ~A() {
        std::cout << "a dtor" << std::endl;
        delete i;
    }

    void get() {
        std::cout << *i << std::endl;
    }

private:
    int* i;
};

const A& foo() {
    return A(32);
}

const A& foo_2() {
    return 6;
}

int main()
{
    A a = foo();
    a.get();
}

I know that returning references to local values ​​is bad. But, on the other hand, a constant reference should extend the temporary lifetime of an object.

This code creates UB output. So no life extension.

Why? I mean, can someone explain what is going on step by step?

Where is the error in my logical chain?

Foo ():

• A (32) - ctor

• return A (32) - creates a const reference to the local object and returns

• A a = foo (); - a is initialized with the return value foo (), the return value goes beyond the scope (outside the expression) and is destroyed, but a is already initialized;

(But actually, the destructor is called before the copy constructor)

foo_2 ():

• return 6 - A , ( )

• A a = foo(); - a foo(), ( ) , a ;

( )