Convert T to &mut T with `.`?

According to my understanding, next needs a &mut Test, but create_test() returns a Test.

Why can this be compiled?

My guess is that . will implicitly convert Test to &mut Test, I am not sure. Can somebody explain more about this?

pub struct Test {
    t: u64,
}

fn create_test() -> Test {
    Test {
        t: 1
    }
} 

impl Test {
    pub fn next(&mut self) {
        self.t = 10;
    }
}

fn main() {
    let mut t = Test { t: 20 };
    t.next();

    create_test().next();  // here
}

>Solution :

This is explained in the Method-call expressions section of the book.

When looking up a method call, the receiver may be automatically dereferenced or borrowed in order to call a method.

This is exactly what is happening here. The rust compiler is automatically borrowing value returned create_test.