Pin::as_mut(&mut self)

Question

From the source forPin::as_mut - pin.rs:

impl<P: DerefMut> Pin<P> {
    ...
    pub fn as_mut(&mut self) -> Pin<&mut P::Target> {    
        unsafe { Pin::new_unchecked(&mut *self.pointer) }
    }
    ...
}

I don't understand why &mut *self.pointer is used instead of self.pointer , What is wrong with the following sentence?

unsafe { Pin::new_unchecked(self.pointer) }

Answer 1

Pin::new_unchecked(self.pointer) would just return a Pin<P> instead of a Pin<&mut P::Target> , deref coercions won't be considered. It will also attempt to move pointer out of self and cause a compiler error.

Using * will dereference self.pointer into a P::Target and &mut makes it return a mutable reference.