How does the mechanism behind the creation of boxed traits work?

However, variant c does not, probably because the new function takes only values of the same type which is not the case since Fooer != i32.

No, it's because there is no new function for Box<dyn Fooer>. In the documentation:

impl<T> Box<T>

pub fn new(x: T) -> Box<T>

Most methods on Box<T> allow T: ?Sized, but new is defined in an impl without a T: ?Sized bound. That means you can only call Box::<T>::new when T is a type with a known size. dyn Fooer is unsized, so there simply isn't a new function to call.

In fact, that function can't exist in today's Rust. Box<T>::new needs to know the concrete type T so that it can allocate memory of the right size and alignment. Therefore, you can't erase T before you send it to Box::new. (It's conceivable that future language extensions may allow functions to accept unsized parameters; however, it's unclear whether even unsized_locals would actually enable Box<T>::new to accept unsized T.)

For the time being, unsized types like dyn Fooer can only exist behind a "fat pointer", that is, a pointer to the object and a pointer to the implementation of Fooer for that object. How do you get a fat pointer? You start with a thin pointer and coerce it. That's what's happening in these two lines:

let d: Box<Fooer> = Box::new(32);        // works, creates a Box<Fooer>
let e: Box<Fooer> = Box::<i32>::new(32); // works, creates a Box<Fooer>

Box::new returns a Box<i32>, which is then coerced to Box<Fooer>. You could consider this a conversion, but the Box isn't changed; all the compiler does is stick an extra pointer on it and forget its original type. rodrigo's answer goes into more detail about the language-level mechanics of this coercion.

Hopefully all of this goes to explain why the answer to

Is there a way to create a Box<Fooer> directly from an i32?

is "no": the i32 has to be boxed before you can erase its type. It's the same reason you can't write let x: Fooer = 10i32.

Related

• Why can't I write a function with the same type as Box::new?
• Are polymorphic variables allowed?
• How do you actually use dynamically sized types in Rust?
• Why is `let ref a: Trait = Struct` forbidden?

I'll try to explain what conversions (coercions) happen in your code.

There is a marker trait named Unsize that, between others:

Unsize is implemented for:

• T is Unsize<Trait> when T: Trait.
• [...]

This trait, AFAIK, is not used directly for coercions. Instead, CoerceUnsized is used. This trait is implemented in a lot of cases, some of them are quite expected, such as:

impl<'a, 'b, T, U> CoerceUnsized<&'a U> for &'b T 
where
    'b: 'a,
    T: Unsize<U> + ?Sized,
    U: ?Sized

that is used to coerce &i32 into &Fooer.

The interesting, not so obvious implementation for this trait, that affects your code is:

impl<T, U> CoerceUnsized<Box<U>> for Box<T> 
where
    T: Unsize<U> + ?Sized,
    U: ?Sized

This, together with the definition of the Unsize marker, can be somewhat read as: if U is a trait and T implements U, then Box<T> can be coerced into Box<U>.

About your last question:

Is there a way to create a Box<Fooer> directly from an i32? If not: why not?

Not that I know of. The problem is that Box::new(T) requires a sized value, since the value passed is moved into the box, and unsized values cannot be moved.

In my opinion, the easiest way to do that is to simply write:

let c = Box::new(42) as Box<Fooer>;

That is, you create a Box of the proper type and then coerce to the unsized one (note it looks quite similar to your d example).