The generic parameters on a type definition can specify default values.
Currently `parse_generics()` cannot handle this though. For example when
parsing the following generics:

    <T: Clone, const N: usize = 0>

The `impl_generics` will be set to `T: Clone, const N: usize = 0` and
`ty_generics` will be set to `T, N`. Now using the `impl_generics` on an
impl block:

    impl<$($impl_generics)*> Foo {}

will result in invalid Rust code, because default values are only
available on type definitions.

Therefore add parsing support for generic parameter default values using
a new kind of generics called `decl_generics` and change the old
behavior of `impl_generics` to not contain the generic parameter default
values.

Now `Generics` has three fields:
- `impl_generics`: the generics with bounds
  (e.g. `T: Clone, const N: usize`)
- `decl_generics`: the generics with bounds and default values
  (e.g. `T: Clone, const N: usize = 0`)
- `ty_generics`:  contains the generics without bounds and without
  default values (e.g. `T, N`)

`impl_generics` is designed to be used on `impl<$impl_generics>`,
`decl_generics` for the type definition, so `struct Foo<$decl_generics>`
and `ty_generics` whenever you use the type, so `Foo<$ty_generics>`.

Here is an example that uses all three different types of generics:

    let (Generics { decl_generics, impl_generics, ty_generics }, rest) = parse_generics(input);
    quote! {
        struct Foo<$($decl_generics)*> {
            // ...
        }

        impl<$impl_generics> Foo<$ty_generics> {
            fn foo() {
                // ...
            }
        }
    }

The next commit contains a fix to the `#[pin_data]` macro making it
compatible with generic parameter default values by relying on this new
behavior.
Signed-off-by: Benno Lossin <benno.lossin@proton.me>
Reviewed-by: Alice Ryhl <aliceryhl@google.com>
Link: https://lore.kernel.org/r/20240309155243.482334-1-benno.lossin@proton.meSigned-off-by: Miguel Ojeda <ojeda@kernel.org>