How to convert Vec<T> to bytes::Bytes without new allocation and copy?

help
1

Currently Bytes only implements From<Vec<u8>>, and bytemuck doesn't provide conversion from Vec<T> of arbitrary alignment to Vec<u8>.

2

I could use a Vec<u8> as a Vec<T>.
Pushing becomes std::io::Write::write(bytemuck::bytes_of(&t)).
And view the Vec<u8> as &[T] using bytemuck::cast_slice().
But it is a little bit inconvenient.

3

This may actually fail in some cases.

4
5

what makes your type not able to use cast_vec()?

6

bytes::Bytes::from_owner takes impl AsRef<[u8]>, so you could write a custom wrapper implementing it to convert Vec<T> into bytes::Bytes like this:

7

Thanks. This is exactly what I need.

8

try_cast_vec() only works when

  • The start and end content type of the Vec must have the exact same alignment.
9

Thanks for replying. But leaking is not what I need.