How to convert bitvec to slice?

[David-OConnor]

Hi! I'm attempting to shift a &[u8] by 4 bits, using the bitvec crate.

Here's my naive guess from the Readme:

let bits  = (data.view_bits::<Msb0>()[4..]).as_raw_slice();

error:

51 |     let bits  = (data.view_bits::<Msb0>()[4..]).as_raw_slice()[..];
   |                                                  ^^^^^^^^^^^^ method cannot be called on `BitSlice<u8, Msb0>` due to unsatisfied trait bounds
   |
  ::: C:\Users\david\.cargo\registry\src\index.crates.io-6f17d22bba15001f\bitvec-1.0.1\src\slice.rs:60:1
   |
60 | pub struct BitSlice<T = usize, O = Lsb0>
   | ----------------------------------------
   | |
   | doesn't satisfy `BitSlice<u8, bitvec::order::Msb0>: BitViewSized`
   | doesn't satisfy `BitSlice<u8, bitvec::order::Msb0>: Sized`
   | doesn't satisfy `_: BitStore`

How would I do this? Thank you! I need to then pass this to a function as a byte array. I've found a workaround, but it's messy:

    let bits  = &data.view_bits::<Msb0>()[4..];

    let mut bit_i = 0;
    for _ in 0..num_commands {
        let byte0 = bits[bit_i..bit_i + 8].load_be();
        let byte1 = bits[bit_i+ 8..bit_i + 16].load_be();
        let byte2 = bits[bit_i + 16..bit_i + 24].load_be();
        let byte3 = bits[bit_i + 14..bit_i + 32].load_be();
        let command = ActuatorCommand::from_buf(&[byte0, byte1, byte2, byte3]);
        
        bit_i += COMMAND_SIZE * 8;

[connorskees]

In your particular case, you could just load a u32 and call u32::to_be_bytes(). Otherwise, there's also BitVec::as_raw_slice. I don't see an equivalent method on BitSlice, however.