I would like to have macro splitting one byte into tuple with 2-8 u8 parts using bitreader
crate.
I managed to achieve that by following code:
use bitreader::BitReader;
trait Tupleprepend<T> {
type ResultType;
fn prepend(self, t: T) -> Self::ResultType;
}
macro_rules! impl_tuple_prepend {
( () ) => {};
( ( $t0:ident $(, $types:ident)* ) ) => {
impl<$t0, $($types,)* T> Tupleprepend<T> for ($t0, $($types,)*) {
type ResultType = (T, $t0, $($types,)*);
fn prepend(self, t: T) -> Self::ResultType {
let ($t0, $($types,)*) = self;
(t, $t0, $($types,)*)
}
}
impl_tuple_prepend! { ($($types),*) }
};
}
impl_tuple_prepend! {
(_1, _2, _3, _4, _5, _6, _7, _8)
}
macro_rules! split_byte (
($reader:ident, $bytes:expr, $count:expr) => {{
($reader.read_u8($count).unwrap(),)
}};
($reader:ident, $bytes:expr, $count:expr, $($next_counts:expr),+) => {{
let head = split_byte!($reader, $bytes, $count);
let tail = split_byte!($reader, $bytes, $($next_counts),+);
tail.prepend(head.0)
}};
($bytes:expr $(, $count:expr)* ) => {{
let mut reader = BitReader::new($bytes);
split_byte!(reader, $bytes $(, $count)+)
}};
);
Now I can use this code as I would like to:
let buf: &[u8] = &[0x72];
let (bit1, bit2, bits3to8) = split_byte!(&buf, 1, 1, 6);
Is there a way to avoid using Tupleprepend
trait and create only 1 tuple instead of 8 in the worst scenario?
Because the number of bit widths directly corresponds to the number of returned values, I'd solve the problem using generics and arrays instead. The macro only exists to remove the typing of the
[]
, which I don't really think is worth it.See also:
If it's ok to target nightly Rust, I'd use the unstable
min_const_generics
feature:See also: