How to convert a &str to a &[u8]

This seems trivial, but I cannot find a way to do it.

For example,

fn f(s: &[u8]) {}

pub fn main() {
    let x = "a";
    f(x)
}

Fails to compile with:

error: mismatched types:
 expected `&[u8]`,
    found `&str`
(expected slice,
    found str) [E0308]

documentation, however, states that:


The actual representation of strs have direct mappings to slices: &str
is the same as &[u8].

You can use the as_bytes method:

fn f(s: &[u8]) {}

pub fn main() {
    let x = "a";
    f(x.as_bytes())
}

or, in your specific example, you could use a byte literal:

let x = b"a";
f(x)

Stack Overflow Q#31289588, score: 144