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I have a [u8; 128] that I would like to print. To do so, I am passing in a borrow of that buffer to println! using debug formatting. However, I am receiving a compiler error.

The compiler says that [u8;128] does not implement std::fmt::Debug. However, I am passing in a borrow to println!, which leads me to question why Rust does not recognize this is a slice. I was able to fix the problem by attaching [#derive(Debug)] to a struct with one member, buf: &[u8]. Passing my buffer in through this struct to println! solves the problem, which tells me that &[u8] does indeed implement std::fmt::Debug. 

use std::io::prelude::*;
use std::net::TcpStream;

#[derive(Debug)]
struct Fah<'a> {
    buf: &'a[u8],
}

fn main() -> std::io::Result<()> {
    let mut stream = TcpStream::connect("216.58.194.206:80")?;

    stream.write(&[1])?;
    let mut buf = [0; 128];
    stream.read(&mut buf)?;
    println!("buf: {:?}", &buf); // This line does not compile
    println!("fah: {:?}", Fah { buf: &buf}); // This line works fine
    Ok(())
}

Playground link: https://play.integer32.com/?version=stable&mode=debug&edition=2018&gist=b95ad4762f23ca35e890d0cbd2c072ce

I expect this to compile just fine. However, the compiler complains that the trait std::fmt::Debug is not implemented for [u8; 128], as I said before.

riley lyman
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1 Answers1

4

&buf is a &[u8; 128] that gets implicitly coerced (by unsizing) to &[u8] a lot of the time. If you’re explicit about wanting the slice, it’ll work:

println!("{:?}", &a as &[u8]);

but also just ask for a slice in the first place:

println!("{:?}", &a[..]);
Ry-
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