In RFC1738, the BNF for domainlabel
is the following:
domainlabel = alphadigit | alphadigit *[ alphadigit | "-" ] alphadigit
That is, it's either an alphadigit, or it's a string where the first/last characters have to be an alphadigit but the intermediate characters can be an alphadigit or a dash.
How do I implement this with nom? Ignoring the single character scenario to simplify the case, my final attempt is:
fn domain_label(s: &[u8]) -> IResult<&[u8], (&[u8], &[u8], &[u8])> {
let left = take_while_m_n(1, 1, is_alphanumeric);
let middle = take_while(|c| is_alphanumeric(c) || c == b'-');
let right = take_while_m_n(1, 1, is_alphanumeric);
let whole = tuple((left, middle, right));
whole(s)
}
The problem with this is that middle
can consume the last character and hence right
fails because there is no character to consume.
println!("{:?}", domain_label(b"abcde"));
Err(Error(([], TakeWhileMN)))
Parsers should be able to attempt all possible consumption paths, but how to do this with nom
?
domainlabel = alphadigit | alphadigit *[ alphadigit | "-" ] alphadigit
It is a series of alphanumeric sequence delimited by any number of character -
. So here is one way to do it:
use nom::bytes::complete::{tag, take_while1};
use nom::character::is_alphanumeric;
use nom::combinator::recognize;
use nom::multi::{many1, separated_list};
use nom::IResult;
fn domain_label(input: &[u8]) -> IResult<&[u8], &[u8]> {
let alphadigits = take_while1(is_alphanumeric);
let delimiter = many1(tag(b"-"));
let parser = separated_list(delimiter, alphadigits);
recognize(parser)(input)
}
fn main() {
let (_, res) = domain_label(b"abcde").unwrap();
assert_eq!(res, b"abcde");
let (_, res) = domain_label(b"abcde-123-xyz-").unwrap();
assert_eq!(res, b"abcde-123-xyz");
let (_, res) = domain_label(b"rust-lang--1---37---0.org").unwrap();
assert_eq!(res, b"rust-lang--1---37---0");
}
Notice, you don't need individual parts of a successful parsing. The result is just the longest input that conforms to the domain label BNF. That's where the recognize
combinator comes in.