Chapter 6: Repeating Parsers
A single parser which repeats a predicate is useful, but more useful still is a combinator that
repeats a parser. Nom has multiple combinators which operate on this principle; the most obvious of
which is many0, which applies a parser as many times as possible; and returns a vector of
the results of those parses. Here is an example:
extern crate nom; use std::error::Error; use nom::IResult; use nom::multi::many0; use nom::bytes::complete::tag; fn parser(s: &str) -> IResult<&str, Vec<&str>> { many0(tag("abc"))(s) } fn main() { assert_eq!(parser("abcabc"), Ok(("", vec!["abc", "abc"]))); assert_eq!(parser("abc123"), Ok(("123", vec!["abc"]))); assert_eq!(parser("123123"), Ok(("123123", vec![]))); assert_eq!(parser(""), Ok(("", vec![]))); }
There are many different parsers to choose from:
| combinator | usage | input | output | comment |
|---|---|---|---|---|
| count | count(take(2), 3) | "abcdefgh" | Ok(("gh", vec!["ab", "cd", "ef"])) | Applies the child parser a specified number of times |
| many0 | many0(tag("ab")) | "abababc" | Ok(("c", vec!["ab", "ab", "ab"])) | Applies the parser 0 or more times and returns the list of results in a Vec. many1 does the same operation but must return at least one element |
| many_m_n | many_m_n(1, 3, tag("ab")) | "ababc" | Ok(("c", vec!["ab", "ab"])) | Applies the parser between m and n times (n included) and returns the list of results in a Vec |
| many_till | many_till(tag( "ab" ), tag( "ef" )) | "ababefg" | Ok(("g", (vec!["ab", "ab"], "ef"))) | Applies the first parser until the second applies. Returns a tuple containing the list of results from the first in a Vec and the result of the second |
| separated_list0 | separated_list0(tag(","), tag("ab")) | "ab,ab,ab." | Ok((".", vec!["ab", "ab", "ab"])) | separated_list1 works like separated_list0 but must returns at least one element |
| fold_many0 | fold_many0(be_u8, \|\| 0, \|acc, item\| acc + item) | [1, 2, 3] | Ok(([], 6)) | Applies the parser 0 or more times and folds the list of return values. The fold_many1 version must apply the child parser at least one time |
| fold_many_m_n | fold_many_m_n(1, 2, be_u8, \|\| 0, \|acc, item\| acc + item) | [1, 2, 3] | Ok(([3], 3)) | Applies the parser between m and n times (n included) and folds the list of return value |
| length_count | length_count(number, tag("ab")) | "2ababab" | Ok(("ab", vec!["ab", "ab"])) | Gets a number from the first parser, then applies the second parser that many times |