nom-both

nom的扩展，提供特殊的both_解析器。

要求

nom必须为5.0.0或更高版本。nom-both只能应用于函数式解析器。

使用

[dependencies]
nom-both = "0.1.1"

示例

nom-both提供both_opt和both_alt解析器。both_opt表示解析器尝试同时进行参数解析器和跳过解析。both_alt表示解析器尝试第一个参数解析器和第二个参数解析器。

use nom::branch::*;
use nom::bytes::complete::*;
use nom::IResult;
use nom_both::both_parser;

#[both_parser]
pub fn both_opt_parser(s: &str) -> IResult<&str, Option<&str>> {
    let (s, _) = tag("1")(s)?;
    let (s, x) = both_opt(tag("2"))(s)?;
    let (s, _) = tag("2")(s)?;
    let (s, _) = tag("3")(s)?;
    Ok((s, x))
}

#[both_parser]
pub fn both_alt_parser(s: &str) -> IResult<&str, &str> {
    let (s, _) = tag("1")(s)?;
    let (s, x) = both_alt(tag("22"), tag("2"))(s)?;
    let (s, _) = tag("2")(s)?;
    let (s, _) = tag("3")(s)?;
    Ok((s, x))
}

#[test]
fn test() {
    let ret = both_opt_parser("123");
    assert_eq!("None", format!("{:?}", ret.unwrap().1));

    let ret = both_alt_parser("1223");
    assert_eq!("\"2\"", format!("{:?}", ret.unwrap().1));
}

both_opt_parser的展开如下

pub fn both_opt_parser(s: &str) -> IResult<&str, Option<&str>> {
    alt((
        { |s| {
            let (s, _) = tag("1")(s)?;
            let (s, x) = nom_both::some(tag("2"))(s)?;
            let (s, _) = tag("2")(s)?;
            let (s, _) = tag("3")(s)?;
            Ok((s, x))
        } },
        { |s| {
            let (s, _) = tag("1")(s)?;
            let (s, x) = nom_both::none(tag("2"))(s)?;
            let (s, _) = tag("2")(s)?;
            let (s, _) = tag("3")(s)?;
            Ok((s, x))
        } },
    ))(s)
}

both_alt_parser的展开如下

pub fn both_opt_parser(s: &str) -> IResult<&str, Option<&str>> {
    alt((
        { |s| {
            let (s, _) = tag("1")(s)?;
            let (s, x) = nom_both::alt_left(tag("22"), tag("2"))(s)?;
            let (s, _) = tag("2")(s)?;
            let (s, _) = tag("3")(s)?;
            Ok((s, x))
        } },
        { |s| {
            let (s, _) = tag("1")(s)?;
            let (s, x) = nom_both::alt_right(tag("22"), tag("2"))(s)?;
            let (s, _) = tag("2")(s)?;
            let (s, _) = tag("3")(s)?;
            Ok((s, x))
        } },
    ))(s)
}

许可证

以下任一许可证下授权：

• Apache许可证2.0版，(LICENSE-APACHE 或 https://apache.ac.cn/licenses/LICENSE-2.0)
• MIT许可证 (LICENSE-MIT 或 http://opensource.org/licenses/MIT)

任选其一。

贡献

除非您明确声明，否则任何提交给工作以包含在内的有意贡献，根据Apache-2.0许可证定义，应双授权如上所述，不附加任何额外条款或条件。