Question

我想将一个字母字符（a-zA-Z）与nom。

匹配

我知道我可以使用take_while!贪婪地匹配这样的事情：

// match one or more alphabetical characters
pub fn alpha_many(input: &[u8]) -> IResult<&[u8], &[u8]> {
    take_while!(input, |c| {
        (c >= 0x41 && c <= 0x5a) || (c >= 0x61 && c <= 0x7a)
    })
}

但我找不到如何只匹配一个字节。有one_of!，但我不能使用闭包，我必须传递整个片段：

// match exactly one alphabetical character
pub fn alpha_one(input: &[u8]) -> IResult<&[u8], u8> {
    one_of!(
        input,
        [
            0x41, 0x42, 0x43,
            // etc until 0x5a and then from 0x61 to 0x7a
            // ...
        ].as_ref()
    )
}

Answer 1

我想出了这个。如果没有人提出更好的解决方案，我会在明天将此标记为已接受的答案：

use nom::{self, ErrorKind, IResult, Needed};

/// Alphabetical characters ([RFC5234 appendix B.1])
///
/// [RFC5234 appendix B.1]: https://tools.ietf.org/html/rfc5234#appendix-B.1
///
/// ```no_rust
/// ALPHA          =  %x41-5A / %x61-7A   ; A-Z / a-z
/// ```
pub struct Alpha;

impl Alpha {
    /// Return true if the given byte represents an alphabetical character
    pub fn is_alpha(c: u8) -> bool {
        (c >= 0x41 && c <= 0x5a) || (c >= 0x61 && c <= 0x7a)
    }

    /// Parse one or more alphabetical characters
    pub fn parse_many(input: &[u8]) -> IResult<&[u8], &[u8]> {
        take_while!(input, Self::is_alpha)
    }

    /// Parse one alphabetical character
    pub fn parse_one(input: &[u8]) -> IResult<&[u8], u8> {
        Self::parse_n(input, 1).map(|res| res[0])
    }

    /// Parse n alphabetical characters
    pub fn parse_n(input: &[u8], n: usize) -> IResult<&[u8], &[u8]> {
        Self::parse_m_n(input, n, n)
    }

    /// Parse between m and n alphabetical characters
    pub fn parse_m_n(input: &[u8], m: usize, n: usize) -> IResult<&[u8], &[u8]> {
        if input.len() < m {
            return IResult::Incomplete(Needed::Size(input.len() - m));
        }
        for i in 0..n {
            if !Self::is_alpha(input[i]) {
                // We were supposed to have at least m printable bytes
                if i < m {
                    return IResult::Error(error_position!(ErrorKind::ManyMN, &input[..]));
                } else {
                    return IResult::Done(&input[i..], &input[0..i]);
                }
            }
        }
        return IResult::Done(&input[n..], &input[0..n]);
    }
}

如何使用nom匹配一个字节？

1 个答案: