Find next char boundary index in string after char

Question

Given the string s, and the index i which is where the 好 character starts:

let s = "abc 好 def";
let i = 4;

What's the best way to get the index after that character, so that I can slice the string and get abc 好? In code:

let end = find_end(s, i);
assert_eq!("abc 好", &s[0..end]);

(Note, + 1 doesn't work because that assumes that the character is only 1 byte long.)

I currently have the following:

fn find_end(s: &str, i: usize) -> usize {
    i + s[i..].chars().next().unwrap().len_utf8()
}

But I'm wondering if I'm missing something and there's a better way?

Chris Emerson · Accepted Answer · 2017-04-07 13:10:41Z

7

You could use char_indices to get the next index rather than using len_utf8 on the character, though that has a special case for the last character.

I would use the handy str::is_char_boundary() method. Here's an implementation using that:

fn find_end(s: &str, i: usize) -> usize {
    assert!(i < s.len());
    let mut end = i+1;
    while !s.is_char_boundary(end) {
        end += 1;
    }
    end
}

Playground link

Normally I would make such a function return Option<usize> in case it's called with an index at the end of s, but for now I've just asserted.

In many cases, instead of explicitly calling find_end it may make sense to iterate using char_indices, which gives you each index along with the characters; though it's slightly annoying if you want to know the end of the current character.

answered Apr 7, 2017 at 13:10

Chris Emerson

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1 Comment

robinst Over a year ago

Thanks for the answer! I didn't include it in the question, but I also had is_char_boundary at some point. With char_indices, if you stop at a char and want to get the next index, you can use i + c.len_utf8(), so that's a good idea too!

E_net4 · Accepted Answer · 2017-04-07 13:34:13Z

1

To serve as a complement to @ChrisEmerson's answer, this is how one could implement a find_end that searches for the end of a character's first occurrence. Playground

fn find_end<'s>(s: &'s str, p: char) -> Option<usize> {
    let mut indices = s.char_indices();
    let mut found = false;
    for (_, v) in &mut indices {
        if v == p {
            found = true;
            break;
        }
    }
    if found {
        Some(indices.next()
                    .map_or_else(|| s.len(), |(i, _)| i))
    } else {
        None
    }
}

Although it avoids the byte boundary loop, it is still not very elegant. Ideally, an iterator method for traversing until a predicate is met would simplify this.

edited Apr 7, 2017 at 13:34

answered Apr 7, 2017 at 13:22

E_net4

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4 Comments

Chris Emerson Over a year ago

I'm a bit surprised there doesn't seem to be a next_char_boundary method!

robinst Over a year ago

Thanks as well! The downside with using next() is that it only works if there is another character after the current one.

robinst Over a year ago

@ChrisEmerson Yes, a next_char_boundary method would be perfect.

E_net4 Over a year ago

@robinst Characters at the end can be addressed easily. I've updated the function.

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