change return type of slice_shift_char #18911
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Contributor
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Thanks for the pull request, and welcome! The Rust team is excited to review your changes, and you should hear from @pcwalton (or someone else) soon. |
Contributor
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@canndrew Looks like a sensible change to me! Can you please add Otherwise, with that and a rebase, I'd be happy to take this. |
`slice_shift_char` splits a `str` into it's leading `char` and the remainder
of the `str`. Currently, it returns a `(Option<char>, &str)` such that:
"bar".slice_shift_char() => (Some('b'), "ar")
"ar".slice_shift_char() => (Some('a'), "r")
"r".slice_shift_char() => (Some('r'), "")
"".slice_shift_char() => (None, "")
This is a little odd. Either a `str` can be split into both a head and a
tail or it cannot. So the return type should be `Option<(char, &str)>`.
With the current behaviour, in the case of the empty string, the `str`
returned is meaningless - it is always the empty string.
This commit changes slice_shift_char so that:
"bar".slice_shift_char() => Some(('b', "ar"))
"ar".slice_shift_char() => Some(('a', "r"))
"r".slice_shift_char() => Some(('r', ""))
"".slice_shift_char() => None
[breaking-change]
canndrew
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All done. Should be building again now. |
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slice_shift_charsplits astrinto it's leadingcharand the remainder of thestr. Currently, it returns a(Option<char>, &str)such that:This is a little odd. Either a
strcan be split into both a head and a tail or it cannot. So the return type should beOption<(char, &str)>. With the current behaviour, in the case of the empty string, thestrreturned is meaningless - it is always the empty string.This PR changes
slice_shift_charso that: