标准库是否提供了一种使用另一个相同类型的片作为分隔符来拆分片[T]的方法？
从Rust 1.9开始，没有，但是你可以实现它：

fn main() {
    let a = [1, 4, 7, 0, 0, 8, 10, 0, 0];
    let b = [0, 0];
    let mut iter = split_subsequence(&a, &b);
    assert_eq!(&[1, 4, 7], iter.next().unwrap());
    assert_eq!(&[8, 10], iter.next().unwrap());
    assert!(iter.next().unwrap().is_empty());
    assert_eq!(None, iter.next());
}

pub struct SplitSubsequence<'a, 'b, T: 'a + 'b> {
    slice: &'a [T],
    needle: &'b [T],
    ended: bool,
}

impl<'a, 'b, T: 'a + 'b + PartialEq> Iterator for SplitSubsequence<'a, 'b, T> {
    type Item = &'a [T];

    fn next(&mut self) -> Option<Self::Item> {
        if self.ended {
            None
        } else if self.slice.is_empty() {
            self.ended = true;
            Some(self.slice)
        } else if let Some(p) = self.slice
                                    .windows(self.needle.len())
                                    .position(|w| w == self.needle) {
            let item = &self.slice[..p];
            self.slice = &self.slice[p + self.needle.len()..];
            Some(item)
        } else {
            self.ended = true;
            let item = self.slice;
            self.slice = &self.slice[self.slice.len() - 1..];
            Some(item)
        }
    }
}

fn split_subsequence<'a, 'b, T>(slice: &'a [T], needle: &'b [T]) -> SplitSubsequence<'a, 'b, T>
    where T: 'a + 'b + PartialEq
{
    SplitSubsequence {
        slice: slice,
        needle: needle,
        ended: false,
    }
}

注意，该实现使用naive算法来查找相等的子序列。