iterator - 前向和当前迭代器

Question

还有别的写法吗？

fn main() {
    let mut a = [1, 2, 3, 4, 5];
    let mut iter1 = a.iter();
    let mut iter2 = a.iter().skip(1).peekable();

    while iter2.peek().is_some() {
        println!("iter1: {:?}, iter2: {:?}", iter1.next(), iter2.next());
    }
}

Answer 1

我不确定确切的问题是什么，但代码本身可以使用 windows 重写迭代器:

fn main() {
    let a = [1, 2, 3, 4, 5];
    for w in a.windows(2) {
        println!("elem1: {:?}, elem2: {:?}", w[0], w[1]);
    }
}