分解Iterator的引用：：在锈蚀中找到

Question

我正在通过示例阅读Rust，并且在搜索迭代器时遇到了本节的问题。

pub trait Iterator {
    // The type being iterated over.
    type Item;

    // `find` takes `&mut self` meaning the caller may be borrowed
    // and modified, but not consumed.
    fn find<P>(&mut self, predicate: P) -> Option<Self::Item> where
        // `FnMut` meaning any captured variable may at most be
        // modified, not consumed. `&Self::Item` states it takes
        // arguments to the closure by reference.
        P: FnMut(&Self::Item) -> bool {}
}

fn main() {
    let vec1 = vec![1, 2, 3];
    let vec2 = vec![4, 5, 6];

    // `iter()` for vecs yields `&i32`.
    let mut iter = vec1.iter();
    // `into_iter()` for vecs yields `i32`.
    let mut into_iter = vec2.into_iter();

    // `iter()` for vecs yields `&i32`, and we want to reference one of its
    // items, so we have to destructure `&&i32` to `i32`
    println!("Find 2 in vec1: {:?}", iter     .find(|&&x| x == 2));
    // `into_iter()` for vecs yields `i32`, and we want to reference one of
    // its items, so we have to destructure `&i32` to `i32`
    println!("Find 2 in vec2: {:?}", into_iter.find(| &x| x == 2));
}

如果iter()产生了&i32，Iterator::find获得了&mut self，为什么我们需要2个符号&&x而不是一个&x？既然如此，为什么&mut x不与函数的签名相匹配呢？

类似地，如果into_iter()生成i32，而Iterator::find获取&mut self，那么为什么我们不在管道之间编写&mut x而不是&x呢？

Answer 1

find()接受&mut self这一事实与此无关。这意味着迭代器需要不断地借用，即iter.find(f)实际上是Iterator::find(&mut iter, f)。

|&&x|/|&x|用于回调，其类型为F where F: FnMut(&Self::Item) -> bool。Self::Item是&i32表示iter，i32表示into_iter，因此我们分别以&&i32和&i32结尾。