迭代器iter的find实现解析

原创已于 2024-01-28 19:15:56 修改 · 474 阅读

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#前端 #服务器 #linux

于 2024-01-28 18:57:05 首次发布

本文解析了rcore代码中find()函数的工作原理，重点在于其接受闭包作为参数，该闭包用于判断元素是否满足条件。find()函数是短路的，一旦找到匹配项就停止执行。同时，讲解了check函数如何将FnMut类型的闭包转换为处理元组并返回ControlFlow的函数。

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在学习rcore代码过程中突发好奇迭代器的find()原理

从源代码注释可知，find 根据闭包作为参数，然后bool类型，闭包捕捉迭代器的参数类型为迭代器成员类型

    /// Searches for an element of an iterator that satisfies a predicate.
    ///
    /// `find()` takes a closure that returns `true` or `false`. It applies
    /// this closure to each element of the iterator, and if any of them return
    /// `true`, then `find()` returns [`Some(element)`]. If they all return
    /// `false`, it returns [`None`].
    ///
    /// `find()` is short-circuiting; in other words, it will stop processing
    /// as soon as the closure returns `true`.
    ///
    /// Because `find()` takes a reference, and many iterators iterate over
    /// references, this leads to a possibly confusing situation where the
    /// argument is a double reference. You can see this effect in the
    /// examples below, with `&&x`.
    ///
    /// If you need the index of the element, see [`position()`].
    ///
    /// [`Some(element)`]: Some
    /// [`position()`]: Iterator::position
    ///
    /// # Examples
    ///
    /// Basic usage:
    ///
    /// ```
    /// let a = [1, 2, 3];
    ///
    /// assert_eq!(a.iter().find(|&&x| x == 2), Some(&2));
    ///
    /// assert_eq!(a.iter().find(|&&x| x == 5), None);
    /// ```
    ///
    /// Stopping at the first `true`:
    ///
    /// ```
    /// let a = [1, 2, 3];
    ///
    /// let mut iter = a.iter();
    ///
    /// assert_eq!(iter.find(|&&x| x == 2), Some(&2));
    ///
    /// // we can still use `iter`, as there are more elements.
    /// assert_eq!(iter.next(), Some(&3));
    /// ```
    ///
    /// Note that `iter.find(f)` is equivalent to `iter.filter(f).next()`.
    #[inline]
    #[stable(feature = "rust1", since = "1.0.0")]
    #[rustc_do_not_const_check]
    fn find<P>(&mut self, predicate: P) -> Option<Self::Item>
    where
        Self: Sized,
        P: FnMut(&Self::Item) -> bool,
    {
        #[inline]
        fn check<T>(mut predicate: impl FnMut(&T) -> bool) -> impl FnMut((), T) -> ControlFlow<T> {
            move |(), x| {
                if predicate(&x) { ControlFlow::Break(x) } else { ControlFlow::Continue(()) }
            }
        }

        self.try_fold((), check(predicate)).break_value()
    }

以上源码主要难点在于：
1，predicate 类型为： FnMut(&Self::Item) -> bool

2，check 函数类型，输入参数为 predicate类型，返回类型为

FnMut((), T) -> ControlFlow<T>

也就是 check函数的原型是将 FnMut(&Self::Item) -> bool转化为

FnMut((), &Self::Item) -> ControlFlow<&Self::Item>

fold函数是成员展开，如果配到true为则返回当前Item, 否则继续下一个成员展开

check函数的难点是函数原型