Implementing Cursors

好了，我们现在讨论 CursorMut。就像我最初的设计一样，它有一个包含 None 的 "幽灵 "元素，用来指示列表的开始/结束，你可以 "跨过它"，绕到列表的另一边。要实现它，我们需要

指向当前节点的指针
指向列表的指针
当前索引

等等，当我们指向 "幽灵 "时，索引是多少？

好吧，游标 (cursors)上的索引返回一个 Option<usize>，这很合理。Std 的实现做了一堆垃圾来避免将其存储为一个 Option，但是...... 我们是一个链接列表，这很好。此外，std 还有 cursor_front/cursor_back 功能，它可以在前/后元素上启动光标，感觉很直观，但当列表为空时，又要做一些奇怪的事情。

如果你愿意，也可以实现这些东西，但我打算减少所有重复的垃圾和角落情况，只做一个从 ghost 处开始的 cursor_mut 方法，人们可以使用 move_next/move_prev 来获取他们想要的元素（如果你真的愿意，也可以将其封装为 cursor_front）。

让我们开始吧：

非常简单直接，上面的需求列表每一项都有一个字段！

#![allow(unused)]
fn main() {
pub struct CursorMut<'a, T> {
    cur: Link<T>,
    list: &'a mut LinkedList<T>,
    index: Option<usize>,
}
}

现在是cursor_mut 方法:

#![allow(unused)]
fn main() {
impl<T> LinkedList<T> {
    pub fn cursor_mut(&mut self) -> CursorMut<T> {
        CursorMut { 
            list: self, 
            cur: None, 
            index: None,
        }
    }
}
}

既然我们从幽灵节点开始，我们所以开始节点都是 None，简单明了！下一个是 move_next：

#![allow(unused)]
fn main() {
impl<'a, T> CursorMut<'a, T> {
    pub fn index(&self) -> Option<usize> {
        self.index
    }

    pub fn move_next(&mut self) {
        if let Some(cur) = self.cur {
            unsafe {
                // We're on a real element, go to its next (back)
                self.cur = (*cur.as_ptr()).back;
                if self.cur.is_some() {
                    *self.index.as_mut().unwrap() += 1;
                } else {
                    // We just walked to the ghost, no more index
                    self.index = None;
                }
            }
        } else if !self.list.is_empty() {
            // We're at the ghost, and there is a real front, so move to it!
            self.cur = self.list.front;
            self.index = Some(0)
        } else {
            // We're at the ghost, but that's the only element... do nothing.
        }
    }
}
}

所以这有4种有趣的情况：

正常情况
正常情况，但我们移动到了幽灵节点
幽灵节点开始，向列表头部节点移动
幽灵节点开始，列表是空的，所以什么都不做

move_prev 的逻辑完全相同，但前后颠倒，索引变化也颠倒：

#![allow(unused)]
fn main() {
pub fn move_prev(&mut self) {
    if let Some(cur) = self.cur {
        unsafe {
            // We're on a real element, go to its previous (front)
            self.cur = (*cur.as_ptr()).front;
            if self.cur.is_some() {
                *self.index.as_mut().unwrap() -= 1;
            } else {
                // We just walked to the ghost, no more index
                self.index = None;
            }
        }
    } else if !self.list.is_empty() {
        // We're at the ghost, and there is a real back, so move to it!
        self.cur = self.list.back;
        self.index = Some(self.list.len - 1)
    } else {
        // We're at the ghost, but that's the only element... do nothing.
    }
}
}

接下来，让我们添加一些方法来查看游标周围的元素：current、peek_next 和 peek_prev。 一个非常重要的注意事项：这些方法必须通过 &mut self 借用我们的游标，并且结果必须与借用绑定。我们不能让用户获得可变引用的多个副本，也不能让他们在持有该引用的情况下使用我们的 insert/remove/split/splice API！

值得庆幸的是，这是 rust 在使用生命周期省略规则时的默认设置，因此我们将默认做正确的事情！

#![allow(unused)]
fn main() {
pub fn current(&mut self) -> Option<&mut T> {
    unsafe {
        self.cur.map(|node| &mut (*node.as_ptr()).elem)
    }
}

pub fn peek_next(&mut self) -> Option<&mut T> {
    unsafe {
        self.cur
            .and_then(|node| (*node.as_ptr()).back)
            .map(|node| &mut (*node.as_ptr()).elem)
    }
}

pub fn peek_prev(&mut self) -> Option<&mut T> {
    unsafe {
        self.cur
            .and_then(|node| (*node.as_ptr()).front)
            .map(|node| &mut (*node.as_ptr()).elem)
    }
}
}

Split

首先是 split_before 和 split_after，它们会将当前元素之前/之后的所有内容以 LinkedList 的形式返回（在幽灵元素处停止，在这种情况下，我们只返回整个 List，光标现在指向一个空 list）：

这个逻辑其实并不复杂，所以我们得一步一步来。

我发现 split_before 有四种潜在的情况：

正常情况
正常情况，但 prev 是幽灵节点
幽灵节点情况，我们返回整个列表，然后变成空列表
幽灵节点情况，但列表是空的，所以什么也不做，返回空列表

让我们先从极端情况开始。我认为第三种情况

#![allow(unused)]
fn main() {
mem::replace(self.list, LinkedList::new())
}

对不对？我们是空的了，并返回了整个列表，而我们的字段都应该是 "None"，所以没什么可更新的。不错。哦，嘿嘿，这在第四种情况下也对！

现在是普通情况......，我需要画下图。最常见的情况是这样的

list.front -> A <-> B <-> C <-> D <- list.back
                          ^
                         cur

我们想变成这样：

list.front -> C <-> D <- list.back
              ^
             cur

return.front -> A <-> B <- return.back

因此，我们需要打破当前数据和前一个数据之间的联系，而且......天哪，需要改变的东西太多了。好吧，我只需要把它分成几个步骤，这样我就能说服自己这是有意义的。虽然有点啰嗦，但我至少能说得通：

#![allow(unused)]
fn main() {
pub fn split_before(&mut self) -> LinkedList<T> {
    if let Some(cur) = self.cur {
        // We are pointing at a real element, so the list is non-empty.
        unsafe {
            // Current state
            let old_len = self.list.len;
            let old_idx = self.index.unwrap();
            let prev = (*cur.as_ptr()).front;
            
            // What self will become
            let new_len = old_len - old_idx;
            let new_front = self.cur;
            let new_idx = Some(0);

            // What the output will become
            let output_len = old_len - new_len;
            let output_front = self.list.front;
            let output_back = prev;

            // Break the links between cur and prev
            if let Some(prev) = prev {
                (*cur.as_ptr()).front = None;
                (*prev.as_ptr()).back = None;
            }

            // Produce the result:
            self.list.len = new_len;
            self.list.front = new_front;
            self.index = new_idx;

            LinkedList {
                front: output_front,
                back: output_back,
                len: output_len,
                _boo: PhantomData,
            }
        }
    } else {
        // We're at the ghost, just replace our list with an empty one.
        // No other state needs to be changed.
        std::mem::replace(self.list, LinkedList::new())
    }
}
}

你可能注意到，我们没有处理 prev 是幽灵节点的情况。但据我所知，其他一切都只是顺便做了正确的事。等我们写测试的时候就知道了！(复制粘贴完成 split_after）。

Splice

还有一个老大难，那就是 splice_before 和 splice_after，我估计这是最容易出错的一个。这两个函数接收一个 LinkedList，并将其内容嫁接到我们的列表中。我们的列表可能是空的，他们的列表也可能是空的，我们还有幽灵节点要处理......叹口气，让我们一步一步来吧，从 splice_before 开始。

如果他们的列表是空的，我们就什么都不用做。
如果我们的列表是空的，那么我们的列表就变成了他们的列表。
如果我们指向的是幽灵节点，则追加到后面（更改 list.back）
如果我们指向的是第一个元素（0），则追加到前面（更改 list.front）
一般情况下，我们会进行大量的指针操作

一般情况：

input.front -> 1 <-> 2 <- input.back

 list.front -> A <-> B <-> C <- list.back
                     ^
                    cur

变成这样：

list.front -> A <-> 1 <-> 2 <-> B <-> C <- list.back

好的，让我们来写一下：

#![allow(unused)]
fn main() {
    pub fn splice_before(&mut self, mut input: LinkedList<T>) {
        unsafe {
            if input.is_empty() {
                // Input is empty, do nothing.
            } else if let Some(cur) = self.cur {
                if let Some(0) = self.index {
                    // We're appending to the front, see append to back
                    (*cur.as_ptr()).front = input.back.take();
                    (*input.back.unwrap().as_ptr()).back = Some(cur);
                    self.list.front = input.front.take();

                    // Index moves forward by input length
                    *self.index.as_mut().unwrap() += input.len;
                    self.list.len += input.len;
                    input.len = 0;
                } else {
                    // General Case, no boundaries, just internal fixups
                    let prev = (*cur.as_ptr()).front.unwrap();
                    let in_front = input.front.take().unwrap();
                    let in_back = input.back.take().unwrap();

                    (*prev.as_ptr()).back = Some(in_front);
                    (*in_front.as_ptr()).front = Some(prev);
                    (*cur.as_ptr()).front = Some(in_back);
                    (*in_back.as_ptr()).back = Some(cur);

                    // Index moves forward by input length
                    *self.index.as_mut().unwrap() += input.len;
                    self.list.len += input.len;
                    input.len = 0;
                }
            } else if let Some(back) = self.list.back {
                // We're on the ghost but non-empty, append to the back
                // We can either `take` the input's pointers or `mem::forget`
                // it. Using take is more responsible in case we do custom
                // allocators or something that also needs to be cleaned up!
                (*back.as_ptr()).back = input.front.take();
                (*input.front.unwrap().as_ptr()).front = Some(back);
                self.list.back = input.back.take();
                self.list.len += input.len;
                // Not necessary but Polite To Do
                input.len = 0;
            } else {
                // We're empty, become the input, remain on the ghost
                *self.list = input;
            }
        }
    }
}

好吧，这个程序真的很可怕，现在真的感觉到 Option 的痛苦了。但我们可以做很多清理工作。首先，我们可以把这段代码拖到最后。

#![allow(unused)]
fn main() {
self.list.len += input.len;
input.len = 0;
}

好了，现在在分支 "we're empty" 中有以下错误。所以我们应该使用 swap:

Use of moved value: input

#![allow(unused)]
fn main() {
// We're empty, become the input, remain on the ghost
std::mem::swap(self.list, &mut input);
}

在我反向思考下面这种情况时，我发现了这个 unwrap 有问题(因为 cur 的 front 在前面已经被设置为其它值了)：

#![allow(unused)]
fn main() {
if let Some(0) = self.index {

} else {
    let prev = (*cur.as_ptr()).front.unwrap();
}
}

这行也是重复的，可以提升:

#![allow(unused)]
fn main() {
*self.index.as_mut().unwrap() += input.len;
}

好了，把上面的问题修改后得到这些：

#![allow(unused)]
fn main() {
if input.is_empty() {
    // Input is empty, do nothing.
} else if let Some(cur) = self.cur {
    // Both lists are non-empty
    if let Some(prev) = (*cur.as_ptr()).front {
        // General Case, no boundaries, just internal fixups
        let in_front = input.front.take().unwrap();
        let in_back = input.back.take().unwrap();

        (*prev.as_ptr()).back = Some(in_front);
        (*in_front.as_ptr()).front = Some(prev);
        (*cur.as_ptr()).front = Some(in_back);
        (*in_back.as_ptr()).back = Some(cur);
    } else {
        // We're appending to the front, see append to back below
        (*cur.as_ptr()).front = input.back.take();
        (*input.back.unwrap().as_ptr()).back = Some(cur);
        self.list.front = input.front.take();
    }
    // Index moves forward by input length
    *self.index.as_mut().unwrap() += input.len;
} else if let Some(back) = self.list.back {
    // We're on the ghost but non-empty, append to the back
    // We can either `take` the input's pointers or `mem::forget`
    // it. Using take is more responsible in case we do custom
    // allocators or something that also needs to be cleaned up!
    (*back.as_ptr()).back = input.front.take();
    (*input.front.unwrap().as_ptr()).front = Some(back);
    self.list.back = input.back.take();

} else {
    // We're empty, become the input, remain on the ghost
    std::mem::swap(self.list, &mut input);
}

self.list.len += input.len;
// Not necessary but Polite To Do
input.len = 0;

// Input dropped here
}

还是不对，下面的代码存在bug：

#![allow(unused)]
fn main() {
    (*back.as_ptr()).back = input.front.take();
    (*input.front.unwrap().as_ptr()).front = Some(back);
}

我们使用 take 拿走了 input.front 的值，然后在下一行将其 unwrap！boom，panic！

#![allow(unused)]
fn main() {
// We can either `take` the input's pointers or `mem::forget`
// it. Using `take` is more responsible in case we ever do custom
// allocators or something that also needs to be cleaned up!
if input.is_empty() {
    // Input is empty, do nothing.
} else if let Some(cur) = self.cur {
    // Both lists are non-empty
    let in_front = input.front.take().unwrap();
    let in_back = input.back.take().unwrap();

    if let Some(prev) = (*cur.as_ptr()).front {
        // General Case, no boundaries, just internal fixups
        (*prev.as_ptr()).back = Some(in_front);
        (*in_front.as_ptr()).front = Some(prev);
        (*cur.as_ptr()).front = Some(in_back);
        (*in_back.as_ptr()).back = Some(cur);
    } else {
        // No prev, we're appending to the front
        (*cur.as_ptr()).front = Some(in_back);
        (*in_back.as_ptr()).back = Some(cur);
        self.list.front = Some(in_front);
    }
    // Index moves forward by input length
    *self.index.as_mut().unwrap() += input.len;
} else if let Some(back) = self.list.back {
    // We're on the ghost but non-empty, append to the back
    let in_front = input.front.take().unwrap();
    let in_back = input.back.take().unwrap();

    (*back.as_ptr()).back = Some(in_front);
    (*in_front.as_ptr()).front = Some(back);
    self.list.back = Some(in_back);
} else {
    // We're empty, become the input, remain on the ghost
    std::mem::swap(self.list, &mut input);
}

self.list.len += input.len;
// Not necessary but Polite To Do
input.len = 0;

// Input dropped here
}

总之，我已经筋疲力尽了，所以 insert 和 remove 以及所有其他应用程序接口就留给读者练习。下面是 Cursor 的最终代码，我做对了吗？我只有在写下一章并测试这个怪东西时才能知道！

#![allow(unused)]
fn main() {
pub struct CursorMut<'a, T> {
    list: &'a mut LinkedList<T>,
    cur: Link<T>,
    index: Option<usize>,
}

impl<T> LinkedList<T> {
    pub fn cursor_mut(&mut self) -> CursorMut<T> {
        CursorMut { 
            list: self, 
            cur: None, 
            index: None,
        }
    }
}

impl<'a, T> CursorMut<'a, T> {
    pub fn index(&self) -> Option<usize> {
        self.index
    }

    pub fn move_next(&mut self) {
        if let Some(cur) = self.cur {
            unsafe {
                // We're on a real element, go to its next (back)
                self.cur = (*cur.as_ptr()).back;
                if self.cur.is_some() {
                    *self.index.as_mut().unwrap() += 1;
                } else {
                    // We just walked to the ghost, no more index
                    self.index = None;
                }
            }
        } else if !self.list.is_empty() {
            // We're at the ghost, and there is a real front, so move to it!
            self.cur = self.list.front;
            self.index = Some(0)
        } else {
            // We're at the ghost, but that's the only element... do nothing.
        }
    }

    pub fn move_prev(&mut self) {
        if let Some(cur) = self.cur {
            unsafe {
                // We're on a real element, go to its previous (front)
                self.cur = (*cur.as_ptr()).front;
                if self.cur.is_some() {
                    *self.index.as_mut().unwrap() -= 1;
                } else {
                    // We just walked to the ghost, no more index
                    self.index = None;
                }
            }
        } else if !self.list.is_empty() {
            // We're at the ghost, and there is a real back, so move to it!
            self.cur = self.list.back;
            self.index = Some(self.list.len - 1)
        } else {
            // We're at the ghost, but that's the only element... do nothing.
        }
    }

    pub fn current(&mut self) -> Option<&mut T> {
        unsafe {
            self.cur.map(|node| &mut (*node.as_ptr()).elem)
        }
    }

    pub fn peek_next(&mut self) -> Option<&mut T> {
        unsafe {
            self.cur
                .and_then(|node| (*node.as_ptr()).back)
                .map(|node| &mut (*node.as_ptr()).elem)
        }
    }

    pub fn peek_prev(&mut self) -> Option<&mut T> {
        unsafe {
            self.cur
                .and_then(|node| (*node.as_ptr()).front)
                .map(|node| &mut (*node.as_ptr()).elem)
        }
    }

    pub fn split_before(&mut self) -> LinkedList<T> {
        // We have this:
        //
        //     list.front -> A <-> B <-> C <-> D <- list.back
        //                               ^
        //                              cur
        // 
        //
        // And we want to produce this:
        // 
        //     list.front -> C <-> D <- list.back
        //                   ^
        //                  cur
        //
        // 
        //    return.front -> A <-> B <- return.back
        //
        if let Some(cur) = self.cur {
            // We are pointing at a real element, so the list is non-empty.
            unsafe {
                // Current state
                let old_len = self.list.len;
                let old_idx = self.index.unwrap();
                let prev = (*cur.as_ptr()).front;
                
                // What self will become
                let new_len = old_len - old_idx;
                let new_front = self.cur;
                let new_back = self.list.back;
                let new_idx = Some(0);

                // What the output will become
                let output_len = old_len - new_len;
                let output_front = self.list.front;
                let output_back = prev;

                // Break the links between cur and prev
                if let Some(prev) = prev {
                    (*cur.as_ptr()).front = None;
                    (*prev.as_ptr()).back = None;
                }

                // Produce the result:
                self.list.len = new_len;
                self.list.front = new_front;
                self.list.back = new_back;
                self.index = new_idx;

                LinkedList {
                    front: output_front,
                    back: output_back,
                    len: output_len,
                    _boo: PhantomData,
                }
            }
        } else {
            // We're at the ghost, just replace our list with an empty one.
            // No other state needs to be changed.
            std::mem::replace(self.list, LinkedList::new())
        }
    }

    pub fn split_after(&mut self) -> LinkedList<T> {
        // We have this:
        //
        //     list.front -> A <-> B <-> C <-> D <- list.back
        //                         ^
        //                        cur
        // 
        //
        // And we want to produce this:
        // 
        //     list.front -> A <-> B <- list.back
        //                         ^
        //                        cur
        //
        // 
        //    return.front -> C <-> D <- return.back
        //
        if let Some(cur) = self.cur {
            // We are pointing at a real element, so the list is non-empty.
            unsafe {
                // Current state
                let old_len = self.list.len;
                let old_idx = self.index.unwrap();
                let next = (*cur.as_ptr()).back;
                
                // What self will become
                let new_len = old_idx + 1;
                let new_back = self.cur;
                let new_front = self.list.front;
                let new_idx = Some(old_idx);

                // What the output will become
                let output_len = old_len - new_len;
                let output_front = next;
                let output_back = self.list.back;

                // Break the links between cur and next
                if let Some(next) = next {
                    (*cur.as_ptr()).back = None;
                    (*next.as_ptr()).front = None;
                }

                // Produce the result:
                self.list.len = new_len;
                self.list.front = new_front;
                self.list.back = new_back;
                self.index = new_idx;

                LinkedList {
                    front: output_front,
                    back: output_back,
                    len: output_len,
                    _boo: PhantomData,
                }
            }
        } else {
            // We're at the ghost, just replace our list with an empty one.
            // No other state needs to be changed.
            std::mem::replace(self.list, LinkedList::new())
        }
    }

    pub fn splice_before(&mut self, mut input: LinkedList<T>) {
        // We have this:
        //
        // input.front -> 1 <-> 2 <- input.back
        //
        // list.front -> A <-> B <-> C <- list.back
        //                     ^
        //                    cur
        //
        //
        // Becoming this:
        //
        // list.front -> A <-> 1 <-> 2 <-> B <-> C <- list.back
        //                                 ^
        //                                cur
        //
        unsafe {
            // We can either `take` the input's pointers or `mem::forget`
            // it. Using `take` is more responsible in case we ever do custom
            // allocators or something that also needs to be cleaned up!
            if input.is_empty() {
                // Input is empty, do nothing.
            } else if let Some(cur) = self.cur {
                // Both lists are non-empty
                let in_front = input.front.take().unwrap();
                let in_back = input.back.take().unwrap();

                if let Some(prev) = (*cur.as_ptr()).front {
                    // General Case, no boundaries, just internal fixups
                    (*prev.as_ptr()).back = Some(in_front);
                    (*in_front.as_ptr()).front = Some(prev);
                    (*cur.as_ptr()).front = Some(in_back);
                    (*in_back.as_ptr()).back = Some(cur);
                } else {
                    // No prev, we're appending to the front
                    (*cur.as_ptr()).front = Some(in_back);
                    (*in_back.as_ptr()).back = Some(cur);
                    self.list.front = Some(in_front);
                }
                // Index moves forward by input length
                *self.index.as_mut().unwrap() += input.len;
            } else if let Some(back) = self.list.back {
                // We're on the ghost but non-empty, append to the back
                let in_front = input.front.take().unwrap();
                let in_back = input.back.take().unwrap();

                (*back.as_ptr()).back = Some(in_front);
                (*in_front.as_ptr()).front = Some(back);
                self.list.back = Some(in_back);
            } else {
                // We're empty, become the input, remain on the ghost
                std::mem::swap(self.list, &mut input);
            }

            self.list.len += input.len;
            // Not necessary but Polite To Do
            input.len = 0;
            
            // Input dropped here
        }        
    }

    pub fn splice_after(&mut self, mut input: LinkedList<T>) {
        // We have this:
        //
        // input.front -> 1 <-> 2 <- input.back
        //
        // list.front -> A <-> B <-> C <- list.back
        //                     ^
        //                    cur
        //
        //
        // Becoming this:
        //
        // list.front -> A <-> B <-> 1 <-> 2 <-> C <- list.back
        //                     ^
        //                    cur
        //
        unsafe {
            // We can either `take` the input's pointers or `mem::forget`
            // it. Using `take` is more responsible in case we ever do custom
            // allocators or something that also needs to be cleaned up!
            if input.is_empty() {
                // Input is empty, do nothing.
            } else if let Some(cur) = self.cur {
                // Both lists are non-empty
                let in_front = input.front.take().unwrap();
                let in_back = input.back.take().unwrap();

                if let Some(next) = (*cur.as_ptr()).back {
                    // General Case, no boundaries, just internal fixups
                    (*next.as_ptr()).front = Some(in_back);
                    (*in_back.as_ptr()).back = Some(next);
                    (*cur.as_ptr()).back = Some(in_front);
                    (*in_front.as_ptr()).front = Some(cur);
                } else {
                    // No next, we're appending to the back
                    (*cur.as_ptr()).back = Some(in_front);
                    (*in_front.as_ptr()).front = Some(cur);
                    self.list.back = Some(in_back);
                }
                // Index doesn't change
            } else if let Some(front) = self.list.front {
                // We're on the ghost but non-empty, append to the front
                let in_front = input.front.take().unwrap();
                let in_back = input.back.take().unwrap();

                (*front.as_ptr()).front = Some(in_back);
                (*in_back.as_ptr()).back = Some(front);
                self.list.front = Some(in_front);
            } else {
                // We're empty, become the input, remain on the ghost
                std::mem::swap(self.list, &mut input);
            }

            self.list.len += input.len;
            // Not necessary but Polite To Do
            input.len = 0;
            
            // Input dropped here
        }        
    }
}
}

Rust语言圣经(Rust Course)

Implementing Cursors

Split

Splice