迭代器方法take_while将闭包作为其参数。

例如:

fn main() {
    let s = "hello!";
    let iter = s.chars();
    let s2 = iter.take_while(|x| *x != 'o').collect::<String>();
    //                       ^^^^^^^^^^^^^
    //                          closure

    println!("{}", s2);   // hell
}

playground link

对于简单的闭包来说这很好,但是如果我想要一个更复杂的谓词,我不想直接在take_while参数中编写它。相反,我想从函数返回闭包。

我似乎很难使它正常工作。这是我天真的尝试:
fn clos(a: char) -> Box<Fn(char) -> bool> {
    Box::new(move |b| a != b)
}

fn main() {
    // println!("{}", clos('a')('b'));   // <-- true
    //                      ^--- Using the closure here is fine
    let s = "hello!";
    let mut iter = s.chars();
    let s2 = iter.take_while( clos('o') ).collect::<String>();
    //                           ^--- This causes lots of issues

    println!("{}", s2);
}

playground link

但是,事实证明它引起的错误很难理解:
error[E0277]: the trait bound `for<'r> Box<std::ops::Fn(char) -> bool>: std::ops::FnMut<(&'r char,)>` is not satisfied
  --> <anon>:11:23
   |
11 |         let s2 = iter.take_while( clos('o') ).collect::<String>();
   |                       ^^^^^^^^^^ trait `for<'r> Box<std::ops::Fn(char) -> bool>: std::ops::FnMut<(&'r char,)>` not satisfied

error[E0277]: the trait bound `for<'r> Box<std::ops::Fn(char) -> bool>: std::ops::FnOnce<(&'r char,)>` is not satisfied
  --> <anon>:11:23
   |
11 |         let s2 = iter.take_while( clos('o') ).collect::<String>();
   |                       ^^^^^^^^^^ trait `for<'r> Box<std::ops::Fn(char) -> bool>: std::ops::FnOnce<(&'r char,)>` not satisfied
   |
   = help: the following implementations were found:
   = help:   <Box<std::boxed::FnBox<A, Output=R> + 'a> as std::ops::FnOnce<A>>
   = help:   <Box<std::boxed::FnBox<A, Output=R> + Send + 'a> as std::ops::FnOnce<A>>

error: no method named `collect` found for type `std::iter::TakeWhile<std::str::Chars<'_>, Box<std::ops::Fn(char) -> bool>>` in the current scope
  --> <anon>:11:47
   |
11 |         let s2 = iter.take_while( clos('o') ).collect::<String>();
   |                                               ^^^^^^^
   |
   = note: the method `collect` exists but the following trait bounds were not satisfied: `Box<std::ops::Fn(char) -> bool> : std::ops::FnMut<(&char,)>`, `std::iter::TakeWhile<std::str::Chars<'_>, Box<std::ops::Fn(char) -> bool>> : std::iter::Iterator`

error: aborting due to 3 previous errors

我尝试了其他一些操作,包括使用FnBox,但没有用。我没有太多使用闭包,所以我真的很想了解问题出在哪里,以及如何解决。

Related

最佳答案

您的代码中有两个问题。

首先, take_while 通过引用该函数传递值(请注意&中的where P: FnMut(&Self::Item) -> bool),而您的闭包期望按值接收它。

fn clos(a: char) -> Box<Fn(&char) -> bool> {
    Box::new(move |&b| a != b)
}

然后就是Box<Fn(&char) -> bool>没有实现FnMut(&char) -> bool的问题。如果我们查看 FnMut 的文档,我们将看到标准库提供了以下实现:
impl<'a, A, F> FnMut<A> for &'a F where F: Fn<A> + ?Sized
impl<'a, A, F> FnMut<A> for &'a mut F where F: FnMut<A> + ?Sized

好的,所以实现了FnMut来引用Fn的实现。我们手中有一个Fn特质对象,它实现了Fn,所以很好。我们只需要将Box<Fn>变成&Fn即可。我们首先需要取消引用产生左值的框,然后引用该左值以产生&Fn
fn main() {
    let s = "hello!";
    let iter = s.chars();
    let s2 = iter.take_while(&*clos('o')).collect::<String>();
    println!("{}", s2);
}

关于iterator - 如何将盒装封口传递给 `take_while`?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/39563998/

10-12 04:34