如何从函数返回过滤器迭代器? [英] How do I return a Filter iterator from a function?

问题描述

我想要这样的东西:

fn filter_one<'a, T: Int>(input: &'a Vec<T>) -> ??? {
    input.iter().filter(|&x| x == Int::one())
}

那个函数的返回类型是什么?(我想返回迭代器)

What's the return type of that function? (I want to return the Iterator)

(我希望这不是太明显，我已经尝试了半个小时，现在开始感到沮丧:p)

(I hope this isn't too obvious, I've been trying for half an hour now and am just starting to get frustrated :p )

我尝试按照这里的说明 => playpen 链接

编译器给了我以下错误:

the compiler gives me the following error:

<anon>:5:1: 7:2 error: the trait `core::kinds::Sized` is not implemented for the type `for<'r> core::ops::Fn(&'r T) -> bool + 'a`
<anon>:5 fn filter_one<'a, T: Int>(input: &'a Vec<T>) -> Filter<&T, Iter<'a, T>, Fn(&T) -> bool>{
<anon>:6     input.iter().filter(|&x| x == Int::one())
<anon>:7 }
<anon>:5:1: 7:2 note: required by `core::iter::Filter`
<anon>:5 fn filter_one<'a, T: Int>(input: &'a Vec<T>) -> Filter<&T, Iter<'a, T>, Fn(&T) -> bool>{
<anon>:6     input.iter().filter(|&x| x == Int::one())
<anon>:7 }
<anon>:5:1: 7:2 error: the trait `for<'r> core::ops::Fn(&'r &'a T) -> bool` is not implemented for the type `for<'r> core::ops::Fn(&'r T) -> bool + 'a`
<anon>:5 fn filter_one<'a, T: Int>(input: &'a Vec<T>) -> Filter<&T, Iter<'a, T>, Fn(&T) -> bool>{
<anon>:6     input.iter().filter(|&x| x == Int::one())
<anon>:7 }
<anon>:5:1: 7:2 note: required by `core::iter::Filter`
<anon>:5 fn filter_one<'a, T: Int>(input: &'a Vec<T>) -> Filter<&T, Iter<'a, T>, Fn(&T) -> bool>{
<anon>:6     input.iter().filter(|&x| x == Int::one())
<anon>:7 }
error: aborting due to 2 previous errors
playpen: application terminated with error code 101

我如何告诉 rustc Fn(&T) ->bool 是大小??

推荐答案

Rust 1.26

fn filter_one(input: &[u8]) -> impl Iterator<Item = &u8> {
    input.iter().filter(|&&x| x == 1)
}

fn main() {
    let nums = vec![1, 2, 3, 1, 2, 3];
    let other: Vec<_> = filter_one(&nums).collect();
    println!("{:?}", other);
}

Rust 1.0

fn filter_one<'a>(input: &'a [u8]) -> Box<Iterator<Item = &'a u8> + 'a> {
    Box::new(input.iter().filter(|&&x| x == 1))
}

fn main() {
    let nums = vec![1, 2, 3, 1, 2, 3];
    let other: Vec<_> = filter_one(&nums).collect();
    println!("{:?}", other);
}

此解决方案需要额外分配.我们创建了一个盒装特征对象.在这里，对象的大小总是已知的(它只是一两个指针)，但不需要知道堆中对象的大小.

This solution requires additional allocation. We create a boxed trait object. Here, the size of the object is always known (it's just a pointer or two), but the size of the object in the heap does not need to be known.

正如Vladimir Matveev 指出的，如果您的谓词逻辑不需要来自环境的任何信息，您可以使用一个函数而不是一个闭包:

As Vladimir Matveev points out, if your predicate logic doesn't need any information from the environment, you can use a function instead of a closure:

use std::{iter::Filter, slice::Iter};

fn filter_one<'a>(input: &'a [u8]) -> Filter<Iter<u8>, fn(&&u8) -> bool> {
    fn is_one(a: &&u8) -> bool {
        **a == 1
    }

    input.iter().filter(is_one)
}

fn main() {
    let nums = vec![1, 2, 3, 1, 2, 3];
    let other: Vec<_> = filter_one(&nums).collect();
    println!("{:?}", other);
}

另见:

• 返回迭代器(或任何其他特征)的正确方法是什么?

这篇关于如何从函数返回过滤器迭代器?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！