为什么在使用 sort_by_key 对向量进行排序时不能使用返回引用的键函数? [英] Why can't I use a key function that returns a reference when sorting a vector with sort_by_key?

问题描述

我正在尝试使用返回对向量中字符串的引用的键函数对 Vec 进行排序.一个人为的例子是使用恒等函数作为关键函数(这当然没用，但它是重现我的问题的最小例子):

fn key(x: &String) ->&字符串{X}

现在给定items: Vec，我希望能够做到

items.sort_by_key(key);

出现以下错误:

error[E0271]:类型不匹配解析`for<'r><fn(&std::string::String) ->&std::string::String {main::key} as std::ops::FnOnce<(&'r std::string::String,)>>::Output == _`-->src/main.rs:19:11|19 |items.sort_by_key(key);|^^^^^^^^^^^ 预期绑定生命周期参数，找到具体生命周期|= 注意:找到的具体生命周期是生命周期 '_#16r

我不明白为什么我会收到这个错误，所以我试图追踪这个错误.我首先实现了我自己的 sort_by_key() 版本:

fn sort_by_key(a: &mut [T], key: fn(&T) -> K) {a.sort_by(|x, y| key(x).cmp(&key(y)));}

当尝试调用这个函数时，我得到看起来像相反"的错误:

错误[E0308]:类型不匹配-->src/main.rs:22:29|22 |sort_by_key(&mut items, key);|^^^ 预期的具体生命周期，找到绑定的生命周期参数|= 注意:预期类型 `fn(&std::string::String) ->_`找到类型 `fn(&std::string::String) ->&std::string::String {main::key}`

我可以通过将键类型固定为 &T 而不是使用通用参数 K 或使用 &K 而不是 K 作为键函数的返回类型:

fn sort_by_key_v2(a: &mut [T], key: fn(&T) -> &T) {a.sort_by(|x, y| key(x).cmp(&key(y)));}fn sort_by_key_v3(a: &mut [T], key: fn(&T) -> &K) {a.sort_by(|x, y| key(x).cmp(&key(y)));}

我也尝试添加生命周期注释，但这只会改变错误而没有解决它.

这是sort_by_key() 操场上的功能.

为什么我会收到这些错误?有没有办法在保持密钥类型 K 完全通用的同时修复它们?

解决方案

现在，你必须使用长"形式:

v.sort_by(|x, y| key(x).cmp(&key(y)));

<块引用>

为什么我会收到这些错误?有什么办法可以解决吗?

原因和修复是一回事:Rust 目前的表现力不足以代表您想要的东西.所需的功能称为通用关联类型 (GATs);以前称为关联类型构造函数 (ATC) 或更高级的类型 (HKT).

来自相关问题:

<块引用>

为了使 sort_by_key 调用正常，需要将输入引用 [...] 的生命周期合并到 B 中以使其返回类型 &'a str，但 B 是一个类型参数.

我不知道 sort_by_key 的签名在实施时是否能够无缝移动到 GAT.

<小时>

在您控制所有类型签名的类似情况下，您可以要求返回引用:

使用 std::cmp::Ordering;结构用户{名称:字符串，}fn compare_keys<T，R>(a:T，b:T，key:impl Fn(&T)->&R)->订购在哪里对于<'a>&'a R:奥德，{让 ak = key(&a);让 bk = key(&b);ak.cmp(&bk)}fn 主(){让爱丽丝 = 用户 {名称:String::from("alice"),};让鲍勃 = 用户 {名称:String::from("bob"),};compare_keys(alice, bob, |u| &u.name);}

这是不理想的，因为现在您不能返回非引用，但在实施 GAT 之前根本没有完整的解决方案.根据您的情况，您可以添加诸如 sort_by 和 sort_by_key 之类的并行方法.

I'm trying to sort a Vec<String> using a key function that returns references to the strings in the vector. A contrived example is to use the identity function as key function (which of course is useless, but it's the minimal example to reproduce my problem):

fn key(x: &String) -> &String {
    x
}

Now given items: Vec<String>, I'd like to be able to do

items.sort_by_key(key);

This gives the following error:

error[E0271]: type mismatch resolving `for<'r> <fn(&std::string::String) -> &std::string::String {main::key} as std::ops::FnOnce<(&'r std::string::String,)>>::Output == _`
  --> src/main.rs:19:11
   |
19 |     items.sort_by_key(key);
   |           ^^^^^^^^^^^ expected bound lifetime parameter, found concrete lifetime
   |
   = note: concrete lifetime that was found is lifetime '_#16r

I don't understand why I get this error, so I tried to track this down. I first implemented my own version of sort_by_key():

fn sort_by_key<T, K: Ord>(a: &mut [T], key: fn(&T) -> K) {
    a.sort_by(|x, y| key(x).cmp(&key(y)));
}

When trying to call this function, I get what looks like the "opposite" error:

error[E0308]: mismatched types
  --> src/main.rs:22:29
   |
22 |     sort_by_key(&mut items, key);
   |                             ^^^ expected concrete lifetime, found bound lifetime parameter
   |
   = note: expected type `fn(&std::string::String) -> _`
              found type `fn(&std::string::String) -> &std::string::String {main::key}`

I can make this code compile by fixing the key type to &T instead of using the generic parameter K, or by using &K instead of K as return type for the key function:

fn sort_by_key_v2<T: Ord>(a: &mut [T], key: fn(&T) -> &T) {
    a.sort_by(|x, y| key(x).cmp(&key(y)));
}
fn sort_by_key_v3<T, K: Ord>(a: &mut [T], key: fn(&T) -> &K) {
    a.sort_by(|x, y| key(x).cmp(&key(y)));
}

I also tried adding lifetime annotations, but that only shifted the error around without resolving it.

Here's the three versions of the sort_by_key() function on the Playground.

Why am I getting these errors? Is there any way to fix them while keeping the key type K completely generic?

解决方案

For now, you have to use the "long" form:

v.sort_by(|x, y| key(x).cmp(&key(y)));

Why am I getting these errors? Is there any way to fix them?

The cause and fix are one-and-the same: Rust is simply not currently expressive enough to represent what you want. The feature needed is called generic associated types (GATs); previously known as associated type constructors (ATCs) or higher-kinded types (HKTs).

From the associated issue:

For the sort_by_key call to be okay, the lifetime of the input reference [...] needs to be incorporated into B to make the return type &'a str, but B is a type parameter.

I don't know if the signature for sort_by_key will be able to be seamlessly moved to a GAT when they are implemented.

---

In similar cases where you control the signature of all the types, you can require that a reference be returned:

use std::cmp::Ordering;

struct User {
    name: String,
}

fn compare_keys<T, R>(a: T, b: T, key: impl Fn(&T) -> &R) -> Ordering
where
    for<'a> &'a R: Ord,
{
    let ak = key(&a);
    let bk = key(&b);
    ak.cmp(&bk)
}

fn main() {
    let alice = User {
        name: String::from("alice"),
    };
    let bob = User {
        name: String::from("bob"),
    };

    compare_keys(alice, bob, |u| &u.name);
}

This is non-ideal because now you cannot return a non-reference, but there's simply no complete solution until GATs are implemented. You may be able to add a parallel methods like sort_by and sort_by_key, depending on your case.

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