无法将`* x`借为可变的，因为它也被借为不可变的 [英] Cannot borrow `*x` as mutable because it is also borrowed as immutable

问题描述

我正在做一个组合优化项目来学习Rust，但遇到一个我无法解决自己的问题...

我有2个功能:

pub fn get_pareto_front_offline<'a>(scheduling_jobs: &'a Vec<Vec<u32>>, costs_vector: &'a Vec<(u32, u32)>) -> Vec<(&'a Vec<u32>, &'a (u32, u32))> {
    // ...
}

和

pub fn pareto_approach_offline<'a>(list_of_jobs: &'a mut Vec<Vec<u32>>, neighborhood: &'a mut Vec<Vec<u32>>, costs: &'a Vec<(u32, u32)>) -> Vec<(&'a Vec<u32>, &'a (u32, u32))> {
    let pareto_front = get_pareto_front_offline(neighborhood, costs);

    loop {
        if pareto_front == vec![] {
            break;
        }

        neighborhood.clear();

        for front in pareto_front.iter() {
            neighborhood.push((front.0).clone());
        }
    }

    pareto_front
}

我遇到了一个问题，因为编译器告诉我:

cannot borrow '*neighborhood' as mutable because it is also borrowed as immutableat line 15 col 9
cannot borrow '*neighborhood' as mutable because it is also borrowed as immutableat line 19 col 13

解决方案

您正在尝试做根本上不可能的事情.

调用get_pareto_front_offline时，会将neighborhood的重新借用传递给该函数.必须重新维护此 ，以使pareto_front保持有效.换句话说，只要pareto_front存在，编译器将 not 不允许您以任何方式访问neighborhood.

这是一件好事，因为您随后尝试尝试清除我们的neighborhood，这肯定会肯定使pareto_front无效，可能导致使用后使用和破坏程序的状态.

目前尚不清楚您要尝试做什么；但是您不能用这种方式.

顺便说一句，即使编译了该循环，该循环也可能永远不会结束运行:因为您永远不会修改pareto_front，所以永远不会满足您的终止条件(pareto_front == vec![])；它会立即停止，或者永远运行.

摆脱借款问题的最简单方法是制作事物的副本，这样您就不需要长期的借款.如果get_pareto_front_offline返回了Vec<(Vec<u32>, (u32, u32))>，则不会出现此问题.或者，将其修改为代码，以在调用get_pareto_front_offline时不触碰neighborhood.

I'm making a Combinatory Optimization project to learn Rust and I've got a problem I cannot resolve myself...

I've got 2 functions :

pub fn get_pareto_front_offline<'a>(scheduling_jobs: &'a Vec<Vec<u32>>, costs_vector: &'a Vec<(u32, u32)>) -> Vec<(&'a Vec<u32>, &'a (u32, u32))> {
    // ...
}

and

pub fn pareto_approach_offline<'a>(list_of_jobs: &'a mut Vec<Vec<u32>>, neighborhood: &'a mut Vec<Vec<u32>>, costs: &'a Vec<(u32, u32)>) -> Vec<(&'a Vec<u32>, &'a (u32, u32))> {
    let pareto_front = get_pareto_front_offline(neighborhood, costs);

    loop {
        if pareto_front == vec![] {
            break;
        }

        neighborhood.clear();

        for front in pareto_front.iter() {
            neighborhood.push((front.0).clone());
        }
    }

    pareto_front
}

I've got a problem because the compiler tells me:

cannot borrow '*neighborhood' as mutable because it is also borrowed as immutableat line 15 col 9
cannot borrow '*neighborhood' as mutable because it is also borrowed as immutableat line 19 col 13

解决方案

You're trying to do something fundamentally impossible.

When you call get_pareto_front_offline, you pass a re-borrowing of neighborhood into that function. This re-borrow must be maintained in order for pareto_front to remain valid. In other words, as long as pareto_front exists, the compiler will not allow you to access neighborhood in any fashion whatsoever.

This is a good thing, because you then proceed to try and clear our neighborhood, which would almost certainly invalidate pareto_front, likely leading to use-after-free and corrupting your program's state.

It's not clear what it is you're attempting to do; but you cannot do it this way.

As an aside, even if it compiled, that loop would probably never finish running: your termination condition (pareto_front == vec![]) will never be satisfied because you never modify pareto_front; it'll either stop immediately, or run forever.

The simplest way to get out from under borrowing problems is to make copies of things, so that you don't need a long-lived borrow; if get_pareto_front_offline returned a Vec<(Vec<u32>, (u32, u32))> instead, you wouldn't have this issue. That, or modify to code to not touch neighborhood once you call get_pareto_front_offline.

这篇关于无法将`* x`借为可变的，因为它也被借为不可变的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！