为什么 Rust 禁止现有类型的实现？

Question

这是怎么...

impl String {
    fn foo(&self) {
        //...
    }
}

...与此有何不同？

fn foo(s: &String) {
    //...
}

然后，如果您在 crate 中定义特征，则可以扩展类型实现。为什么？

Answer 1

下面 source 指出了几种不同的论点，说明为什么人们无法实现包装箱之外的现有类型。

本地实现可能会被未来的实现破坏。例如，考虑 "you've locally defined Add on Vec<T> as a concat operator, ... , and then ... after years of debate ... some mathy operation [is] to be performed instead. If you delete your impl and upgrade, your code will be ... broken²."
代码的可读性也将受到此更改的影响，即它可能使 "value of that reading far more transient³."
还有一个安全问题。考虑以下在技术上可行的场景，如果允许的话，即 "an attacker [could] find a change in an impl in [some] library, a call site in an application they wish to backdoor, and send a "重构“拉取请求 "accidentally" 将新的 impl 替换为旧的 impl 以创建漏洞，但他们的拉取可以从库中引用旧代码。他们可以将恶意实现嵌入到另一个 create⁴."

假设如果允许本地实现，则本地实现将是首选实现。这将“违反一致性属性 [目前正在维护]⁵." This point can be further clarified through what is called the 'HashTable' problem⁵。

mod foo {
    impl Hash for i32 { ... }

    fn f(mut table: HashMap<i32, &'static str>) {
        table.insert(0, "hello");
        ::bar::f(&table);
    }
}

mod bar {
    impl Hash for i32 { ... }

    fn f(table: &HashMap<i32, &'static str>) {
        assert_eq!(table.get(0), Some("hello"));
    }
}

为什么 Rust 禁止现有类型的实现？

Why does Rust forbid implementations of existing types?

methods

implementation

types

rust