有没有办法区分具有相同值的不同`Rc`? [英] Is there a way to distinguish between different `Rc`s of the same value?

问题描述

这是一个例子:

use std::rc::Rc;

#[derive(PartialEq, Eq)]
struct MyId;

pub fn main() {
    let rc_a_0 = Rc::new(MyId);
    let rc_a_1 = rc_a_0.clone();
    let rc_b_0 = Rc::new(MyId);
    let rc_b_1 = rc_b_0.clone();

    println!("rc_a_0 == rc_a_1: {:?}", rc_a_0 == rc_a_1);
    println!("rc_a_0 == rc_b_0: {:?}", rc_a_0 == rc_b_0);
}

两个println!都打印true.有没有办法区分rc_a_*和rc_b_*指针?

Both println!s above print true. Is there a way distinguish between the rc_a_* and rc_b_* pointers?

推荐答案

您可以将&*rc强制转换为*const T以获得指向基础数据的指针并比较这些指针的值:

You can cast &*rc to *const T to get a pointer to the underlying data and compare the value of those pointers:

use std::rc::Rc;

#[derive(PartialEq, Eq)]
struct MyId;

pub fn main() {
    let rc_a_0 = Rc::new(MyId);
    let rc_a_1 = rc_a_0.clone();
    let rc_b_0 = Rc::new(MyId);
    let rc_b_1 = rc_b_0.clone();

    println!(
        "rc_a_0 == rc_a_1: {:?}",
        &*rc_a_0 as *const MyId == &*rc_a_1 as *const MyId
    );
    println!(
        "rc_a_0 == rc_b_0: {:?}",
        &*rc_a_0 as *const MyId == &*rc_b_0 as *const MyId
    );
}

打印

rc_a_0 == rc_a_1: true
rc_a_0 == rc_b_0: false

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