无法将 `&Thing` 与 `Thing` 进行比较

Question

我知道错误的含义，但我无法修复它。我正在使用mockers 来测试我的工作，并且在尝试验证提供给模拟特征函数的结构参数时遇到了困难。简化代码：

#[cfg(test)]
extern crate mockers;
#[cfg(test)]
extern crate mockers_derive;

#[cfg(test)]
use mockers_derive::mocked;

#[derive(Ord, PartialOrd, Eq, PartialEq, Debug)]
pub struct Thing {
    pub key: String,
    pub class: String,
}

#[cfg_attr(test, mocked)]
pub trait DaoTrait {
    fn get(&self, thing: &Thing) -> String;
}

struct DataService {
    dao: Box<DaoTrait>,
}

impl DataService {
    pub fn get(&self, thing: &Thing) -> String {
        self.dao.get(thing)
    }
}

#[cfg(test)]
mod test {
    use super::*;
    use mockers::matchers::eq;
    use mockers::Scenario;

    #[test]
    fn my_test() {
        use mockers::matchers::check;
        let scenario = Scenario::new();
        let mut dao = scenario.create_mock_for::<DaoTrait>();
        let thing = Thing {
            key: "my test".to_string(),
            class: "for test".to_string(),
        };

        scenario.expect(
            dao.get_call(check(|t: &Thing| t.to_owned() == thing))
                .and_return("hello".to_string()),
        );
        let testee = DataService { dao: Box::new(dao) };

        let rtn = testee.get(&thing);
        assert_eq!(rtn, "hello");
    }
}

我得到了错误：

warning: unused import: `mockers::matchers::eq`
  --> src/main.rs:33:9
   |
33 |     use mockers::matchers::eq;
   |         ^^^^^^^^^^^^^^^^^^^^^
   |
   = note: #[warn(unused_imports)] on by default

error[E0277]: can't compare `&Thing` with `Thing`
  --> src/main.rs:47:57
   |
47 |             dao.get_call(check(|t: &Thing| t.to_owned() == thing))
   |                                                         ^^ no implementation for `&Thing == Thing`
   |
   = help: the trait `std::cmp::PartialEq<Thing>` is not implemented for `&Thing`

error[E0277]: the trait bound `mockers::matchers::BoolFnMatchArg<Thing, [closure@src/main.rs:47:32: 47:65 thing:_]>: mockers::MatchArg<&Thing>` is not satisfied
  --> src/main.rs:47:17
   |
47 |             dao.get_call(check(|t: &Thing| t.to_owned() == thing))
   |                 ^^^^^^^^ the trait `mockers::MatchArg<&Thing>` is not implemented for `mockers::matchers::BoolFnMatchArg<Thing, [closure@src/main.rs:47:32: 47:65 thing:_]>`
   |
   = help: the following implementations were found:
             <mockers::matchers::BoolFnMatchArg<T, F> as mockers::MatchArg<T>>

我查看了check的源代码：

pub fn check<T, F: Fn(&T) -> bool>(f: F) -> BoolFnMatchArg<T, F> {
    BoolFnMatchArg { func: f, _phantom: PhantomData }
}

我认为我给出的关闭|t: &Thing| t.to_owned() == thing 是正确的。我也尝试了以下闭包，但都没有奏效。

|t: &Thing| t == &thing

|t: &Thing| *t == thing

|t: Thing| t == thing

Cargo.toml：

[dev-dependencies]
mockers = "0.12.1"
mockers_derive = "0.12.1"

Answer 1

您无法使用默认派生 PartialEq 将 Thing 与 &Thing 进行比较：

#[derive(Debug, PartialEq)]
struct Thing(String);

fn main() {
    let t_val = Thing(String::new());
    let t_ref = &t_val;

    t_val == t_ref;
}

error[E0308]: mismatched types
 --> src/main.rs:8:14
  |
8 |     t_val == t_ref;
  |              ^^^^^ expected struct `Thing`, found &Thing
  |
  = note: expected type `Thing`
             found type `&Thing`

要修复该错误，您需要执行以下两项操作之一：

1. 匹配参考水平：

   • t_val == *t_ref

   • &t_val == t_ref

2. 对不匹配数量的引用实现相等：

   impl<'a> PartialEq<&'a Thing> for Thing {
       fn eq(&self, other: &&'a Thing) -> bool {
           self == *other
       }
   }
   
   impl<'a> PartialEq<Thing> for &'a Thing {
       fn eq(&self, other: &Thing) -> bool {
           *self == other
       }
   }
   

---

但是，这些都不能解决您的实际问题。你误解了 mockers 库是如何工作的；您的闭包使用了错误的参考级别，它需要拥有要比较的值：

let expected_thing = thing.clone();
scenario.expect(
    dao.get_call(check(move |t: &&Thing| t == &&expected_thing))
        .and_return("hello".to_string()),
);

Answer 2

首先要注意的是t.to_owned() 生成&Thing，而不是您可能预期的Thing。那是因为Thing 没有实现Clone，因此它也没有实现ToOwned（因为有一个全面的实现为所有Clone 类型实现ToOwned），它提供了to_owned方法。但是为什么通话仍然有效？因为引用实现了Clone，所以&Thing 实现了ToOwned！这给了to_owned这个签名：

fn to_owned(self: &&Thing) -> &Thing;

您可以通过为Thing 派生Clone 来解决此问题。

但是，您不需要克隆Thing 来比较它。您可以改为比较对 Thing 的两个引用（例如，通过编写 |t: &Thing| t == &thing）。 PartialEq::eq（== 运算符转换为）通过引用获取参数，references implement PartialEq 通过剥离一层引用（即它们不比较指针值，与原始指针类型不同）。