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如何为需要引用“静态生存期”并实现同步的函数提供模拟值?

type-conversion rust mocking unit-testing

llxxbb · 技术社区 · 7 年前

我用过 mocker 来测试我的项目。我不知道需要什么类型的类型转换 say_hello_brother . 简化代码列表如下:

lib。卢比

#![feature(plugin, custom_derive)]
#![plugin(mockers_macros)]
#[cfg(test)]
extern crate mockers;

use mockers::Scenario;

#[derive(Mock)]
trait SayHello {
    fn hello(&self);
}

// assume `SayHello` is a service and  worked on multiple threads
fn say_hello_brother<T: SayHello + Sync>(brother: &'static T) {
    brother.hello()
}

#[test]
fn test_sya_hello() {
    let scenario = Scenario::new();
    let mock = scenario.create_mock_for::<SayHello>();
    say_hello_brother(&mock)
}

货物汤姆

[package]
name = "mock"
version = "0.1.0"
authors = ["llxxb"]

[dependencies]
mockers = "0.9.4"
mockers_macros = "0.9.4"

和错误信息:

error[E0277]: the trait bound `std::rc::Rc<std::cell::RefCell<mockers::ScenarioInternals>>: std::marker::Sync` is not satisfied in `SayHelloMock`
  --> src\lib.rs:22:5
   |
22 |     say_hello_brother(&mock)
   |     ^^^^^^^^^^^^^^^^^ `std::rc::Rc<std::cell::RefCell<mockers::ScenarioInternals>>` cannot be shared between threads safely
   |
   = help: within `SayHelloMock`, the trait `std::marker::Sync` is not implemented for `std::rc::Rc<std::cell::RefCell<mockers::ScenarioInternals>>`
   = note: required because it appears within the type `SayHelloMock`

更新经过一番尝试,我成功地 Sync 到 SayHelloMock . 新库。卢比:

#![feature(plugin, custom_derive)]
#![plugin(mockers_macros)]
#[cfg(test)]
extern crate mockers;

use mockers::Scenario;

trait SayHello {
    fn hello(&self);
}


mock! {
    SayHelloMock,
    self,
    trait SayHello {
        fn hello(&self);
    }
}

unsafe impl Sync for SayHelloMock {}


// assume `SayHello` is a service and  worked on multiple threads
fn say_hello_brother<T: SayHello + Sync>(brother: &'static T) {
    brother.hello()
}

#[test]
fn test_sya_hello() {
    let scenario = Scenario::new();
// not work
//    let mock = scenario.create_mock::<SayHelloMock>();
//    static MOCK: SayHelloMock = || { mock };
//    say_hello_brother(&MOCK)

    // not work yet
    let mock : &'static SayHelloMock = &(scenario.create_mock::<SayHelloMock>());
    say_hello_brother(mock)
}

但我仍然无法将其转换为“静态错误信息:

error[E0597]: borrowed value does not live long enough
--> src\lib.rs:38:41
|
38 |     let mock : &'static SayHelloMock = &(scenario.create_mock::<SayHelloMock>());
|                                         ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ temporary value does not live long enough
39 |     say_hello_brother(mock)
40 | }
| - temporary value only lives until here
|
= note: borrowed value must be valid for the static lifetime...

2 回复 | 直到 7 年前

Shepmaster Tim Diekmann 7 年前

你不能为此,请使用mocker库,因为它生成的Mock是 not thread safe .

也就是说,没有什么限制您创建自己的线程安全模拟:

use std::sync::atomic::{AtomicUsize, Ordering};

#[derive(Debug, Default)]
struct MyMock {
    call_count: AtomicUsize,
}

impl SayHello for MyMock {
    fn hello(&self) {
        self.call_count.fetch_add(1, Ordering::SeqCst);
    }
}

但是,您的签名需要 &'static T ,这可能真的是个糟糕的主意。您需要使用类似惰性静态的东西:

#[macro_use]
extern crate lazy_static;

#[test]
fn test_say_hello() {
    lazy_static! { 
        static ref MOCK: MyMock = MyMock::default();
    };
    say_hello_brother(&*MOCK);
    assert_eq!(MOCK.call_count.load(Ordering::SeqCst), 1);
}

Shepmaster Tim Diekmann 7 年前

这个问题有点模糊,但如果我理解正确,您想知道如何将模拟传递给函数 say_hello_brother . 问题是该函数需要一个实现 Sync 还有特质。因此,您无法转换 mock 以获取要编译的代码。

根据 documentation of mocker ,你可以试着同时模仿两个特征。下面是一些伪代码,说明了这一想法:

mock! {
    SayHelloMock,
    self,
    trait SayHello {
        // trait methods here
    },
    self,
    trait Sync {
        // trait methods here
    }
}

然后在测试中,您将创建如下模拟:

let mut mock = scenario.create_mock::<SayHelloMock>();