这是错误消息：
error[E0038]: the trait `room::RoomInterface` cannot be made into an object
--> src\permanent\registry\mod.rs:12:33
|
12 | pub fn load(id : String) -> Result<Box<dyn RoomInterface>, String>{
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ `room::RoomInterface` cannot be made into an object
|
note: for a trait to be "object safe" it needs to allow building a vtable to allow the call to be resolvable dynamically; for more information visit <https://doc.rust-lang.org/reference/items/traits.html#object-safety>
--> src\permanent\rooms\mod.rs:36:31
|
36 | pub trait RoomInterface : Copy + Sized {
| ------------- ^^^^ ^^^^^ …是否可能,如果是,那么怎么做,创建一个类的新对象会返回除对象本身之外的东西？
假设,我希望每个新创建的对象都以包含自身的列表开头.
>> class A:
*magic*
>> a = A()
>> print a
[<__main__.A instance at 0x01234567>]
可能它可以通过__new__某种方式覆盖方法来完成,但是如何？
我希望有这个漂亮,干净的方式来描述具有函数和模式匹配的对象的属性:
data Animal = Cat | Dog | Cow
isBig :: Animal -> Bool
isLoyal :: Animal -> Bool
-- many more possible properties, including complicated non-bools, methods, and whatnot
--- Describing Cat
isBig Cat = False
isLoyal Cat = False
--- more properties
--- Describing Dog
isBig Dog = False
isLoyal Dog = True
--- more properties
--- Describing Cow
isBig Cow = True
isLoyal Cow = False
--- more properties
但是,这会产生有关多个声明的错误.因为,显然,通过模式匹配的函数定义必须在连续的行中完成.
这表明我的方法是错误的,不像Haskell一样吗？或者它只是语言中的一个缺陷？或者我误解了什么？
看看这个小片段：
pub trait TestTrait {
fn test_function(&self) {
generic_function(self);
}
}
fn generic_function<T : TestTrait> (x : T) {
//do something
}
这会产生以下错误消息：
error[E0277]: the trait bound `&Self: TestTrait` is not satisfied
--> src\main.rs:10:26
|
10 | generic_function(self);
| ---------------- ^^^^ the trait `TestTrait` is not implemented for `&Self`
| |
| required by a bound introduced by this call
|
note: required by a bound in `generic_function`
--> src\main.rs:14:25
|
14 | fn generic_function<T : TestTrait> (x : T) …