我是trying to implement a flexible type system at runtime in Rust。到目前为止,这是我所能使用的:
use std::borrow::Cow;
pub struct Float {
pub min: f64,
pub max: f64,
pub value: f64,
}
pub struct Text<'a> {
pub value: Cow<'a, str>
}
pub enum Value<'a> {
None,
Float(Float),
Text(Text<'a>),
}
这可以按我想要的方式工作,现在我需要一个向量(下一步就是地图),所以我添加了:
pub struct Vector<'a> {
pub value: Cow<'a, Vec<Value<'a>>>,
}
并将枚举扩展为:
pub enum Value<'a> {
None,
Float(Float),
Text(Text<'a>),
Vector(Vector<'a>),
}
现在我收到一条错误消息:
error[E0277]: the trait bound `Value<'a>: std::clone::Clone` is not satisfied
--> src/lib.rs:14:5
|
14 | pub value: Cow<'a, Vec<Value<'a>>>,
| ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ the trait `std::clone::Clone` is not implemented for `Value<'a>`
|
= note: required because of the requirements on the impl of `std::clone::Clone` for `std::vec::Vec<Value<'a>>`
= note: required because of the requirements on the impl of `std::borrow::ToOwned` for `std::vec::Vec<Value<'a>>`
= note: required by `std::borrow::Cow`
error[E0277]: the trait bound `Value<'a>: std::clone::Clone` is not satisfied
--> src/lib.rs:21:12
|
21 | Vector(Vector<'a>),
| ^^^^^^^^^^ the trait `std::clone::Clone` is not implemented for `Value<'a>`
|
= note: required because of the requirements on the impl of `std::clone::Clone` for `std::vec::Vec<Value<'a>>`
= note: required because of the requirements on the impl of `std::borrow::ToOwned` for `std::vec::Vec<Value<'a>>`
= note: required because it appears within the type `Vector<'a>`
= note: no field of an enum variant may have a dynamically sized type
我尝试了几种实现Clone
的方法,但是作为一个初学者,我最终遇到了无数其他错误消息。如何将Value
的向量导入该系统?
为什么要这样做?
我有以下代码来简化Value
的使用:
impl<'a> Value<'a> {
pub fn float_val(&self) -> f64 {
match self {
Value::None => 0.0,
Value::Float(f) => f.value,
Value::Text(t) => t.value.parse().unwrap_or(0.0),
}
}
pub fn str_val(&'a self) -> Cow<'a, str> {
match self {
Value::None => Cow::Owned("".to_string()),
Value::Float(f) => Cow::Owned(f.value.to_string()),
Value::Text(t) => Cow::Borrowed(&t.value),
}
}
}
这使我可以在以下功能中使用它:
fn append_string(s1: &Value, s2: &Value) {
Value::Text(Text {
value: format!("{}{}", s1.str_val(), s2.str_val()),
})
}
我想要一个向量,我想这应该是这样的:
pub fn vec(&'a self) -> Vec<Value> {
match self {
Value::Vector(v) => v.value,
_ => Cow::Owned(Vector { value: vec![] }),
}
}
答案 0 :(得分:3)
我想要一个向量,我想这应该是这样的:
pub fn vec(&'a self) -> Vec<Value> { match self { Value::Vector(v) => v.value, _ => Cow::Owned(Vector { value: vec![] }), } }
首先,从函数返回Cow
并不意味着您必须将数据存储为Cow
。您可以这样做:
pub struct Vector<'a> {
pub value: Vec<Value<'a>>,
}
pub enum Value<'a> {
None,
Float(Float),
Text(Text<'a>),
Vector(Vector<'a>),
}
然后您的to_vec
会看起来像这样:
impl<'a> Value<'a> {
pub fn to_vec(&'a self) -> Cow<'a, [Value<'a>]> {
match self {
Value::Vector(v) => Cow::Borrowed(&v.value),
_ => Cow::Owned(Vec::new()),
}
}
}
除了对于ToOwned
实施Value<'a>
仍然会有一些问题,因此这将不会立即起作用。
但是,我仍然不明白为什么这里需要Cow
。 Cow
对借用vs拥有类型进行抽象,但是您拥有的值始终为空,那么在两种情况下返回借用切片的危害是什么?看起来像这样:
impl<'a> Value<'a> {
pub fn to_vec_slice(&'a self) -> &'a [Value<'a>] {
match self {
Value::Vector(v) => &v.value,
_ => &[],
}
}
}