我希望 Rust 中的枚举可以像 Haskell 的生产类型一样使用。我要
- 直接访问字段的值
- 直接分配一个字段的值或使用变化的值创建一个克隆。
直接就是不使用太长的模式匹配代码,只是可以像let a_size = a.size
那样访问。
在 Haskell 中:
data TypeAB = A {size::Int, name::String} | B {size::Int, switch::Bool} deriving Show
main = do
let a = A 1 "abc"
let b = B 1 True
print (size a) -- could access a field's value directly
print (name a) -- could access a field's value directly
print (switch b) -- could access a field's value directly
let aa = a{size=2} -- could make a clone directly with the changing value
print aa
我尝试了两种风格的 Rust 枚举定义,例如
样式 A:
#[derive(Debug)]
enum EntryType {
A(TypeA),
B(TypeB),
}
#[derive(Debug)]
struct TypeA {
size: u32,
name: String,
}
#[derive(Debug)]
struct TypeB {
size: u32,
switch: bool,
}
fn main() {
let mut ta = TypeA {
size: 3,
name: "TAB".to_string(),
};
println!("{:?}", &ta);
ta.size = 2;
ta.name = "TCD".to_string();
println!("{:?}", &ta);
let mut ea = EntryType::A(TypeA {
size: 1,
name: "abc".to_string(),
});
let mut eb = EntryType::B(TypeB {
size: 1,
switch: true,
});
let vec_ab = vec![&ea, &eb];
println!("{:?}", &ea);
println!("{:?}", &eb);
println!("{:?}", &vec_ab);
// Want to do like `ta.size = 2` for ea
// Want to do like `ta.name = "bcd".to_string()` for ea
// Want to do like `tb.switch = false` for eb
// ????
println!("{:?}", &ea);
println!("{:?}", &eb);
println!("{:?}", &vec_ab);
}
样式 B:
#[derive(Debug)]
enum TypeCD {
TypeC { size: u32, name: String },
TypeD { size: u32, switch: bool },
}
fn main() {
// NOTE: Rust requires representative struct name before each constructor
// TODO: Check constructor name can be duplicated
let mut c = TypeCD::TypeC {
size: 1,
name: "abc".to_string(),
};
let mut d = TypeCD::TypeD {
size: 1,
switch: true,
};
let vec_cd = vec![&c, &d];
println!("{:?}", &c);
println!("{:?}", &d);
println!("{:?}", &vec_cd);
// Can't access a field's value like
// let c_size = c.size
let c_size = c.size; // [ERROR]: No field `size` on `TypeCD`
let c_name = c.name; // [ERROR]: No field `name` on `TypeCD`
let d_switch = d.switch; // [ERROR]: No field `switch` on `TypeCD`
// Can't change a field's value like
// c.size = 2;
// c.name = "cde".to_string();
// d.switch = false;
println!("{:?}", &c);
println!("{:?}", &d);
println!("{:?}", &vec_cd);
}
我无法以任何样式直接访问/分配值。我是否必须实现功能或特征才能访问字段的值?是否有某种方法可以帮助解决这种情况?
最佳答案
风格C呢:
#[derive(Debug)]
enum Color {
Green { name: String },
Blue { switch: bool },
}
#[derive(Debug)]
struct Something {
size: u32,
color: Color,
}
fn main() {
let c = Something {
size: 1,
color: Color::Green {
name: "green".to_string(),
},
};
let d = Something {
size: 2,
color: Color::Blue { switch: true },
};
let vec_cd = vec![&c, &d];
println!("{:?}", &c);
println!("{:?}", &d);
println!("{:?}", &vec_cd);
let _ = c.size;
}
如果所有变体都有共同点,为什么要将它们分开?
Of course, I need to access not common field too.
这意味着 Rust 应该定义当运行时的实际类型不包含您需要的字段时要做什么。所以,我认为 Rust 不会有一天添加这个。
你可以自己做。它将需要一些代码行,但这与您的 Haskell 代码的行为相匹配。但是,我认为这不是最好的做法。 Haskell 就是 Haskell,我认为你应该用 Rust 编写代码,而不是尝试使用 Rust 编写 Haskell。一般规则,Rust 的某些功能直接来自 Haskell,但在我看来,你想要的 Rust 代码非常奇怪。
#[derive(Debug)]
enum Something {
A { size: u32, name: String },
B { size: u32, switch: bool },
}
impl Something {
fn size(&self) -> u32 {
match self {
Something::A { size, .. } => *size,
Something::B { size, .. } => *size,
}
}
fn name(&self) -> &String {
match self {
Something::A { name, .. } => name,
Something::B { .. } => panic!("Something::B doesn't have name field"),
}
}
fn switch(&self) -> bool {
match self {
Something::A { .. } => panic!("Something::A doesn't have switch field"),
Something::B { switch, .. } => *switch,
}
}
fn new_size(&self, size: u32) -> Something {
match self {
Something::A { name, .. } => Something::A {
size,
name: name.clone(),
},
Something::B { switch, .. } => Something::B {
size,
switch: *switch,
},
}
}
// etc...
}
fn main() {
let a = Something::A {
size: 1,
name: "Rust is not haskell".to_string(),
};
println!("{:?}", a.size());
println!("{:?}", a.name());
let b = Something::B {
size: 1,
switch: true,
};
println!("{:?}", b.switch());
let aa = a.new_size(2);
println!("{:?}", aa);
}
关于enums - 有没有办法在没有模式匹配的情况下直接访问枚举结构中的字段值?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54061281/