解构枚举时是否可以向上转换为特征?
Is it possible to upcast to a trait when destructuring enums?
我希望能够 destructure/upcast 枚举常见特征的变体。这是设置:
trait SomeTrait {
fn some_func(&self);
}
struct SomeStruct1;
impl SomeTrait for SomeStruct1 {
fn some_func(&self) {}
}
struct SomeStruct2;
impl SomeTrait for SomeStruct2 {
fn some_func(&self) {}
}
enum SomeEnum {
Value1(SomeStruct1),
Value2(SomeStruct2),
}
这是我尝试过的一种可能性:
fn call_some_func(some_enum: SomeEnum) {
match some_enum {
SomeEnum::Value1(ref some_trait: &SomeTrait) |
SomeEnum::Value2(ref some_trait: &SomeTrait) => some_trait.some_func()
}
}
它导致了这个错误:
error: expected one of `)`, `,`, or `@`, found `:`
--> src/main.rs:22:40
|
22 | SomeEnum::Value1(ref some_trait: &SomeTrait) |
| ^ expected one of `)`, `,`, or `@` here
这是我的另一个想法:
fn call_some_func2(some_enum: SomeEnum) {
match some_enum {
_(ref some_trait: &SomeTrait) => some_trait.some_func()
}
}
也失败了:
error: expected one of `=>`, `if`, or `|`, found `(`
--> src/main.rs:22:10
|
22 | _(ref some_trait: &SomeTrait) => some_trait.some_func()
| ^ expected one of `=>`, `if`, or `|` her
有谁知道完成这种解构的方法吗?
不,你不能同时解构。你可以在 之后完成它:
// My preferred
fn call_some_func(some_enum: SomeEnum) {
let x: &SomeTrait = match some_enum {
SomeEnum::Value1(ref x) => x,
SomeEnum::Value2(ref x) => x,
};
x.some_func();
}
// Also works
fn call_some_func(some_enum: SomeEnum) {
let x = match some_enum {
SomeEnum::Value1(ref x) => x as &SomeTrait,
SomeEnum::Value2(ref x) => x,
};
x.some_func();
}
我鼓励您将其提取到一个方法中以供重用:
impl SomeEnum {
fn as_trait(&self) -> &SomeTrait {
match *self {
SomeEnum::Value1(ref x) => x,
SomeEnum::Value2(ref x) => x,
}
}
}
fn call_some_func(some_enum: SomeEnum) {
some_enum.as_trait().some_func();
}
并且可能在枚举本身上实现特征,如果所有变体都实现的话。这样消费者就不用关心了:
impl SomeTrait for SomeEnum {
fn some_func(&self) {
self.as_trait().some_func()
}
}
我希望能够 destructure/upcast 枚举常见特征的变体。这是设置:
trait SomeTrait {
fn some_func(&self);
}
struct SomeStruct1;
impl SomeTrait for SomeStruct1 {
fn some_func(&self) {}
}
struct SomeStruct2;
impl SomeTrait for SomeStruct2 {
fn some_func(&self) {}
}
enum SomeEnum {
Value1(SomeStruct1),
Value2(SomeStruct2),
}
这是我尝试过的一种可能性:
fn call_some_func(some_enum: SomeEnum) {
match some_enum {
SomeEnum::Value1(ref some_trait: &SomeTrait) |
SomeEnum::Value2(ref some_trait: &SomeTrait) => some_trait.some_func()
}
}
它导致了这个错误:
error: expected one of `)`, `,`, or `@`, found `:`
--> src/main.rs:22:40
|
22 | SomeEnum::Value1(ref some_trait: &SomeTrait) |
| ^ expected one of `)`, `,`, or `@` here
这是我的另一个想法:
fn call_some_func2(some_enum: SomeEnum) {
match some_enum {
_(ref some_trait: &SomeTrait) => some_trait.some_func()
}
}
也失败了:
error: expected one of `=>`, `if`, or `|`, found `(`
--> src/main.rs:22:10
|
22 | _(ref some_trait: &SomeTrait) => some_trait.some_func()
| ^ expected one of `=>`, `if`, or `|` her
有谁知道完成这种解构的方法吗?
不,你不能同时解构。你可以在 之后完成它:
// My preferred
fn call_some_func(some_enum: SomeEnum) {
let x: &SomeTrait = match some_enum {
SomeEnum::Value1(ref x) => x,
SomeEnum::Value2(ref x) => x,
};
x.some_func();
}
// Also works
fn call_some_func(some_enum: SomeEnum) {
let x = match some_enum {
SomeEnum::Value1(ref x) => x as &SomeTrait,
SomeEnum::Value2(ref x) => x,
};
x.some_func();
}
我鼓励您将其提取到一个方法中以供重用:
impl SomeEnum {
fn as_trait(&self) -> &SomeTrait {
match *self {
SomeEnum::Value1(ref x) => x,
SomeEnum::Value2(ref x) => x,
}
}
}
fn call_some_func(some_enum: SomeEnum) {
some_enum.as_trait().some_func();
}
并且可能在枚举本身上实现特征,如果所有变体都实现的话。这样消费者就不用关心了:
impl SomeTrait for SomeEnum {
fn some_func(&self) {
self.as_trait().some_func()
}
}