1

我不知道如何将闭包对象存储在结构中。闭包对象的参数和返回值是已知的。这是我的精简代码：

struct Instr<F>
    where F: Fn([i32;4],[i32;3]) -> [i32;4]
{
    name: String,
    op: F
}

fn main()
{
    // Simple example showing the difficulty:
    let tmp : Instr<Fn([i32;4],[i32;3]) -> [i32;4]> = Instr { name: "asd".to_string(), op: |a,b| a};

    // What I really want is something more like this:
    // let instrs = vec![
    //     Instr { name: "asdf", op: |a,b| a },
    //     Instr { name: "qwer", op: |a,b| a }
    // ];
}

坦率地说，我不明白这些错误是什么意思。在我看来，这很简单。闭包具有类型和已知尺寸。将其存储在相同类型的类型字段中应该很简单。正确的？

尝试F: ?Sized按照错误消息提示进行添加并不能修复“编译时大小未知”错误。

有人可以帮我正确编译吗？

error[E0277]: the size for values of type `dyn Fn([i32; 4], [i32; 3]) -> [i32; 4]` cannot be known at compilation time
  --> a.rs:11:15
   |
1  | struct Instr<F>
   |              - required by this bound in `Instr`
...
11 |     let tmp : Instr<Fn([i32;4],[i32;3]) -> [i32;4]> = Instr { name: "asd".to_string(), op: |a,b| a};
   |               ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ doesn't have a size known at compile-time
   |
   = help: the trait `Sized` is not implemented for `dyn Fn([i32; 4], [i32; 3]) -> [i32; 4]`
help: you could relax the implicit `Sized` bound on `F` if it were used through indirection like `&F` or `Box<F>`
  --> a.rs:1:14
   |
1  | struct Instr<F>
   |              ^ this could be changed to `F: ?Sized`...
2  |     where F : Fn([i32;4],[i32;3]) -> [i32;4]
   |           - ...if indirection was used here: `Box<F>`
...
5  |     op : F
   |          - ...if indirection was used here: `Box<F>`

error[E0277]: the size for values of type `dyn Fn([i32; 4], [i32; 3]) -> [i32; 4]` cannot be known at compilation time
  --> a.rs:11:55
   |
1  | / struct Instr<F>
2  | |     where F : Fn([i32;4],[i32;3]) -> [i32;4]
3  | | {
4  | |     name : String,
5  | |     op : F
6  | | }
   | |_- required by `Instr`
...
11 |       let tmp : Instr<Fn([i32;4],[i32;3]) -> [i32;4]> = Instr { name: "asd".to_string(), op: |a,b| a};
   |                                                         ^^^^^ doesn't have a size known at compile-time
   |
   = help: the trait `Sized` is not implemented for `dyn Fn([i32; 4], [i32; 3]) -> [i32; 4]`

error[E0308]: mismatched types
  --> a.rs:11:92
   |
11 |     let tmp : Instr<Fn([i32;4],[i32;3]) -> [i32;4]> = Instr { name: "asd".to_string(), op: |a,b| a};
   |                                                                                            ^^^^^^^ expected trait object `dyn Fn`, found closure
   |
   = note: expected trait object `dyn Fn([i32; 4], [i32; 3]) -> [i32; 4]`
                   found closure `[closure@a.rs:11:92: 11:99]`
4

2 回答 2

5

不同的闭包具有不同的大小，因此您不能在结构中存储“原始闭包”或“原始特征对象”，它们必须位于指针后面，因此您可以将它们放在Box这样的位置：

struct Instr {
    name: String,
    op: Box<dyn Fn([i32; 4], [i32; 3]) -> [i32; 4]>,
}

fn main() {
    let instrs = vec![
         Instr { name: "asdf".into(), op: Box::new(|a,b| a) },
         Instr { name: "qwer".into(), op: Box::new(|a,b| a) }
     ];
}

操场

于 2021-01-17T00:26:31.230 回答
3

接受的答案完美地解决了您的用例的解决方案，但我想澄清“未调整大小”的错误消息以及“简单示例”无法正常工作。

Rust 非常有能力Instr按照问题中的定义存储闭包，但是您的类型规范混淆了它。每个闭包的类型都是匿名的，你不能命名它。您尝试通过拼写 trait 来指定闭包类型是Fn(ARGS...) -> RESULT错误的，因为在 Rust 中，当您使用预期类型的​​ trait 时，它指的是 trait 的动态实现，也就是trait object。它是未调整大小的特征对象，必须通过引用或智能指针访问。

所以，如果你让 Rust 推断它的类型，你可以在其中创建一个Instr带有任意闭包的：

struct Instr<F>
    where F: Fn([i32;4],[i32;3]) -> [i32;4]
{
    name: String,
    op: F
}

fn main()
{
    // Simple example
    let tmp : Instr<_> = Instr { name: "asd".to_string(), op: |a,b| a};
}

但这不允许您创建一个Instrs 的向量，每个向量都有不同的闭包，因为这些Instrs 将具有不同的类型。为此，您需要使用参考或 aBox如接受的答案所示。

于 2021-01-17T09:33:30.717 回答