如何在Rust结构中存储闭包？

html5 • • 问答 

struct User {
    id: i8,
    name: &'static str
}

struct UserMethods {
    add_user: Fn(User) -> (),
}

fn main() {
    UserMethods {
        add_user: |user| {

        }
    };
}

编译器说，

error[E0277]: the size for values of type `(dyn Fn(User) + 'static)` cannot be known at compilation time
  --> src/main.rs:11:5
   |
11 | /     UserMethods {
12 | |         add_user: |user| {},
13 | |     };
   | |_____^ doesn't have a size known at compile-time

回答

Fn是一个特征，因此它是未定大小的。为了将它存储在结构体或变量中，您必须使用实现的东西，Sized否则编译器无法知道要分配多少内存。

有两种方法可以解决此问题：

  1. 使用泛型：
struct UserMethods<T: Fn(User)> {
    add_user: T,
}

fn main() {
    UserMethods {
        add_user: |user| {},
    };
}
  1. 使用Box或任何其他智能指针：
struct UserMethods {
    add_user: Box<dyn Fn(User)>,
}

fn main() {
    UserMethods {
        add_user: Box::new(|user| {}),
    };
}

您也可以使用普通函数指针（感谢@user2722968），但它不如闭包灵活（无法从环境中捕获任何内容）：

struct UserMethods {
    add_user: fn(User),
}

fn main() {
    UserMethods {
        add_user: |user| {},
    };
}

  • The third possibility (which is implicit in the `generic`-one) is to use simple function pointers. That is, `fn(User)` instead of `Fn(User)`; `fn` is `Sized`, because free-standing functions can't capture their environment.
以上是如何在Rust结构中存储闭包？的全部内容。
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