是否可以将包含闭包的 Rust struct 存储在不同的 struct 中

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pub struct Command<P, T, CMD>
where
    T: Send,
    CMD: Fn(P) -> Result<T, Box<CommandError>> + Sync + Send,
{
    pub config: Option<Config>,
    pub cmd: CMD,
    phantom_data: PhantomData<P>,
}

是否可以将Command的实例存储为另一个 struct 中的字段？

我开始try 从

/// This function constructs a simple instance of `Command<P, T, CMD>` with the
/// types set to:
///
///     P ~ u8
///     T ~ u8
///     CMD: Fn(u8) -> Result<u8, Box<CommandError>> + Send + Sync
///
/// This function compiles fine. However, there is no *concrete* type
/// for `CMD`. In compiler output it will be referred to as an
/// "anonymous" type looking like this:
///
///    Command<u8, u8, [closure@src/lib.rs:19:21: 19:38]>
fn simple_command_instance() {
    let _ = Command::define(|n: u8| Ok(n * 2));
}

当为

fn return_command_instance() -> Command<u8, u8, ???> {
                                                ^
                                                |
                          What goes here? -------

    Command::define(|n: u8| Ok(n * 2))
}

编译器推断出的类型是匿名的，无法放入

在具有新impl Trait功能的Rust版本中(例如

/// Use new `impl Trait` syntax as a type parameter in the return
/// type:
fn impl_trait_type_param() -> Command<u8, u8, impl Fn(u8) -> Result<u8, Box<CommandError>>> {
    Command::define(|n: u8| Ok(n * 2))
}

这在稳定Rust 的情况下不起作用，impl Trait只能

试图传播泛型类型的结果是

fn return_cmd_struct<F>() -> Command<u8, u8, F>
where
    F: Fn(u8) -> Result<u8, Box<CommandError>> + Send + Sync,
{
    Command::define(|n: u8| Ok(n * 2))
}

但这并没有编译:

error[E0308]: mismatched types
  --> src/lib.rs:33:21
   |
33 |     Command::define(|n: u8| Ok(n * 2))
   |                     ^^^^^^^^^^^^^^^^^ expected type parameter, found closure
   |
   = note: expected type `F`
              found type `[closure@src/lib.rs:33:21: 33:38]`

再说一遍，我不知道有什么方法可以在

即使将类型作为泛型参数传播有效，它也会

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