相关疑难解决方法(0)

我有以下代码(游乐场):

// Two dummy functions, both with the signature `fn(u32) -> bool`
fn foo(x: u32) -> bool {
    x % 2 == 0
}
fn bar(x: u32) -> bool {
    x == 27
}


fn call_both<F>(a: F, b: F)
where
    F: Fn(u32) -> bool,
{
    a(5);
    b(5);
}

fn main() {
    call_both(foo, bar);  // <-- error
}

对我来说,它看起来这应该是编译foo和bar具有相同的签名:fn(u32) -> bool.然而,我收到以下错误:

error[E0308]: mismatched types
  --> src/main.rs:20:20
   |
20 |     call_both(foo, bar);
   |                    ^^^ …

我试图创建闭包矢量:

fn main() {
    let mut vec = Vec::new();

    vec.push(Box::new(|| 10));
    vec.push(Box::new(|| 20));

    println!("{}", vec[0]());
    println!("{}", vec[1]());
}

这产生了以下错误报告:

error[E0308]: mismatched types
 --> src/main.rs:5:23
  |
5 |     vec.push(Box::new(|| 20));
  |                       ^^^^^ expected closure, found a different closure
  |
  = note: expected type `[closure@src/main.rs:4:23: 4:28]`
             found type `[closure@src/main.rs:5:23: 5:28]`
  = note: no two closures, even if identical, have the same type
  = help: consider boxing your closure and/or using it as a trait object

我通过明确指定类型来修复它:

let mut vec: Vec<Box<Fn() -> i32>> …

我有两个虚函数:

fn foo() -> u32 { 3 }
fn bar() -> u32 { 7 }

我想创建一个盒装的函数指针切片:Box<[fn() -> u32]>.我想用box语法来做(我知道它不是两个元素所必需的,但我的实际用例是不同的).

我尝试了几件事(游乐场):

// Version A
let b = box [foo, bar] as Box<[_]>;

// Version B
let tmp = [foo, bar];
let b = box tmp as Box<[_]>;

// Version C
let b = Box::new([foo, bar]) as Box<[_]>;

版本B和C工作正常(C虽然不会对我有用Box::new),但版本A错误:

error[E0605]: non-primitive cast: `std::boxed::Box<[fn() -> u32; 2]>` as `std::boxed::Box<[fn() -> u32 {foo}]>`
 --> src/main.rs:8:13
  |
8 …