带有未装箱关闭的错误消息 [英] Error message with unboxed closures

问题描述

这个使用未装箱闭包的小型 FizzBu​​zz 程序给出了一个相当神秘的错误信息.

This small FizzBuzz program using unboxed closures gives a rather mysterious error message.

fn fizzbuzz<F: Fn(i64) -> bool>(n: i64, f: F, fs: &str, b: F, bs: &str) {
    for i in range(1i64, n+1) {
        match (f(i), b(i)) {
            (true, true) => println!("{:3}: {}{}", i, fs, bs),
            (true, _)    => println!("{:3}: {}", i, fs),
            (_, true)    => println!("{:3}: {}", i, bs),
            _            => (),
        }
    }
}

fn main() {
    fizzbuzz(30,
             |&: i: i64| { i % 3 == 0 }, "fizz",
             |&: j: i64| { j % 5 == 0 }, "buzz");
}

错误信息:

<anon>:15:14: 15:40 error: mismatched types: expected `closure[<anon>:14:14: 14:40]`, found `closure[<anon>:15:14: 15:40]` (expected closure, found a different closure)
<anon>:15              |&: j: i64| { j % 5 == 0 }, "buzz");
                   ^~~~~~~~~~~~~~~~~~~~~~~~~~
error: aborting due to previous error

有人可以描述一下错误吗?谢谢.

Could someone describe the error please? Thanks.

推荐答案

这段代码说明了问题的本质:

This piece of code demonstrates the essence of the problem:

fn show_both<S: Show>(x: S, y: S) {
    println!("{} {}", x, y);
}

您只能使用相同类型的两个参数来调用它，即这是允许的:

You can only call it with both arguments of the same type, i.e. this is allowed:

let x: i32 = 10;
let y: i32 = 20;
show_both(x, y);

但这不是:

let x: i32 = 10;
let y: f64 = 20.0;
show_both(x, y);

这是相当自然的:您指定两个参数必须是同一类型，即使该类型可以是任意的，只要它实现了Show.

This is fairly natural: you specified that both of the parameters must be of the same type, even though this type can be arbitrary as long as it implements Show.

您的代码中具有基本相同的内容:

You have essentially the same thing in your code:

fn fizzbuzz<F: Fn(i64) -> bool>(n: i64, f: F, fs: &str, b: F, bs: &str)

您声明 f 和 b 必须具有相同的类型.然而，对于每个未装箱的闭包，编译器会生成不同的类型——这也很自然，因为不同的闭包可以捕获不同的变量.

You declared that both f and b must have the same type. However, for each unboxed closure the compiler generates a different type - this is quite natural as well, given that different closures can capture different variables.

你需要指定不同的类型参数才能传递不同的闭包:

You need to specify different type parameters in order to be able to pass different closures:

fn fizzbuzz<F1, F2>(n: i64, f: F1, fs: &str, b: F2, bs: &str)
    where F1: Fn(i64) -> bool,
          F2: Fn(i64) -> bool

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