Question

如何迭代一系列闭包，依次调用每个闭包？

使用函数，我发现我可以通过迭代数组并取消引用生成的值来实现这一点：

fn square(x: int) -> int { x * x }

fn add_one(x: int) -> int { x + 1 }

fn main() {
    let funcs  = [square, add_one];
    for func in funcs.iter() {
        println!("{}", (*func)(5i));
    }
}

但是，当我尝试对闭包执行相同操作时，我收到错误：

fn main() {
    let closures = [|x: int| x * x, |x| x + 1];
    for closure in closures.iter() {
        println!("{}", (*closure)(10i));
    }
}

产地：

<anon>:4:24: 4:34 error: closure invocation in a `&` reference
<anon>:4         println!("{}", (*closure)(10i));
                                ^~~~~~~~~~
note: in expansion of format_args!
<std macros>:2:23: 2:77 note: expansion site
<std macros>:1:1: 3:2 note: in expansion of println!
<anon>:4:9: 4:41 note: expansion site
<anon>:4:24: 4:34 note: closures behind references must be called via `&mut`
<anon>:4         println!("{}", (*closure)(10i));
                                ^~~~~~~~~~
note: in expansion of format_args!
<std macros>:2:23: 2:77 note: expansion site
<std macros>:1:1: 3:2 note: in expansion of println!
<anon>:4:9: 4:41 note: expansion site

如果我尝试声明迭代变量ref mut，它仍然不起作用：

fn main() {
    let closures = [|x: int| x * x, |x| x + 1];
    for ref mut closure in closures.iter() {
        println!("{}", (*closure)(10i));
    }
}

结果：

<anon>:4:24: 4:39 error: expected function, found `&|int| -> int`
<anon>:4         println!("{}", (*closure)(10i));
                                ^~~~~~~~~~~~~~~
note: in expansion of format_args!
<std macros>:2:23: 2:77 note: expansion site
<std macros>:1:1: 3:2 note: in expansion of println!
<anon>:4:9: 4:41 note: expansion site

如果我添加另一个解除引用：

fn main() {
    let closures = [|x: int| x * x, |x| x + 1];
    for ref mut closure in closures.iter() {
        println!("{}", (**closure)(10i));
    }
}

我回到原来的错误：

<anon>:4:24: 4:35 error: closure invocation in a `&` reference
<anon>:4         println!("{}", (**closure)(10i));
                                ^~~~~~~~~~~
note: in expansion of format_args!
<std macros>:2:23: 2:77 note: expansion site
<std macros>:1:1: 3:2 note: in expansion of println!
<anon>:4:9: 4:42 note: expansion site
<anon>:4:24: 4:35 note: closures behind references must be called via `&mut`
<anon>:4         println!("{}", (**closure)(10i));
                                ^~~~~~~~~~~
note: in expansion of format_args!
<std macros>:2:23: 2:77 note: expansion site
<std macros>:1:1: 3:2 note: in expansion of println!
<anon>:4:9: 4:42 note: expansion site

我在这里缺少什么？是否有某些文档描述了它是如何工作的？

Answer 1

vector的.iter()方法产生不可变引用，你需要可变的引用来调用闭包，因此你应该使用.iter_mut()：

fn main() {
    let mut closures = [|x: int| x * x, |x| x + 1];
    for closure in closures.iter_mut() {
        println!("{}", (*closure)(10i));
    }
}

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11

从Rust中的数组调用闭包

1 个答案: