从Rust中的数组调用闭包

Question

如何迭代一系列闭包，然后依次调用每个闭包？

用的功能，我发现我可以只遍历数组过来，并取消引用所产生的价值做到这一点：

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)); 
    } 
} 

然而，当我试图做同样的封锁，我得到一个错误：

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瓦里能够ref mut，它仍然不能正常工作：

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 

如果我添加其他非关联：

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 

我在这里错过了什么？是否有文档介绍这是如何工作的？

Answer 1

矢量产量不变引用的.iter()方法，你需要可变的人打电话给关闭，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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