從Rust中的數組調用閉包

Question

如何迭代一系列閉包，然後依次調用每個閉包？

用的功能，我發現我可以只遍歷數組過來，並取消引用所產生的價值做到這一點：

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，當我試圖做同樣的封鎖，我得到一個錯誤：

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

產地：

<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 

如果我嘗試聲明迭代瓦里能夠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)); 
    } 
} 

結果：

<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 

如果我添加其他非關聯：

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

我回到原來的錯誤：

<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 

我在這裏錯過了什麼？是否有文檔介紹這是如何工作的？

Answer 1

矢量產量不變引用的.iter()方法，你需要可變的人打電話給關閉，因此，你應該使用.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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