Rust：如何从函数返回一个迭代器并使用它？

Question

我试图通过在一个小日历项目上工作来教自己锈。

在这里，我试图生成一个约定在一个给定日期三个整月的日期列表。 我想返回一个可以迭代这些日期的迭代器。 这是我第一次尝试：

fn three_months_range(tm: time::Tm) -> std::iter::Iterator<Item=time::Tm> { 
    let fpm: time::Tm = first_of_previous_month(&tm); 
    (0..) 
     .map(|i| fpm + time::Duration::days(i)) 
     .take_while(|&t| t.tm_mon != (tm.tm_mon + 2) % 12) 
} 

不幸的是，这并不编译，我得到一个错误。

src/main.rs:49:40: 49:75 error: the trait `core::marker::Sized` is not implemented for the type `core::iter::Iterator<Item=time::Tm> + 'static` [E0277] 
src/main.rs:49 fn three_months_range(tm: time::Tm) -> std::iter::Iterator <Item=time::Tm> { 
                 ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 
src/main.rs:49:40: 49:75 help: run `rustc --explain E0277` to see a detailed explanation 
src/main.rs:49:40: 49:75 note: `core::iter::Iterator<Item=time::Tm> + 'static` does not have a constant size known at compile-time 
src/main.rs:49:40: 49:75 note: the return type of a function must have a statically known size 

“函数的返回类型必须具有静态已知大小”。好的，经过一番研究，似乎解决方案是通过Box返回迭代器。我仍然想知道标准库map,filter，take_while ...方法如何管理返回迭代器而不是盒子）。

嗯，这里是第二次尝试，成功编译：

fn three_months_range(tm: time::Tm) -> Box<iter::Iterator<Item=time::Tm>> { 
    let fpm: time::Tm = first_of_previous_month(&tm); 
    Box::new(
     (0..) 
     .map(move |i| fpm + time::Duration::days(i)) 
     .take_while(move |&t| t.tm_mon != (tm.tm_mon + 2) % 12) 
    ) 
} 

不幸的是，我不管理使用这个迭代器。 例如，假设我想构建一个包含每个日期的月份日期（1,2,3，...，31,1,2，...，30,1,2，...）的向量。 31）：

let days_vec: Vec<u64> = 
    (*three_months_range(time::now_utc())) 
    .map(|&t: &time::Tm| t.tm_mday) 
    .collect(); 

→

src/main.rs:14:10: 14:42 error: the trait `core::marker::Sized` is not implemented for the type `core::iter::Iterator<Item=time::Tm> + 'static` [E0277] 
src/main.rs:14   .map(|&t: &time::Tm| t.tm_mday) 
         ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 
src/main.rs:14:10: 14:42 help: run `rustc --explain E0277` to see a detailed explanation 
src/main.rs:14:10: 14:42 note: `core::iter::Iterator<Item=time::Tm> + 'static` does not have a constant size known at compile-time 
src/main.rs:15:10: 15:19 error: no method named `collect` found for type `core::iter::Map<core::iter::Iterator<Item=time::Tm> + 'static, [closure@src/main.rs:14:15: 14:40]>` in the current scope 
src/main.rs:15   .collect(); 
         ^~~~~~~~~ 
src/main.rs:15:10: 15:19 note: the method `collect` exists but the following trait bounds were not satisfied: `core::iter::Iterator<Item=time::Tm> : core::marker::Sized`, `[closure@src/main.rs:14:15: 14:40] : core::ops::FnMut<(time::Tm,)>`, `core::iter::Map<core::iter::Iterator<Item=time::Tm> + 'static, [closure@src/main.rs:14:15: 14:40]> : core::iter::Iterator` 
src/main.rs:14:10: 14:42 error: type mismatch: the type `[closure@src/main.rs:14:15: 14:40]` implements the trait `for<'r> core::ops::FnMut<(&'r time::Tm,)>`, but the trait `core::ops::FnMut<(time::Tm,)>` is required (expected struct `time::Tm`, found &-ptr) [E0281] 
src/main.rs:14   .map(|&t: &time::Tm| t.tm_mday) 
         ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 
src/main.rs:14:10: 14:42 help: run `rustc --explain E0281` to see a detailed explanation 
src/main.rs:14:10: 14:42 error: type mismatch: the type `[closure@src/main.rs:14:15: 14:40]` implements the trait `for<'r> core::ops::FnOnce<(&'r time::Tm,)>`, but the trait `core::ops::FnOnce<(time::Tm,)>` is required (expected struct `time::Tm`, found &-ptr) [E0281] 
src/main.rs:14   .map(|&t: &time::Tm| t.tm_mday) 
         ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 
src/main.rs:14:10: 14:42 help: run `rustc --explain E0281` to see a detailed explanation 
error: aborting due to 4 previous errors 

这是一个很大的错误。

那么我在这里做错了什么？

有没有一种相对简单的方法来转换Rust中的迭代器和/或从函数中返回它们？

Answer 1

问题是您正试图将Iterator从Box中删除（因为它不是特征对象，因此不是Sized）。但Box是透明的，可以直接使用就可以了map：

let days_vec: Vec<u64> = 
    three_months_range(time::now_utc()) 
    .map(|&t: &time::Tm| t.tm_mday) 
    .collect(); 

注意，map函数需要按值，而不是采取论证的参考。因此，呼叫可以是这样的：

.map(|t| t.tm_mday)