小编Qua*_*rew的帖子

我想写这个结构:

struct A {
    b: B,
    c: C,
}

struct B {
    c: &C,
}

struct C;

本B.c应借A.c.

A ->
  b: B ->
    c: &C -- borrow from --+
                           |
  c: C  <------------------+

这是我尝试过的:struct C;

struct B<'b> {
    c: &'b C,
}

struct A<'a> {
    b: B<'a>,
    c: C,
}

impl<'a> A<'a> {
    fn new<'b>() -> A<'b> {
        let c = C;
        A {
            c: c,
            b: B { c: &c },
        }
    }
}

fn …

我想定义一个类型,包括Rc<Fn(T)>,T 不要求Clone特性,例如代码:

use std::rc::Rc;


struct X;

#[derive(Clone)]
struct Test<T> {
    a: Rc<Fn(T)>
}


fn main() {
    let t: Test<X> = Test {
        a: Rc::new(|x| {})
    };
    let a = t.clone();
}

无法编译,错误信息是:

test.rs:16:15: 16:22 note: the method `clone` exists but the following trait bounds were not satisfied: `X : core::clone::Clone`, `X : core::clone::Clone`
test.rs:16:15: 16:22 help: items from traits can only be used if the trait is implemented and in scope; the following trait …

我可以创建一个单独的线程来充当I/O队列,但我不确定这是否是最好的方法.它看起来是最好的.

我不知道如何用mio加载本地文件.

rustc 1.0.0-夜间(be9bd7c93 2015-04-05)(建于2015-04-05)

extern crate rand;

fn test(a: isize) {
    println!("{}", a);
}

fn main() {
    test(rand::random() % 20)
}

此代码在Rust beta之前编译,但现在它没有:

src/main.rs:8:10: 8:22 error: unable to infer enough type information about `_`; type annotations required [E0282]
src/main.rs:8     test(rand::random() % 20)
                       ^~~~~~~~~~~~

我必须编写这段代码来编译:

extern crate rand;

fn test(a: isize) {
    println!("{}", a);
}

fn main() {
    test(rand::random::<isize>() % 20)
}

如何让编译器推断出类型？