一个函数如何要求一个类型实现一个特征而不删除现有的特征边界?

How can a function require that a type implement a trait without removing the existing trait bound?

我正在尝试 main_func returns 具有 SrObject 特征

的类型 T 结构的向量
struct TestA {
    value: u8,
}

pub trait SrObject {
    fn myfunc(&mut self);
}
impl SrObject for TestA {
    fn myfunc(&mut self) {
        unimplemented!();
    }
}
impl Default for TestA {
    fn default() -> TestA {
        TestA { value: 3u8 }
    }
}

fn main_func<T: SrObject>(t: T) -> Vec<T> {
    let mut v = Vec::<T>::new();
    for i in 0..10 {
        v.push(T::default());
        //v[i].myfunc();
    }
    return v;
}

它给出:

error[E0599]: no function or associated item named `default` found for type `T` in the current scope
  --> src/main.rs:22:16
   |
22 |         v.push(T::default());
   |                ^^^^^^^^^^ function or associated item not found in `T`
   |
   = help: items from traits can only be used if the trait is implemented and in scope
   = note: the following trait defines an item `default`, perhaps you need to implement it:
           candidate #1: `std::default::Default`

我知道我在 fn main_func<T: SrObject> 中没有 Default 特征,但是我如何在不删除 SrObject 特征的情况下实现这一点?

我鼓励你回去重读 The Rust Programming Language。这是 Rust 社区创建的一本免费在线书籍,涵盖了成为一名成功的 Rust 程序员所需了解的广泛内容。

在这种情况下,chapter on traits mentions this about trait bounds:

We can specify multiple trait bounds on a generic type by using +. If we needed to be able to use display formatting on the type T in a function as well as the summary method, we can use the trait bounds T: Summarizable + Display. This means T can be any type that implements both Summarizable and Display.

对于你的情况:

fn main_func<T: SrObject + Default>() -> Vec<T> {
    (0..10).map(|_| T::default()).collect()
}

fn main_func<T>() -> Vec<T>
where
    T: SrObject + Default,
{
    (0..10).map(|_| T::default()).collect()
}

使它符合习惯的其他更改:

  • 调用Vec::new时不指定v的类型;它将被推断。
  • 不要在函数末尾使用显式 return
  • 使用Iterator::map and Iterator::collect将迭代器转换为集合,而不是手动推送元素。

另请参阅: