"expected item, found let" 是什么意思吝啬的? [英] What does "expected item, found let" mean?

问题描述

我的代码

pub struct MyStorage {
    name: Vec<u8>,
}

impl Storage for MyStorage {
    //let mut name: Vec<u8> = [0x11];
    fn get(&mut self) -> Vec<u8> {
        self.name
    }
}

let my_storage = MyStorage { name = [0x11] };

返回错误

error: expected item, found keyword `let`
  --> src/lib.rs:12:1
   |
12 | let my_storage = MyStorage { name = [0x11] };
   | ^^^ expected item

这是什么意思?

推荐答案

这段代码有很多问题，但你得到的错误是因为你试图执行代码，但不是从在函数内:

There's a number of issues with this code, but the error you are getting is because you are trying to execute code but not from within a function:

let my_storage = MyStorage { name = [0x11] };

你需要把它放在某处.在这里，我已将其添加到 main:

You need to put that in something. Here, I've added it to main:

pub struct MyStorage {
    name: Vec<u8>,
}

impl MyStorage {
    fn get(self) -> Vec<u8> {
        self.name
    }
}

fn main() {
    let my_storage = MyStorage { name: vec![0x11] };
}

我还必须:

• 修复向量构建 (vec!)
• 删除不存在的特征的使用(Storage)
• 改变get中self的类型
• 从 = 更改为 :

• fix the vector construction (vec!)
• remove the usage of a trait that doesn't exist (Storage)
• change the type of self in get
• change from = to :

这样，代码就可以编译了.

With all that, the code compiles.

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