struct - 我是否需要始终将基本结构放在 main() 之外以便它们是全局的？

Question

我正在阅读 Rust 文档，但在创建具有函数的结构时遇到了问题:

fn main() {
    let s1 = String::from("bob");
    let s2 = String::from("bob@aol.com");

    struct User {
        name: String,
        email: String,
    }

    let user1 = build_user(s1, s2); //or &s1, &s2
}

fn build_user(email: String, name: String) -> User {
    //or &String, &String
    User { email, name }
}

错误说:

error[E0412]: cannot find type `User` in this scope
  --> src/main.rs:13:47
   |
13 | fn build_user(email: String, name: String) -> User {
   |                                               ^^^^ not found in this scope

error[E0422]: cannot find struct, variant or union type `User` in this scope
  --> src/main.rs:15:5
   |
15 |     User { email, name }
   |     ^^^^ not found in this scope

如果我想构建一个带有函数的结构，我是否也必须通过引用传递基本结构？

Answer 1

你不必总是把它们放在something 之外，但你必须在足够高的级别声明它们，以便所有想要使用它的东西都可以看到它，可见性.

通过在 main 函数内部定义类型，只有 main 函数的主体可以访问它。在这种情况下，是的，您应该将结构的定义放在 main 之外，因为 main 和 build_user 都需要知道它:

struct User {
    name: String,
    email: String,
}

fn build_user(email: String, name: String) -> User {
    User { email, name }
}

当您继续阅读时，您会发现编写此代码的惯用方式:

fn main() {
    let s1 = String::from("bob");
    let s2 = String::from("bob@aol.com");

    let user1 = User::new(s1, s2);
}

struct User {
    name: String,
    email: String,
}

impl User {
    fn new(email: String, name: String) -> User {
        User { email, name }
    }
}