有没有更快/更短的方法来初始化 Rust 结构中的变量? [英] Is there a faster/shorter way to initialize variables in a Rust struct?

问题描述

在下面的示例中，我更愿意在字段声明中为结构中的每个字段分配一个值.或者，它实际上需要为每个字段添加一个语句来为这些字段分配一个值.我想要做的就是在结构体实例化时分配默认值.

In the following example, I would much prefer to assign a value to each field in the struct in the declaration of the fields. Alternatively, it effectively takes one additional statement for each field to assign a value to the fields. All I want to be able to do is to assign default values when the struct is instantiated.

有没有更简洁的方法来做到这一点?

Is there a more succinct way of doing this?

struct cParams {
    iInsertMax: i64,
    iUpdateMax: i64,
    iDeleteMax: i64,
    iInstanceMax: i64,
    tFirstInstance: bool,
    tCreateTables: bool,
    tContinue: bool,
}

impl cParams {
    fn new() -> cParams {
        cParams {
            iInsertMax: -1,
            iUpdateMax: -1,
            iDeleteMax: -1,
            iInstanceMax: -1,
            tFirstInstance: false,
            tCreateTables: false,
            tContinue: false,
        }
    }
}

推荐答案

你可以通过实现 Default trait 来为你的结构提供默认值.default 函数看起来像您当前的 new 函数:

You can provide default values for your struct by implementing the Default trait. The default function would look like your current new function:

impl Default for cParams {
    fn default() -> cParams {
        cParams {
            iInsertMax: -1,
            iUpdateMax: -1,
            iDeleteMax: -1,
            iInstanceMax: -1,
            tFirstInstance: false,
            tCreateTables: false,
            tContinue: false,
        }
    }
}

然后您可以通过仅提供非默认值来实例化结构:

You can then instantiate the struct by giving only the non-default values:

let p = cParams { iInsertMax: 10, ..Default::default() };

通过对数据结构进行一些细微更改，您可以利用自动派生的默认实现.如果您在数据结构上使用 #[derive(Default)]，编译器将自动为您创建一个默认函数，用其默认值填充每个字段.默认布尔值为 false，默认整数值为 0.

With some minor changes to your data structure, you can take advantage of an automatically derived default implementation. If you use #[derive(Default)] on a data structure, the compiler will automatically create a default function for you that fills each field with its default value. The default boolean value is false, the default integral value is 0.

整数的默认值为 0 是一个问题，因为您希望整数字段默认为 -1.您可以定义一个实现默认值 -1 的新类型，并在结构中使用它代替 i64.(我还没有测试过，但它应该可以工作).

An integer's default value being 0 is a problem here since you want the integer fields to be -1 by default. You could define a new type that implements a default value of -1 and use that instead of i64 in your struct. (I haven't tested that, but it should work).

但是，我建议稍微更改您的数据结构并使用 Option 而不是 i64.我不知道您的代码的上下文，但看起来您正在使用 -1 的特殊值来表示无限"或没有最大值"的特殊含义.在 Rust 中，我们使用 Option 来表示可选的当前值.不需要 -1 hack.一个选项可以是 None 或 Some(x)，其中 x 是你的 i64.如果 -1 是唯一的负值，它甚至可能是一个无符号整数.默认的 Option 值为 None，因此根据建议的更改，您的代码可能如下所示:

However, I'd suggest to slightly change your data structure and use Option<i64> instead of i64. I don't know the context of your code, but it looks like you're using the special value of -1 to represent the special meaning "infinite", or "there's no max". In Rust, we use an Option to represent an optionally present value. There's no need for a -1 hack. An option can be either None or Some(x) where x would be your i64 here. It might even be an unsigned integer if -1 was the only negative value. The default Option value is None, so with the proposed changes, your code could look like this:

#[derive(Default)]
struct cParams {
    iInsertMax: Option<u64>,
    iUpdateMax: Option<u64>,
    iDeleteMax: Option<u64>,
    iInstanceMax: Option<u64>,
    tFirstInstance: bool,
    tCreateTables: bool,
    tContinue: bool,
}

let p = cParams { iInsertMax: Some(10), ..Default::default() };

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