是否可以有条件地启用像“derive"这样的属性? [英] Is it possible to conditionally enable an attribute like `derive`?

问题描述

我在我的 crate 中添加了一个功能，它增加了 serde 支持.但是，我不太明白如何正确使用它:

//#[derive(Debug, Serialize, Deserialize, Clone)]//转到:#[派生(调试，克隆)]#[cfg(feature = "serde_support")]#[派生(序列化，反序列化)]酒吧结构MyStruct;

此代码将 cfg(feature) 下的所有内容视为有条件编译，因此如果没有我的 serde_support 功能，我的 crate 也没有 MyStruct.

我试图用大括号把它包裹起来，但它给出了另一个错误:

代码:

#[derive(Debug, Clone)]#[cfg(feature = "serde_support")] {#[派生(序列化，反序列化)]}酒吧结构MyStruct;

错误:

错误:属性后的预期项目-->mycrate/src/lib.rs:65:33|65 |#[cfg(feature = "serde_support")] {|^

那么如何做到这一点?

解决方案

您可以使用 cfg_attr(a, b) 属性:

#[derive(Debug, Clone)]#[cfg_attr(feature = "serde_support", 派生(序列化，反序列化))]酒吧结构MyStruct;

它在 关于条件编译"的 Rust 参考资料中有描述:

<块引用>

#[cfg_attr(a, b)]物品

与#[b] item相同，如果a由cfg设置，item否则.

I have added a feature in my crate which adds serde support. However, I don't quite understand how to use it properly:

// #[derive(Debug, Serialize, Deserialize, Clone)] // goes to:

#[derive(Debug, Clone)]
#[cfg(feature = "serde_support")]
#[derive(Serialize, Deserialize)]
pub struct MyStruct;

This code treats everything below cfg(feature) as conditionally compiled, so without my serde_support feature my crate does not have MyStruct also.

I have tried to wrap it with braces but it gives another error:

Code:

#[derive(Debug, Clone)]
#[cfg(feature = "serde_support")] {
#[derive(Serialize, Deserialize)]
}
pub struct MyStruct;

Error:

error: expected item after attributes
  --> mycrate/src/lib.rs:65:33
   |
65 | #[cfg(feature = "serde_support")] {
   |                                 ^

So how to do this?

解决方案

You can use the cfg_attr(a, b) attribute:

#[derive(Debug, Clone)]
#[cfg_attr(feature = "serde_support", derive(Serialize, Deserialize))]
pub struct MyStruct;

It's described in the Rust reference about "conditional compilation":

#[cfg_attr(a, b)]
item

Will be the same as #[b] item if a is set by cfg, and item otherwise.

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