是否可以有条件地编译函数内的代码块? [英] Is it possible to conditionally compile a code block inside a function?

问题描述

我想知道这样的事情是否可能

I'm wondering if something like this is possible

fn main() {
    #[cfg(foo)] {
        println!("using foo config");
    }

}

上下文是一些无法仅通过单元测试进行充分测试的代码.我经常需要运行一个演示"cfg 来显示信息.我正在寻找替代手动注释输入/注释部分代码的方法.

The context is some code that cannot adequately be tested with just unit tests. I'll often have to run a "demo" cfg which displays information. I'm looking for alternatives to manually commenting in/out some portions of code.

推荐答案

至少从 Rust 1.21.1 开始，可以完全按照您所说的来执行此操作:

As of at least Rust 1.21.1, it's possible to do this as exactly as you said:

fn main() {
    #[cfg(foo)]
    {
        println!("using foo config");
    }
}

<小时>

在此之前，不可能完全有条件地做到这一点(即避免块完全被类型检查)，这样做由 RFC #16.但是，您可以使用 cfg 宏，它根据 --cfg 标志评估为 true 或 false:

---

Before this, it isn't possible to do this completely conditionally (i.e. avoiding the block being type checked entirely), doing that is covered by RFC #16. However, you can use the cfg macro which evaluates to either true or false based on the --cfg flags:

fn main() {
    if cfg!(foo) { // either `if true { ... }` or `if false { ... }`
        println!("using foo config");
    }
}

if 的主体总是运行名称解析和类型检查，因此可能并不总是有效.

The body of the if always has name-resolution and type checking run, so may not always work.

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