Rust函数没有静态寿命吗? [英] Rust function does not have static lifetime?

问题描述

我正在尝试编译以下简单代码:

I am trying to make this simple code compile:

fn dox(x: u8) -> u8 { x*2 }

fn main() {
    let cb: &'static (Fn(u8) -> u8) = &dox;
}

但是在Rust 1.9中失败:

But it fails with Rust 1.9:

x.rs:4:40: 4:43 error: borrowed value does not live long enough
x.rs:4     let cb: &'static (Fn(u8) -> u8) = &dox;
                                              ^~~
note: reference must be valid for the static lifetime...
x.rs:4:44: 5:2 note: ...but borrowed value is only valid for the block suffix following statement 0 at 4:43
x.rs:4     let cb: &'static (Fn(u8) -> u8) = &dox;
x.rs:5 }
error: aborting due to previous error

自由函数没有静态寿命的可能性如何?这段代码怎么会不安全?

How is it possible that a free function does not have static lifetime? How could this code be unsafe?

推荐答案

从Rust 1.21开始，将自动执行静态升级"，并且原始代码将按原样编译.

As of Rust 1.21, "static promotion" is performed automatically and your original code compiles as-is.

此代码也可以编译:

fn main() {
    let a: &'static i32 = &5;
}

此外，可以将不会从其环境中捕获任何内容的闭包自动转换为函数指针，因此您也无需创建单独的函数:

Additionally, closures which do not capture anything from their environment can be automatically converted to function pointers, so you don't need to create a separate function, either:

fn main() {
    let cb: fn(u8) -> u8 = |x| x * 2;
}

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