How do I change a function's qualifiers via conditional compilation?

问题

I have a function that is capable of being implemented as a const:

#![feature(const_fn)]

// My crate would have:

const fn very_complicated_logic(a: u8, b: u8) -> u8 {
    a * b
}

// The caller would have:

const ANSWER: u8 = very_complicated_logic(1, 2);

fn main() {}

I'd like to continue to support stable Rust where it's not possible to define such functions. These stable consumers would not be able to use the function in a const or static, but should be able to use the function in other contexts:

// My crate would have:

fn very_complicated_logic(a: u8, b: u8) -> u8 {
    a * b
}

// The caller would have:    

fn main() {
    let answer: u8 = very_complicated_logic(1, 2);
}

How can I conditionally compile my code so that adventurous users of my crate could enable const fn support, stable users would still be able to use my code, and I don't have to write every function twice?

The same question should apply to the other modifiers of a function, but I'm not sure of concrete cases where these modifiers would change based on some condition:

default
unsafe
extern
pub (and other visibility modifiers)

回答1:

Macros to the rescue!

#![cfg_attr(feature = "const-fn", feature(const_fn))]

#[cfg(not(feature = "const-fn"))]
macro_rules! maybe_const_fn {
    ($($tokens:tt)*) => {
        $($tokens)*
    };
}

#[cfg(feature = "const-fn")]
macro_rules! maybe_const_fn {
    ($(#[$($meta:meta)*])* $vis:vis $ident:ident $($tokens:tt)*) => {
        $(#[$($meta)*])* $vis const $ident $($tokens)*
    };
}

maybe_const_fn! {
    #[allow(unused)] // for demonstration purposes
    pub fn very_complicated_logic(a: u8, b: u8) -> u8 {
        internally_complicated_logic(a, b)
    }
}

maybe_const_fn! {
    fn internally_complicated_logic(a: u8, b: u8) -> u8 {
        a * b
    }
}

#[cfg(test)]
mod tests {
    use super::*;

    #[cfg(feature = "const-fn")]
    #[test]
    fn use_in_const() {
        const ANSWER: u8 = very_complicated_logic(1, 2);
        drop(ANSWER);
    }

    #[test]
    fn use_in_variable() {
        let answer: u8 = very_complicated_logic(1, 2);
        drop(answer);
    }
}

Along with this in Cargo.toml:

[features]
const-fn = []

Since macros can only expand to complete pieces of syntax (i.e. a macro cannot simply expand to const), we have to wrap the whole function in the macro and leave some parts of it unparsed so that we can inject const in the appropriate place. Then, the parser can parse the whole thing as a function definition.

Attributes and visibility qualifiers need special treatment, because they must appear before const. I am using the vis matcher (available since Rust 1.30.0) to simplify the macro's implementation.

来源：https://stackoverflow.com/questions/51884807/conditional-visibility-for-unit-tests

标签

rust

conditional-compilation