how to generate const array by macro in Rust?

Question

I want to generate this const array:

const EXPS: [i64; 9] = [1,
    10_i64.pow(1), 10_i64.pow(2), 10_i64.pow(3), 10_i64.pow(4),
    10_i64.pow(5), 10_i64.pow(6), 10_i64.pow(7), 10_i64.pow(8)]

by macro:

define_exps!(9);

where the argument 9 specifies the length of array.

Can I make this by macro?

score 4 · Answer 1 · answered May 10 '23 at 14:36

4

Yes. You can do that as a procedural macro, but also as a declarative macro using dtolnay's seq-macro:

macro_rules! define_exps {
    ( $num:literal ) => {
        const EXPS: [i64; $num] = ::seq_macro::seq!(N in 0..$num {
            [
                #(
                    10_i64.pow(N),
                )*
            ]
        });
    };
}

answered May 10 '23 at 14:36

Chayim Friedman

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How come `EXPS` can be used outside of `define_exps!`? Doesn't this break macro hygiene? – jthulhu May 10 '23 at 16:40
3

@jthulhu Hygiene only applies to local variables, not to items. – Chayim Friedman May 10 '23 at 17:26
1

@jthulhu You might be thinking of declarative macros 2.0 (i.e. `feature(decl_macro)`), in which `EXPS` would *not* be accessible from the outside (due to hygiene). – Filipe Rodrigues May 10 '23 at 20:00

how to generate const array by macro in Rust?

1 Answers1