英文:
how to generate const array by macro in Rust?
问题
你想通过宏生成这个常量数组：
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)]
使用宏的方式如下：
define_exps!(9);
其中参数 9 指定了数组的长度。可以通过宏来实现吗？
英文:
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?
答案1
得分: 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),)*]});};}
英文:
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),)*]});};}
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