Arbitrary Ext

自 1.2.0 Arbitrary 支持为 derive 字段提供自定义任意实现。

但是如果一个没有实现 Arbitrary 的类型被封装在一个泛型类型中，这仍然很棘手。

此 crate 提供了一组函数组合器，以支持标准库中的容器和集合。

请参见下面的示例。

用法

use arbitrary_ext::{arbitrary_option, arbitrary_vec, arbitrary_hash_map};
use std::collections::HashMap;

// Imagine this is a foreign type, that by some reason does not implement Arbitrary trait.
#[derive(Debug, PartialEq, Eq, PartialOrd, Ord, Hash)]
struct Number(u32);

// Our custom function to generate arbitrary Number out of Unstructured.
fn arbitrary_number(u: &mut Unstructured) -> arbitrary::Result<Number> {
    let value = u.int_in_range(0..=1000)?;
    Ok(Number(value))
}

#[derive(Debug, Arbitrary)]
struct Example {
    #[arbitrary(with = arbitrary_number)]
    number: Number,

    #[arbitrary(with = arbitrary_option(arbitrary_number))]
    option: Option<Number>,

    #[arbitrary(with = arbitrary_vec(arbitrary_number))]
    vec: Vec<Number>,

    #[arbitrary(
        with = arbitrary_hash_map(
            arbitrary_number,
            arbitrary_vec(arbitrary_option(arbitrary_number))
        )
    )]
    hash_map: HashMap<Number, Vec<Option<Number>>>,
}

在没有 arbitrary_option、arbitrary_vec、arbitrary_hash_map 组合器的情况下，将被迫实现自定义函数来生成任意的 Option<Number>、Vec<Number> 和 HashMap<Number, Vec<Option<Number>>>。例如。

fn arbitrary_option_number(u: &mut Unstructured) -> arbitrary::Result<Option<Number>>;
fn arbitrary_vec_number(u: &mut Unstructured) -> arbitrary::Result<Vec<Number>>;
fn arbitrary_hash_map_of_numbers(u: &mut Unstructured) -> arbitrary::Result<HashMap<Number, Vec<Option<Number>>>>;

但这很快就会变得繁琐。

crate 的历史

最初创建此 crate 是为了解决这个 Arbitrary 问题，但后来在 这个 PR 中得到了解决。