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types - 如何比较 Rust 中拥有的向量和静态向量的内容?

转载 作者:行者123 更新时间:2023-11-29 07:43:24 25 4
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作为测试的一部分,我想断言函数返回具有正确内容的向量。因此,我将预期数据作为静态变量提供。但是,我找不到将托管向量的内容与静态向量变量进行比较的正确方法。

#[test]
fn test_my_data_matches_expected_data () {
static expected_data: [u8, ..3] = [1, 2, 3];
let my_data: ~[u8] = ~[1, 2, 3]; // actually returned by the function to test

// This would be obvious, but fails:
// -> mismatched types: expected `~[u8]` but found `[u8 * 3]`
assert_eq!(my_data, expected_data);

// Static vectors are told to be available as a borrowed pointer,
// so I tried to borrow a pointer from my_data and compare it:
// -> mismatched types: expected `&const ~[u8]` but found `[u8 * 3]`
assert_eq!(&my_data, expected_data);

// Dereferencing also doesn't work:
// -> type ~[u8] cannot be dereferenced
assert_eq!(*my_data, expected_data);

// Copying the static vector to a managed one works, but this
// involves creating a copy of the data and actually defeats
// the reason to declare it statically:
assert_eq!(my_data, expected_data.to_owned());
}

更新:在比较之前分配对静态向量的引用解决了这个问题,所以我最终得到了一个小宏来断言向量相等:

macro_rules! assert_typed_eq (($T: ty, $given: expr, $expected: expr) => ({
let given_val: &$T = $given;
let expected_val: &$T = $expected;
assert_eq!(given_val, expected_val);
}))

用法:assert_typed_eq([u8], my_data, expected_data);

最佳答案

实际上有两种静态向量:固定长度的([u8, .. 3])和静态切片(&'static [u8])。前者与其他类型的载体的相互作用不是很好。后者在这里最有用:

fn main() {
static x: &'static [u8] = &[1,2,3];

let y = ~[1u8,2,3];
assert_eq!(y.as_slice(), x);
}

关于types - 如何比较 Rust 中拥有的向量和静态向量的内容?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16427159/

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