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作为测试的一部分，我想断言一个函数返回一个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：在比较静态向量之前分配对它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的相等性：

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);

4

1 回答 1

7

实际上有两种静态向量：固定长度向量 ( [u8, .. 3]) 和静态切片 ( &'static [u8])。前者不能与其他类型的向量很好地交互。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);
}
于 2013-05-08T14:08:21.267 回答