如何迭代HashMap<Vec<u8>，Vec<u8>>

Question

正如标题所说，我正在尝试迭代 2 个向量的 HashMap。我的代码如下：

pub fn serialize_hashmap(data: &HashMap<Vec<u8>, Vec<u8>>) -> Result<Vec<u8>, String> {
    let mut serialized_data = Vec::new();
    for (hash, password_data) in &data {
        serialized_data.extend_from_slice(&hash);
        let password_data_length: u64 = password_data.len().try_into().unwrap();
        serialized_data.write_u64::<NativeEndian>(password_data_length).unwrap(); // TODO add error handling
        serialized_data.extend_from_slice(&password_data);
    }

    return Ok(serialized_data);
}

但是当我运行此代码时，出现以下错误：

error[E0277]: the size for values of type `[_]` cannot be known at compilation time
 --> src/main.rs:8:10
  |
8 |     for (hash, password_data) in &data {
  |          ^^^^ doesn't have a size known at compile-time
  |
  = help: the trait `Sized` is not implemented for `[_]`
  = note: all local variables must have a statically known size
  = help: unsized locals are gated as an unstable feature

error[E0277]: the size for values of type `[_]` cannot be known at compilation time
 --> src/main.rs:8:34
  |
8 |     for (hash, password_data) in &data {
  |                                  ^^^^^ doesn't have a size known at compile-time
  |
  = help: the trait `Sized` is not implemented for `[_]`
  = note: only the last element of a tuple may have a dynamically sized type

问题似乎是我的哈希图是动态分配的，并且其大小在运行时未知。即使 HashMap 是动态分配的，有什么方法可以迭代它吗？像这样的事情通常如何在 Rust 中完成？

Answer 1

我很确定你的问题只是

data

已经是一个引用，所以当你尝试迭代

&data

时，你正在迭代

&&HashMap

，这是不可迭代的。如果您删除 & 符号，使其成为

for ... in data

，我认为循环应该可以工作，因为它只是迭代

&HashMap

，这完全没问题。