When a type implements Borrow<T>, it should provide consistent behavior between the borrowed and owned values, especially
regarding Eq, Ord, and Hash. However, str and [u8] have different Hash
implementations, leading to inconsistent hash values for the same underlying data when accessed through different Borrow implementations. This
violates the principle that hash(x) == hashx as Borrow<[u8]>).borrow( == hashx as Borrow<str>).borrow( must hold, and can
cause issues with hash-based collections and comparisons.
Code examples
Noncompliant code example
use std::borrow::Borrow;
use std::hash::{Hash, Hasher};
struct ExampleType {
data: String,
}
impl Hash for ExampleType {
fn hash<H: Hasher>(&self, state: &mut H) {
self.data.hash(state); // Noncompliant: Inconsistent Hash
}
}
impl Borrow<str> for ExampleType {
fn borrow(&self) -> &str {
&self.data
}
}
impl Borrow<[u8]> for ExampleType {
fn borrow(&self) -> &[u8] {
self.data.as_bytes()
}
}
