use generic_array::*; // 0.12.3
use num::{Float, Zero}; // 0.2.0
#[derive(Clone, Debug)]
struct Vector<T, N: ArrayLength<T>> {
data: GenericArray<T, N>,
}
impl<T, N: ArrayLength<T>> Vector<T, N>
where
T: Float + Zero,
{
fn dot(&self, other: Self) -> T {
self.data
.iter()
.zip(other.data.iter())
.fold(T::zero(), |acc, x| acc + *x.0 * *x.1)
}
fn length_sq(&self) -> T {
self.dot(self.clone())
}
}
error[E0308]: mismatched types
--> src/lib.rs:21:18
|
21 | self.dot(self.clone())
| ^^^^^^^^^^^^ expected struct `Vector`, found reference
|
= note: expected type `Vector<T, N>`
found type `&Vector<T, N>`
Why does this happen? Why does clone
return &T
instead of T
?
Why does this work if I implement Clone
myself?
use generic_array::*; // 0.12.3
use num::{Float, Zero}; // 0.2.0
#[derive(Debug)]
struct Vector<T, N: ArrayLength<T>> {
data: GenericArray<T, N>,
}
impl<T: Float, N: ArrayLength<T>> Clone for Vector<T, N> {
fn clone(&self) -> Self {
Vector::<T, N> {
data: self.data.clone(),
}
}
}
impl<T, N: ArrayLength<T>> Vector<T, N>
where
T: Float + Zero,
{
fn dot(&self, other: Self) -> T {
self.data
.iter()
.zip(other.data.iter())
.fold(T::zero(), |acc, x| acc + *x.0 * *x.1)
}
fn length_sq(&self) -> T {
self.dot(self.clone())
}
}
Eli Friedman
You get this error when your type doesn't implement Clone
:
struct Example;
fn by_value(_: Example) {}
fn by_reference(v: &Example) {
by_value(v.clone())
}
error[E0308]: mismatched types
--> src/lib.rs:6:14
|
6 | by_value(v.clone())
| ^^^^^^^^^ expected struct `Example`, found &Example
|
= note: expected type `Example`
found type `&Example`
This is due to the auto-referencing rules: the compiler sees that Example
doesn't implement Clone
, so it instead tries to use Clone
on &Example
, and immutable references always implement Clone
.
The reason your Vector
type doesn't implement Clone
is because the derived Clone
implementation doesn't have the right bounds on the type parameters (Rust issue #26925). Try explicitly writing self.dot(Self::clone(self))
to get an error message along these lines.
See also:
来源:https://stackoverflow.com/questions/37765586/why-does-cloning-my-custom-type-result-in-t-instead-of-t