How do I serialize/deserialize &[f64] using serde in Rust? - rust

Question

This compiles successfully:
#[derive(Serialize, Deserialize)]
struct Person {
name: String,
value: f64
}
#[tokio::main]
async fn main() -> Result<(), Box<dyn std::error::Error>> {
let perspn = Person { name: "me".into(), value: 30_f64};
let serialized = serde_json::to_string(&perspn).unwrap();
println!("{:?}", serialized);
Ok(())
}
But on changing a member of the struct to &[f64], then compilation fails
#[derive(Serialize, Deserialize)]
struct Person<'a> {
name: String,
value: &'a[f64]
}
#[tokio::main]
async fn main() -> Result<(), Box<dyn std::error::Error>> {
let perspn = Person { name: "me".into(), value: &[30_f64]};
let serialized = serde_json::to_string(&perspn).unwrap();
println!("{:?}", serialized);
Ok(())
}
fails with error:
error[E0277]: the trait bound `&[f64]: Deserialize<'_>` is not satisfied
--> src/main.rs:168:5
|
168 | value: &'a[f64]
| ^^^^^ the trait `Deserialize<'_>` is not implemented for `&[f64]`
|
= help: the following other types implement trait `Deserialize<'de>`:
&'a [u8]
[T; 0]
[T; 10]
[T; 11]
[T; 12]
[T; 13]
[T; 14]
[T; 15]
and 26 others
Why is this? Why does serde have a default deserialiser for f64 but not for &[f64]?
Looking at the error it looks like it is only [u8] that serde can handle by default. Why is this?
Now the question is, how do I deserialize &[f64]? I know I will need to write a custom deserializer for it, but given that a custom one was not given, I am wondering if there are edge case in the implementation that made that the case and which I need to be aware of?