Reputation: 25
I have a struct A which implements BorshDeserialize and BorshSerialize as follows
#[derive(Clone, PartialEq, Eq, BorshSerialize, BorshDeserialize)]
struct A {
a : i32,
b: String,
}
I know I can serialize or deserialize A when I do the following:-
let s = A {a: 1 , b: "a".to_string()};
// Serialize
let serialzed_data = s.try_to_vec().unwrap()
// Deserialize
deserialized_struct = A::try_from_slice(&serialzed_data).unwrap();
I am trying to wrap over these two methods by creating two general traits on main.rs where I import this struct from another file a.rs.
pub trait Serializable<T: BorshSerialize> {
fn serialize(s: T) -> Vec<u8> {
s.try_to_vec().unwrap()
}
}
pub trait Deserializable<T : BorshDeserialize> {
fn deserialize(s: &[u8]) -> Result<T, ()> {
let deserialized_val = match T::try_from_slice(&s) {
Ok(val) => {val},
Err(_) => {return Err(());},
};
Ok(deserialized_val)
}
}
where I implement Serialize and Deserialize for A as follows on a.rs
impl Serializable<A> for A {}
impl Deserializable<A> for A {}
But in the source code when I call the method serialize on A for this instruction,
A::serialize(&some_instance_of_A).unwrap()
I get the following error
A::serialize(&some_instance_of_A).unwrap()
^^^^^^^^^ multiple `serialize` found
|
= note: candidate #1 is defined in an impl of the trait `Serializable` for the type `A`
= note: candidate #2 is defined in an impl of the trait `BorshSerialize` for the type `A`
help: disambiguate the associated function for candidate #1
|
46 | <&Self as A::Serializable>::serialize(&some_instance_of_A), // TODO. See issue #64
| ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
help: disambiguate the associated function for candidate #2
|
46 | <&A as BorshSerialize>::serialize(&some_instance_of_A).concat(), // TODO. See issue #64
|
I understand that the compiler gets confused by the two instances of serialization schemes created(one due to Borsh from derive macro and the other from Serialize trait on main.rs). Is there a way to directly make the serialize call fall to BorshSerialize by default when A::serialize is called.
Upvotes: 2
Views: 828
Reputation: 2114
By preventing to import the BorshSerialize
trait you can get rid of the error.
With only the Serializable
trait in scope, only one method can be called:
#[derive(Clone, PartialEq, Eq, borsh::BorshSerialize, borsh::BorshDeserialize)]
struct A {
a : i32,
b: String,
}
impl Serializable for A {}
fn main() {
let a = A { a: 0, b: String::from("") };
let vec = a.serialize();
eprintln!("{:?}", vec);
}
pub trait Serializable: borsh::BorshSerialize {
fn serialize(&self) -> Vec<u8> {
self.try_to_vec().unwrap()
}
}
pub trait Deserializable<T : borsh::BorshDeserialize> {
fn deserialize(s: &[u8]) -> Result<T, ()> {
let deserialized_val = match T::try_from_slice(&s) {
Ok(val) => {val},
Err(_) => {return Err(());},
};
Ok(deserialized_val)
}
}
Upvotes: 1