Is it possible to simplify the following using - https://hg-edge.mozilla.org/mozilla-central/file/aa05011617e022b3a0668da9fdc8ddc107dcd225/testing/mozbase/rust/mozprofile/src/preferences.rs - https://docs.rs/yaml-rust/latest/yaml_rust/yaml/enum.Yaml.html ?
```rust match value { PrefValue::Bool(bool) => Yaml::Boolean(bool), PrefValue::Int(integer) => Yaml::Integer(integer), PrefValue::String(string) => Yaml::String(string), }; ```
Jun 7, 12:21am
Yes, you can simplify the match expression using the From or Into traits, or by leveraging the yaml-rust library's From implementations for basic types. However, in this case, since PrefValue is a custom enum and Yaml is from the yaml-rust crate, you can use the Into trait if PrefValue implements it for Yaml.
But looking at the provided links and the code, there is no direct From or Into implementation between PrefValue and Yaml in the mozprofile or yaml-rust crates. Therefore, the simplest way to achieve this is to use a map operation or a helper function to convert PrefValue to Yaml.
Simplified Version Using a Helper Function
If you frequently convert PrefValue to Yaml, you can define a helper function or a trait implementation:
rust
impl From<PrefValue> for Yaml {
fn from(value: PrefValue) -> Self {
match value {
PrefValue::Bool(b) => Yaml::Boolean(b),
PrefValue::Int(i) => Yaml::Integer(i),
PrefValue::String(s) => Yaml::String(s),
}
}
}
Now, you can use it like this:
rust
let yaml_value: Yaml = pref_value.into();
// or
let yaml_value = Yaml::from(pref_value);
Simplified Version Using a Macro
If you want to avoid writing a match block every time, you could also use a macro: