Function build_trait_substitution_map

src/inheritance.rs:1094–1126

Build a substitution map for a trait based on `@use` generics and the trait's `@template` parameters. If the using class declares `@use HasFactory ` and the trait `HasFactory` has `@template TFactory`, the returned map is `{TFactory => UserFactory}`.

(
    trait_name: &str,
    trait_info: &ClassInfo,
    use_generics: &[(Atom, Vec<PhpType>)],
)

/// trait `HasFactory` has `@template TFactory`, the returned map is
/// `{TFactory => UserFactory}`.
fn build_trait_substitution_map(
    trait_name: &str,
    trait_info: &ClassInfo,
    use_generics: &[(Atom, Vec<PhpType>)],
) -> HashMap<String, PhpType> {
    if trait_info.template_params.is_empty() || use_generics.is_empty() {
        return HashMap::new();
    }

    let trait_short = short_name(trait_name);

    // Find the @use entry that matches this trait.
    let type_args = use_generics
        .iter()
        .find(|(name, _)| {
            let name_short = short_name(name);
            name_short == trait_short
        })
        .map(|(_, args)| args);

    let type_args = match type_args {
        Some(args) => args,
        None => return HashMap::new(),
    };

    let mut map = HashMap::new();
    for (i, param_name) in trait_info.template_params.iter().enumerate() {
        if let Some(arg) = type_args.get(i) {
            map.insert(param_name.to_string(), arg.clone());
        }
    }
    map
}

/// Build a substitution map for a parent class based on the child's
/// `@extends` generics and the parent's `@template` parameters.