Trait LinkedResource

pub trait LinkedResource<'a, T: Display> {
    // Required methods
    fn child(&self) -> Option<&Self>;
    fn parent(&self) -> Option<&Self>;
    fn is_child(&self) -> bool;
    fn is_root(&self) -> bool;
    fn is_tail(&self) -> bool;
    fn with_child(
        &mut self,
        child: &mut ApiResource<'a, T>,
    ) -> Result<Box<Self>>;
    fn with_parent(
        &mut self,
        parent: &mut ApiResource<'a, T>,
    ) -> Result<Box<Self>>;
}

Allows resources to set their child and parent nodes.

Required Methods

fn child(&self) -> Option<&Self>

The child Resource node.

fn parent(&self) -> Option<&Self>

The parent Resource node.

fn is_child(&self) -> bool

If this is a child of another resource.

Initialy created object should produce a non-child node.

use uri_resources::{ApiResource, LinkedResource};
let resource = ApiResource::<String>::new("resource");
assert_eq!(resource.is_child(), false)

Try to create an instance of two nodes where one is related to the other as the parent.

use uri_resources::{ApiResource, LinkedResource};
let mut child = ApiResource::<String>::new("child_resource");
let parent = ApiResource::<String>::new("parent_resource")
    .with_child(&mut child);
assert_eq!(child.is_child(), true)

fn is_root(&self) -> bool

If this is the first resource of the path.

Initialy created object should produce a root node.

use uri_resources::{ApiResource, LinkedResource};
let resource = ApiResource::<String>::new("resource");
assert_eq!(resource.is_root(), true)

Subsequent objects should not be a root node.

use uri_resources::{ApiResource, LinkedResource};
let mut child = ApiResource::<String>::new("child_resource");
let parent = ApiResource::<String>::new("parent_resource")
    .with_child(&mut child);
assert_ne!(child.is_root(), true)

fn is_tail(&self) -> bool

If this is the last resource of the path.

Root node can be a tail node if it is the only resource node.

use uri_resources::{ApiResource, LinkedResource};
let resource = ApiResource::<String>::new("resource");
assert!(resource.is_tail())

If there are otherwise child nodes, a root node cannot be the ‘tail’.

use uri_resources::{ApiResource, LinkedResource};
let mut child0 = ApiResource::<String>::new("child_resource0");
let mut child1 = ApiResource::<String>::new("child_resource1");

child0 = *child0.with_child(&mut child1).expect("resource node");
let parent = ApiResource::<String>::new("parent_resource")
    .with_child(&mut child0);
assert!(!parent.expect("parent node").is_tail())

The middle child cannot be the tail.

use uri_resources::{ApiResource, LinkedResource};
let mut child0 = ApiResource::<String>::new("child_resource0");
let mut child1 = ApiResource::<String>::new("child_resource1");

child0 = *child0.with_child(&mut child1).expect("resource node");
let parent = ApiResource::<String>::new("parent_resource")
    .with_child(&mut child0);
assert!(child0.is_child() && !child0.is_tail());

The last child should be the tail.

use uri_resources::{ApiResource, LinkedResource};
let mut child0 = ApiResource::<String>::new("child_resource0");
let mut child1 = ApiResource::<String>::new("child_resource1");

child0 = *child0.with_child(&mut child1).expect("resource node");
let parent = ApiResource::<String>::new("parent_resource")
    .with_child(&mut child0);
assert!(child1.is_child() && child1.is_tail())

fn with_child(&mut self, child: &mut ApiResource<'a, T>) -> Result<Box<Self>>

Adds a child node to this resource. Fails if the child is already set.

fn with_parent(&mut self, parent: &mut ApiResource<'a, T>) -> Result<Box<Self>>

Adds the parent node to this resource. Fails if the parent is already set.

Dyn Compatibility

This trait is not dyn compatible.

In older versions of Rust, dyn compatibility was called "object safety", so this trait is not object safe.

Implementors

impl<'a, T: Debug + Display + Clone> LinkedResource<'a, T> for ApiResource<'a, T>