queue-typescript

Simple Typescript Queue with generics type templating and support for iterator and iterable protocols.

This queue uses the linked-list-typescript as the underlying datastructure.

See Also:

• linked-list-typescript
• stack-typescript
• hashlist-typescript

Installation

npm:

npm install --save queue-typescript

yarn:

yarn add queue-typescript

Building from source

install dev dependencies. There are no production dependencies.

yarn
npm install

build using the options in tsconfig.json

yarn|npm run build

run all package tests

yarn|npm run test

see the test coverage report

yarn|npm run coverage
yarn|npm run coverage:report

Usage

Importing:

import { Queue } from 'queue-typescript';
const { Queue } = require('queue-typescript')

API

Queue(...values: T[])

Queue()

Create an empty queue by omitting any arguments during instantiation.

let queue = new Queue<number>()

Queue(...values: T[])

Create a new queue and initialize it with values. Values will be added from front to back. i.e. the first argument will be at the front of the queue and the last argument will be at the back of the queue.

Specify the type using the typescript templating to enable type-checking of all values going into and out of the queue.

let items: number[] = [4, 5, 6, 7];
let queue = new Queue<number>(...items);

let items: string[] = ['one', 'two', 'three', 'four'];
let queue = new Queue<string>(...items);

Typescript will check if the values match the type given to the template when initializing the new queue.

let items: = ['one', 'two', 'three', 4];
let queue = new Queue<string>(...items); // arguments are not all strings

Queue(...values: Foo[])

Create a new queue using custom types or classes. All values are retained as references and not copies so removed values can be compared using strict comparison.

class Foo {
  private val:number;
  constructor(val: number) {
    this.val = val;
  }
  get bar(): number { return this.val }
}

let foo1 = new Foo(1);
let foo2 = new Foo(2);
let foo3 = new Foo(3);

let fooQueue = new Queue<Foo>(foo1, foo2, foo3)

fooQueue.front.bar // => 1
let val = queue.dequeue()
val // => foo1

Queue(...values: any[])

Specify any to allow the queue to take values of any type.

let queue = new Queue<any>(4, 'hello' { hello: 'world' })
queue.length // => 3
queue.front // => 4

Queue#[Symbol.iterator]

The queue supports both iterator and iterable protocols allowing it to be used with the for...of and ...spread operators and with deconstruction.

for...of:

let items: number[] = [4, 5, 6];
let queue = new Queue<number>(...items);

for (let item of queue) {
  console.log(item)
}
//4
//5
//6

...spread:

let items: number[] = [4, 5, 6];
let queue = new Queue<number>(...items);

function manyArgs(...args) {
  for (let i in args) {
    console.log(args[i])
  }
}
manyArgs(...queue);
//4
//5
//6

deconstruction:

let items: number[] = [4, 5, 6, 7];
let queue = new Queue<number>(...items);

let [a, b, c] = queue;
//a => 4
//b => 5
//c => 6

Queue#front :T

Peek at the front of the queue. This will not remove the value from the queue.

let items: number[] = [4, 5, 6, 7];
let queue = new Queue<number>(...items);
queue.front // => 4

Queue#length :number

Query the length of the queue. An empty queue will return 0.

let items: number[] = [4, 5, 6, 7];
let queue = new Queue<number>(...items);
queue.length // => 4

Queue#enqueue(val: T): boolean

Enqueue an item at the back of the queue. The new item will replace the previous last item.

let items: number[] = [4, 5, 6, 7];
let queue = new Queue<number>(...items);
queue.length // => 4
queue.enqueue(8)
queue.length // => 5

Queue#dequeue(): T

Removes the item from the front of the queue and returns the item.

let items: number[] = [4, 5, 6, 7];
let queue = new Queue<number>(...items);
queue.length // => 4
let val = queue.dequeue()
queue.length // => 3
queue.front // => 5
val // => 4

Queue#toArray(): T[]

This method simply returns [...this].

Converts the queue into an array and returns the array representation. This method does not mutate the queue in any way.

Objects are not copied, so all non-primitive items in the array are still referencing the queue items.

let items: number[] = [4, 5, 6, 7];
let queue = new Queue<number>(...items);
let result = queue.toArray()
result // => [4, 5, 6, 7]

License

MIT © Michael Sutherland