Iterable.appendAll — API

Iterable.appendAllconsteffect/Iterable.ts:344

<B>(that: Iterable<B>): <A>(self: Iterable<A>) => Iterable<A | B>
<A, B>(self: Iterable<A>, that: Iterable<B>): Iterable<A | B>

Concatenates two iterables, combining their elements.

When to use

Use to lazily concatenate two iterables while preserving order, yielding all elements from self before that.

Details

The result is lazy. The iterator for that is not created or read until self is exhausted.

Gotchas

If self is infinite or never completes, that is never reached.

Example (Concatenating iterables)

import { Iterable } from "effect"

const first = [1, 2, 3]
const second = [4, 5, 6]
const combined = Iterable.appendAll(first, second)
console.log(Array.from(combined)) // [1, 2, 3, 4, 5, 6]

// Works with different iterable types
const numbers = [1, 2]
const letters = "abc"
const mixed = Iterable.appendAll(numbers, letters)
console.log(Array.from(mixed)) // [1, 2, "a", "b", "c"]

// Lazy evaluation - only consumes what's needed
const infinite = Iterable.range(1)
const finite = [0, -1, -2]
const result = Iterable.take(Iterable.appendAll(finite, infinite), 5)
console.log(Array.from(result)) // [0, -1, -2, 1, 2]

combiningappend prependAll

Source effect/Iterable.ts:34427 lines

export const appendAll: {
  <B>(that: Iterable<B>): <A>(self: Iterable<A>) => Iterable<A | B>
  <A, B>(self: Iterable<A>, that: Iterable<B>): Iterable<A | B>
} = dual(
  2,
  <A, B>(self: Iterable<A>, that: Iterable<B>): Iterable<A | B> => ({
    [Symbol.iterator]() {
      const iterA = self[Symbol.iterator]()
      let doneA = false
      let iterB: Iterator<B>
      return {
        next() {
          if (!doneA) {
            const r = iterA.next()
            if (r.done) {
              doneA = true
              iterB = that[Symbol.iterator]()
              return iterB.next()
            }
            return r
          }
          return iterB.next()
        }
      }
    }
  })
)

Referenced by 2 symbols

Iterable.append Iterable.prependAll