CODEWITHSUNDEEP
javascriptimmutable.js
dorachan2010
dorachan2010

Reputation: 1091

Does ImmutableJS skip unused code block?

I am playing around with Immutable.js code and noticed something funky. Does Immutable.js skip code that saves to variables that won't be used?

const Immutable = require('immutable')

function transformErrors(errors) {
    let key = errors.keySeq()
    let mapped = key.map((v, keystr) => {
      console.log(v, keystr)
      return keystr
    })
  // If I enable the console log below, console log above works
  // console.log('mapped', mapped) 
};
const result = transformErrors(Immutable.fromJS([1, 2]));

For the above code, if

console.log('mapped', mapped)

is disabled, the mapping code doesn't get called. I looked through the documentation but couldn't find any remarks on it

Upvotes: 1

Views: 43

Answers (1)

Matt Watson
Matt Watson

Reputation: 5317

The line: let key = errors.keySeq() will return a Seq object, which in immutable.js is lazy.

The documentation provides the following details (https://facebook.github.io/immutable-js/docs/#/Seq):

Seq is lazy — Seq does as little work as necessary to respond to any method call. Values are often created during iteration, including implicit iteration when reducing or converting to a concrete data structure such as a List or JavaScript Array. For example, the following performs no work, because the resulting Seq's values are never iterated:

const { Seq } = require('immutable')
const oddSquares = Seq([ 1, 2, 3, 4, 5, 6, 7, 8 ])
  .filter(x => x % 2 !== 0)
  .map(x => x * x)

So in your example, immutable.js isn't going to actually evaluate your map function until mapped is used somewhere.

Upvotes: 1

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