Why this snippet is not pure?

Question

When I watch Google Chrome Developers, https://youtu.be/qaGjS7-qWzg?t=636

They said this snippet is not pure. I don't know why.

const g = new Map();

for (const [u,v] of edges) {
    if (!g.has(u))
        g.set(u, []);
    if (!g.has(v))
        g.set(v, []);
    g.get(u).push(v)
}

And they also mentioned this is pure,

const startPoints = new Set(edges.map(([u, v]) => u));

const g = new Map(
  [...startPoints].map(startPoint =>
    edges.filter(([u, v]) => u == startPoint).map(([u, v]) => v)
  )
);

score 4 · Accepted Answer · edited Feb 08 '20 at 17:30

4

When we talk about purity in a system we are talking about a system that doesn't change the observable state.

In the first script at line 1 g is an empty Map and by the last line it is not. Its value was mutated.
In the second script startPoints is the same thing at line one and at the end.

To make it a little easy, let me explain it with a function. Let's suppose you have a function that giving an array of numbers it multiply it by 2. So if:

var a = [1,2,3];
multiply(a)
// a = [2, 4, 6]

This function is not pure. It has changed the observable state of the system that in this case is a. But if:

var a = [1,2,3];
var b = multiply(a)
// a = [1, 2, 3]
// b = [2, 4, 6]

This function is pure. a didn't change.

edited Feb 08 '20 at 17:30

Paulo Freitas

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answered Feb 08 '20 at 15:57

Johnny Zabala

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So you can't use something like `pop`, `push` in the pure world? Seems You have to assign a new variable when you want to mutate an Object/Array every time. – foxiris Feb 09 '20 at 12:53
Exactly in a pure world (that JavaScript is not), you create new copies every time you want to mutate the state. In JS there is a lib that does that https://immutable-js.github.io/immutable-js/. – Johnny Zabala Feb 09 '20 at 12:58

Why this snippet is not pure?

1 Answers1