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Comments on How this recursive treewalker works?

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How this recursive treewalker works?

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Credit for User:Meriton for developing the following code (first published here).

function replaceIn(e) {
  if (e.nodeType == Node.TEXT_NODE) {
    e.nodeValue = e.nodeValue.replaceAll("a", "");
  } else {
    for (const child of e.childNodes) {
      replaceIn(child);
    }
  }
}

replaceIn(document.body);

How this recursive treewalker works?

As a side question which I grasp as important, can there be an even simpler and more direct version without the else?

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2 comment threads

As explained [here](https://software.codidact.com/posts/286333), there are different types of nodes. ... (8 comments)
What is the goal that you want to achieve? Why do you want to avoid using an if-else? (6 comments)
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Note: This answer was written for a version of the question that focused on getting rid of the if-else, rather than on an explanation of the workings of the recursive tree-traversal.

One way of getting rid of if-elses to achieve uniform handling of different objects is to hide the distinction in lower levels of abstraction:

Imagine a bowl of apples and bananas. Basically, you just want to eat all of them. However, if you one by one pick the bowl contents, you first have to check if you have a banana - which you need to peel before you bite into it. That is, your algorithm is "for each fruit: if banana, peel then eat, if apple, eat". You can abstract it by introducing a function "prepare_for_eating". Then your algorithm is "for each fruit: prepare_for_eating(fruit) then eat(fruit)". Seems as if the if-else has disappeared. But, if you look inside the "prepare_for_eating" function, you will again have the original if-else statement where you decide to peel or not to peel.

With object oriented techniques you can hide if-elses behind method calls: A Fruit base class would provide a member function "prepare_for_eating". For banana, this would contain the "peel" statement, for the apple, this would be empty. The algorithm would be "for each fruit: fruit.prepare_for_eating then fruit.eat". If fruit happens to be a banana, the banana member function doing the peel would be called. For an apple object, the apple member function would be called that does nothing.

You can use other tricks to get rid of if-elses by computing boolean expressions and use their result as zero or one in multiplications or table lookups. For example:

if x == 42 then x = -1

can be re-written as (assuming true can be used as 1 and false as 0):

x = (x == 42) * -1 + (x != 42) * x;

Or, using such a result for table lookups:

action_table = [peel, do_nothing]
for each fruit: call action_table[fruit is apple] then eat;

Whether any of the above ideas can be applied meaningfully to your problem and is suited to fulfill your expectations is up to you.

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2 comment threads

A question (3 comments)
Hello ! I understood everything in the answer besides this: x = (x == 42) * -1 + (x != 42) * x;... (8 comments)
A question
deleted user wrote almost 2 years ago

Would it be correct to say that however we try to do it, even if there isn't else, there would always be some if? (even if not explicitly written)?

Dirk Herrmann‭ wrote almost 2 years ago

I would phrase it slightly differently, namely that in your example scenario there would always be two different cases that at run-time would have to be distinguished. Keep in mind that the distinction could also be made by some indirect call through a virtual function table, where, depending on the type of node, the function tables would point to different functions being called. Maybe your statement about "there would always be some if" is meant to also include such ways of distinction? In this case your understanding is right.

deleted user wrote almost 2 years ago

Yes, I meant to distinctions in the context of how the human mind would understand the different codes leading to the same goal.