Recursive function: the process of storing numerous values in variables

There is a code snippet showcasing a recursive function designed to take a number, for example n=5, and produce an array counting down from n to 1 (i.e., [5,4,3,2,1]).

The point of confusion arises just before the numbers/values of n are added to countArray. The query centers around how countup(n - 1) generates the sequence of numbers like 5, 4, 3, 2, 1. It seems puzzling where or how these values get stored. One might expect n to simply become its last defined value, such as n=1, or result in a blank array. However, the reality is that all these values get added into an array that seemingly comes out of nowhere, without prior definition. This is what needs clarification regarding those two lines with comments.

tl;dr: (1) How are the values 5 through 1 saved without getting overwritten by the final value of 1 before being pushed into the array? (2) When and how was countArray actually established as an array before we start adding elements to it?

function countup(n) {
  if (n < 1) {
    return [];
  } else {
    const countArray = countup(n - 1);  //The storage logic behind 5, 4, 3, 2, 1 utterly confuses me
    countArray.push(n); //I'm perplexed about when countArray became initialized as an array
    return countArray;
  }
}
console.log(countup(5)); // [ 1, 2, 3, 4, 5 ]

Edit: Perhaps this post's title should focus more on arrays than variables or similar topics.

Answer №1

Consider enhancing the clarity of the code by incorporating logging.

To achieve this, we can introduce a basic logging mechanism to display the values in the following manner:

Calling with 5
|   Calling with 4
|   |   Calling with 3
|   |   |   Calling with 2
|   |   |   |   Calling with 1
|   |   |   |   |   Calling with 0
|   |   |   |   |   Returning [] (base case)
|   |   |   |   Pushing 1 to []
|   |   |   |   Returning [1]
|   |   |   Pushing 2 to [1]
|   |   |   Returning [1,2]
|   |   Pushing 3 to [1,2]
|   |   Returning [1,2,3]
|   Pushing 4 to [1,2,3]
|   Returning [1,2,3,4]
Pushing 5 to [1,2,3,4]
Returning [1,2,3,4,5]

Initially, the array was initialized in the base case. Subsequently, as we traversed back up the call stack, elements were added incrementally. Although there are alternate approaches, the method illustrated here is commonly adopted and logical.

The implementation of logging is evident in the snippet below:

const log = (depth, message) => 
  console .log ('|   '.repeat (depth - 1)  + message)


function countup(n, depth = 1) {
  log(depth, `Calling with ${n}`)
  if (n < 1) {
    log(depth, `Returning [] (base case)`)
    return [];
  } else {
    const countArray = countup(n - 1, depth + 1);  
    log(depth, `Pushing ${n} to [${countArray}]`)
    countArray.push(n); 
    log(depth, `Returning [${countArray}]`)
    return countArray;
  }
}

countup(5)
.as-console-wrapper {min-height: 100% !important; top: 0}

Update

A potentially clearer representation could be achieved through the following result:

/ Calling with 5
|   / Calling with 4
|   |   / Calling with 3
|   |   |   / Calling with 2
|   |   |   |   / Calling with 1
|   |   |   |   |   / Calling with 0
|   |   |   |   |   \ Returning [] (base case)
|   |   |   |   | Pushing 1 to []
|   |   |   |   \ Returning [1]
|   |   |   | Pushing 2 to [1]
|   |   |   \ Returning [1,2]
|   |   | Pushing 3 to [1,2]
|   |   \ Returning [1,2,3]
|   | Pushing 4 to [1,2,3]
|   \ Returning [1,2,3,4]
| Pushing 5 to [1,2,3,4]
\ Returning [1,2,3,4,5]

This modified version only entails minor adjustments to the existing logging statements:

const log = (depth, message) => 
  console .log ('|   '.repeat (depth - 1)  + message)
  
function countup(n, depth = 1) {
  log(depth, `/ Calling with ${n}`)
  if (n < 1) {
    log(depth, `\\ Returning [] (base case)`)
    return [];
  } else {
    const countArray = countup(n - 1, depth + 1);  
    log(depth, `| Pushing ${n} to [${countArray}]`)
    countArray.push(n); 
    log(depth, `\\ Returning [${countArray}]`)
    return countArray;
  }
}

countup(5)
.as-console-wrapper {min-height: 100% !important; top: 0}

Answer №2

In any recursive function that eventually terminates, there must always be a stopping condition.

For instance, in the case of your function:

if (n < 1) {
 return [];
}

The next essential component of a recursive function is the recursion itself.

This occurs here:

const countArray = countup(n - 1);

You are invoking the function with an n value reduced by one.

As you follow through this process, the function will continue to recursively call itself until reaching a point where n < 1. It is at this moment that the array will be created and returned.

Subsequently, you will have the opportunity to add values to this array.

Remember that it is crucial to return countArray;, ensuring that the array is passed up to the calling functions.

An important aspect to grasp about recursive functions is that each function call is completed before the program can move on to the next step.

To better comprehend recursive functions, visualize the function calls as a stack and track how they interact with one another.

Answer №3

(It's frustrating that I can't add code directly as a comment).

If you're struggling to grasp where the array is initiated, try walking through the process yourself. This hands-on approach often provides better clarity than any written explanation could offer.

Here's how you can proceed:

  1. Launch your browser console by pressing F12 and navigating to the console tab
  2. Insert the following code snippet with a debugger statement:

-

function countup(n) {
  debugger;
  if (n < 1) {
    return [];
  } else {
    const countArray = countup(n - 1);  //the part where 5,4,3,2,1 get stored confuses me
    countArray.push(n); //I'm not sure when countArray was initialized as an array
    return countArray;
  }
}
console.log(countup(5)); // [ 1, 2, 3, 4, 5 ]
  1. Utilize the step in and step over functions
  2. Enjoy the learning process!

Answer №4

As we reach this juncture

const countArray = countup(n - 1);

The function now initiates a "wait" and triggers another call to itself, only concluding when it reaches the stop condition if (n < 1) at countup(0). This results in an empty array being returned before moving on to the previous call of countup(1). As a result, countArray is assigned the value from countup(0), which is [], signifying the validity of future push operations.

During this time, variables are retained in memory. Whenever a function is called, its parameters, local variables, and return address are all stored within the memory stack.

To delve deeper into recursion and the call stack, it would be beneficial to explore further resources.

Answer №5

In a Nutshell

The code snippet where an array is created and returned is return [];.

With each recursive call, the parameter (n) decreases until it reaches 0, at which point the array is created.


Imagine you invoke countup(2).

This leads to the else condition triggering countup(1). The process repeats until countup(0) is called.

At countup(0), an empty array is generated and returned instead of going into the else branch.

Following this, countup(1) adds 1 to the array (and returns the updated result), then countup(2) appends 2.

Let's try to illustrate the recursion:

countup(2)
  2<1 --> false --> else
  countArray=countup(1)
    1<1 --> false --> else
      countArray=countup(0)
        0<1 --> true --> if
        return []
      push 1 --> [1]
      return countArray --> [1]
  push 1 -->[1,2]
  return countArray -->[1,2]

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