Why am I returning [startNum]

JavaScript
1

Tell us what’s happening:
I was able to work through this problem only after finding out I was supposed to return [startNum] and not [endNum]…if we keep .push()ing endNums to the variable then why are we not returning the variable or [endNum]?

Your code so far


function rangeOfNumbers(startNum, endNum) {
if (endNum === startNum) {
  return [startNum];
} else {
  var numbers = rangeOfNumbers(startNum, endNum - 1);
  numbers.push(endNum);
  return numbers;
}
}

Your browser information:

User Agent is: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_15_7) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/86.0.4240.80 Safari/537.36.

Challenge: Use Recursion to Create a Range of Numbers

Link to the challenge:

2

Because [startNum] is only returned startNum === endNum, that means it is exactly the same result as [endNum].

1 Like
3

I guess my question then is how after we reach the base case, how do all the endNums make it into the same array as the starNum.

4

With this line:

numbers.push(endNum);
5

This video might help you understand the code.How-to-execute-javascript-line-b

6

Let me walk through what happens in this function with a short example: rangeOfNumbers(1,3)

  • is 1 equal to 3? No.
  • call rangeOfNumbers(1,2). The next line can’t execute until rangeOfNumbers(1,2) returns, so let’s look at what happens in rangeOfNumbers(1,2):
    • is 1 equal to 2? No.
    • call rangeOfNumbers(1,1). The next line can’t execute until rangeOfNumbers(1,1) returns, so let’s look at what happens in rangeOfNumbers(1,1):
      • is 1 equal to 1? Yes.
      • return [1]
    • Now that rangeOfNumbers(1,1) has returned, numbers in the rangeOfNumbers(1,2)is equal to[1]
    • Push 2 onto numbers.
    • return [1, 2]
  • Now that rangeOfNumbers(1,2) has returned, numbers in rangeOfNumbers(1,3) is equal to [1, 2].
  • push 3 onto numbers
  • return [1, 2, 3]
1 Like