Explain this recursive

JavaScript
1 
function count(n) {
  if (n === 1) {
    return [1];
  } else {
    var numbers = count(n - 1); 
    numbers.push(n);
    return numbers;
  }
}

If you call function count(5) it gives 1,2,3,4,5. However somethings are unclear in the count function. (1) Var numbers is never declared as an array so i don’t know why we are able to push into it like an array. (2) I expect the n-1 that first run to be 4, so i expect the end result to be something like 4,3,2 and not 1,2,3,4,5.

1 Like
2

the first value returned is an array because when the following is reached then the function doesn’t call itself anymore but return a value.

The second to last function call has a value for its numbers variable, which is an array (as that is what count(1) has returned). So now n which in this iteration is 2 is pushed, the array is returned and count(2) has an output, and so on till the process reaches the first function call

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3

This is where a console message within a function can be very informative about what is going on. Try putting console.table(numbers) just before the function returns numbers;.

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4

Hello!

Do the following exercise:

count(3);

n = 3;
n === 1? false // return [1] skipped.
// Here count is called again, hence this steps is paused.
var numbers = count(n - 1); // n would be 2
# STEP1

n = 2;
n === 1? false // return [1] skipped.
// Here count is called again, hence this steps is paused.
var numbers = count(n - 1); // n would be 1
// Here the function is paused while the count(n-1) is executed
#STEP2

n = 1;
n === 1? true // an array with one is returned.

// Now the function call retraces its steps (yep, backwards now)
#Continue with STEP2
// Remember, n = 2
numbers = [1];
numbers.push(2);
return [1, 2];

#Continue with STEP1
// Remember, n = 3
numbers = [1, 2];
numbers.push(3);
return [1, 2, 3];

It helps me understand some algorithms, I hope it helps You too :slight_smile:.

8 Likes
5

Thanks man, it really helps !

1 Like