Javascript recursion count forward and backward

Question

I am trying to understand recursion by writing a recursive count function. One variation of the function is to count forward and other backward. I understood the backward one but not the forward one because just by changing order of statements in the else block I am able to count forward.

function countforward(num) {
    if (num == 0) {
        return 0;
    }
    else if (num == 1) {
        return console.log(1);
    }
    else {
        countforward(num - 1);
        console.log(num);
    }
}


function countbackward(num) {
    if (num == 0) {
        return 0;
    }
    else if (num == 1) {
        return console.log(1);
    }
    else {
        console.log(num);
        countbackward(num - 1);
    }
}

The backward function is clear to me because every time the control hits the else block first the number is displayed and it is reduced and called again, but i didn't get the forward function

Where is the recursion? No function is calling itself. They're calling count(), which is undefined. — Jeremy Thille
– Jeremy Thille, Commented Jul 17, 2020 at 17:31
the line count(num - 1); should be return count(num - 1); — epascarello
– epascarello, Commented Jul 17, 2020 at 17:39

Scott Hunter · Accepted Answer · 2020-07-17 17:35:36Z

1

Suppose we wrote a description of what each function does:

countforward(n) prints the values from 1 to n in ascending order
countbackward(n) prints the values from n to 1 in descending order

Placing these descriptions in for the respective recursive calls, we get:

else {
    countforward(n-1); // prints the values from 1 to n-1 in ascending order
    console.log(n);    // prints n
}

and

else {
    console.log(n);     // prints n
    countbackward(n-1); // prints the values from n-1 to 1 in descending order
}

which (hopefully) makes things clear.

answered Jul 17, 2020 at 17:35

Scott Hunter

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3 Comments

Bala Krishna Kesani Over a year ago

sure I edited my code, (it was a typo)but I am having a hard time to understand the forward function because when i call it with countforward(3), control hits the else block where the first statement reduces it to countforward(2) and below it the console.log statement should print n value which is 3, but i get 1

Scott Hunter Over a year ago

countfoward(3) prints 1,2,3 by (calling countforward(2) which prints 1,2 by (calling countforward(1) which prints 1) then prints 2) then prints 3).

Bala Krishna Kesani Over a year ago

Thank you. I get it now

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