Recursive function Array to string javascript

Question

I am a beginner in javascript. I have to perform a recursive function allowing me to output the concatenated names of the array as a string. Has very little, I am there, except my last value is given to me in undefined

const userNames = ['Elon', 'Susan', 'Bill', 'Marika', 'Ryan'];

function name(i) {
  if (i < userNames.length) {
  var names = userNames[i] + ' - ' + name(++i);
   return names;
 } else {
   console.log(names);
 };
}
name(0);

thx all

Probably because you're not returning anything in the else case. — Carcigenicate
– Carcigenicate, Commented May 16, 2017 at 18:01
@NiettheDarkAbsol This looks like an assignment, so that would entirely defeat the purpose. — Carcigenicate
– Carcigenicate, Commented May 16, 2017 at 18:02
.join is a "forbidden" function to me. Strictly asked to do a recursive function! : d — Goa_From_Gaia
– Goa_From_Gaia, Commented May 16, 2017 at 19:56

Nina Scholz · Accepted Answer · 2017-05-16 18:27:59Z

1

First of all, you can use the join() method:

const userNames = ['Elon', 'Susan', 'Bill', 'Marika', 'Ryan'];
console.log(userNames.join(' - '));

Your recursive function is interesting anyway. The problem is console.log(names); is reached when you call name() with argument 5, and in that function call it is undefined.

You want to console.log() the result of the first call;

To avoid the last call is undefined, just return an empty string

const userNames = ['Elon', 'Susan', 'Bill', 'Marika', 'Ryan'];

function name(i) {
  if (i < userNames.length) {
   var names = userNames[i];
   let n = name(++i);
   return names + (n ? ' - ' + n : '');
 } else {
   return '';
 }
}
console.log(name(0));

edited May 16, 2017 at 18:27

Nina Scholz

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answered May 16, 2017 at 18:04

rpadovani

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4 Comments

Carcigenicate Over a year ago

This will unfortunately leave a trailing dash though. But I agree with this solution anyways. This is the same code I was about to post.

rpadovani Over a year ago

@Carcigenicate well, I think the scope of the OP is understanding how the recursion works - there are a lot of easier way to achieve what he is trying to do. Anyway you are right, I update the function to take in consideration that.

Carcigenicate Over a year ago

Ya, I just thought I'd point it out. I tried to fix it by messing around on my phone in the JSRun app, but fixing it made the code more convoluted and detracted from the original problem.

Goa_From_Gaia Over a year ago

It is a cunning method, but I opt for this solution. I would not have thought of this ternary. Thank you ladies and gentlemens! :D

Stephen L · Accepted Answer · 2017-05-16 18:40:45Z

The problem is you're mixing your recursion code with the code that you use to print the string at the end. The recursive call hits the 'else' block and falls through to the end of the function. Since there is no 'return' defined at the end of the function, the last 'name(++i)' call evaluates to 'undefined' and this is concatenated to the original string. Here are two solutions:

const userNames = ['Elon', 'Susan', 'Bill', 'Marika', 'Ryan'];

function name(i) {
  if (i === userNames.length) {
      return '';
  }
  else {
   return userNames[i] + ' - ' + name(++i);
 }
}

function printList() {
    console.log(name(0))
}

printList();

or

const namesList = ['Elon', 'Susan', 'Bill', 'Marika', 'Ryan'];

function name(i) {
  if (i < userNames.length) {
  var names = userNames[i] + ' - ' + name(++i);
   return names;
 } else {
   console.log(names);
   return '';
 };
}
name(0);

Mulan · Accepted Answer · 2017-05-17 01:55:41Z

First think about the complete domain of your function. Your function accepts an array of strings. The array can contain 0, 1, or more strings and we need your function to work for all cases of that

// contract
joinStrings([])      == ''
joinStrings([a])     == a
joinStrings([a,b])   == a + '-' + b
joinStrings([a,b,c]) == a + '-' + b + '-' + c

Encoding this in a recursive function is trivial in JavaScript – notice there's no need to keep track of array indexes or worry about incrementing an iterator – just think about fulfilling that contract

const joinStrings = xs => {
  switch (xs.length) {
    case 0:   return ''
    case 1:   return xs[0]
    default:  return xs[0] + '-' + joinStrings(xs.slice(1))
  }
}

console.log(joinStrings([]))               // ''
console.log(joinStrings(['a']))            // 'a'
console.log(joinStrings(['a', 'b']))       // 'a-b'
console.log(joinStrings(['a', 'b', 'c']))  // 'a-b-c'

Or course our function could be vastly improved if we allow the user to specify the joining string

const joinStrings = (y, xs) => {
  switch (xs.length) {
    case 0:   return ''
    case 1:   return xs[0]
    default:  return xs[0] + y + joinStrings(y, xs.slice(1))
  }
}

console.log(joinStrings('&', []))               // ''
console.log(joinStrings('&', ['a']))            // 'a'
console.log(joinStrings('&', ['a', 'b']))       // 'a&b'
console.log(joinStrings('&', ['a', 'b', 'c']))  // 'a&b&c'

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