Script•3y ago

Script – 22-55 Feb 1

I'm trying to run the function till b.length === 1 and then return b What am I doing wrong? it returns undefined

function superDigit(n, k) {
  if (k === 1) return n;

  n = n.toString();
  let b = 0;
  for (let i = 0; i < k; i++) {
    n = n.concat(n);
  }
  for (let i in n) {
    b += parseInt(n[i]);
  }

  function recall(bb) {
    bb = bb.toString();
    bbb = 0;
    for (let i in bb) {
      bbb += parseInt(bb[i]);
    }
    b = bbb;
    if (b.length === 1) {
      return b;
    }
  }

 recall(b);
}

function superDigit(n, k) {
  if (k === 1) return n;

  n = n.toString();
  let b = 0;
  for (let i = 0; i < k; i++) {
    n = n.concat(n);
  }
  for (let i in n) {
    b += parseInt(n[i]);
  }

  function recall(bb) {
    bb = bb.toString();
    bbb = 0;
    for (let i in bb) {
      bbb += parseInt(bb[i]);
    }
    b = bbb;
    if (b.length === 1) {
      return b;
    }
  }

 recall(b);
}

30 Replies

ScriptyChris•3y ago

because you don't return recall(b) call?

ghardin137•3y ago

wow you could probably stand to name your variables a bit better 🙂

Script•3y ago

function recall(bb) {
    bb = bb.toString();
    bbb = 0;
    for (let i in bb) {
      bbb += parseInt(bb[i]);
    }
    b = bbb;
    console.log("2", b);
    return b;
  }
  if (b.length === 1) {
    return b;
  } else return recall(b);

function recall(bb) {
    bb = bb.toString();
    bbb = 0;
    for (let i in bb) {
      bbb += parseInt(bb[i]);
    }
    b = bbb;
    console.log("2", b);
    return b;
  }
  if (b.length === 1) {
    return b;
  } else return recall(b);

ScriptyChris•3y ago

i meant this

Script•3y ago

😂 I was just rushing and giving it similar var names I tried that but still got undefined

ScriptyChris•3y ago

is b.length === 1condition ever met?

Script•3y ago

yes

ScriptyChris•3y ago

so recall at that case should return non-undefined b variable, because it surely is array/string with length property equal to 1

Script•3y ago

I'll look at my code

ScriptyChris•3y ago

to be clear, you mean that superDigit returns undefined or recall? and are you sure that recursion ends sometime, not throws call stack exceeded error?

S3BAS•3y ago

Return the last line return recall(b) @Script

Script•3y ago

Okay so I am finally exceeding the call stack

function superDigit(n, k) {
  //   if (k === 1) return n;

  n = n.toString();
  let n2 = n;
  let b = 0;
  for (let i = 0; i < k - 1; i++) {
    n = n.concat(n2);
  }
  for (let i in n) {
    b += parseInt(n[i]);
  }

  function recall(newVal) {
    newVal = newVal.toString();
    point0 = 0;
    for (let i in newVal) {
      point0 += parseInt(newVal[i]);
    }
    b = point0;
    if (b.length == 1) {
      return b;
    } else {
      recall(b);
    }
  }

  return recall(b);
}

function superDigit(n, k) {
  //   if (k === 1) return n;

  n = n.toString();
  let n2 = n;
  let b = 0;
  for (let i = 0; i < k - 1; i++) {
    n = n.concat(n2);
  }
  for (let i in n) {
    b += parseInt(n[i]);
  }

  function recall(newVal) {
    newVal = newVal.toString();
    point0 = 0;
    for (let i in newVal) {
      point0 += parseInt(newVal[i]);
    }
    b = point0;
    if (b.length == 1) {
      return b;
    } else {
      recall(b);
    }
  }

  return recall(b);
}

How do I manage this

S3BAS•3y ago

Yeah then your base condition is never being met

Script•3y ago

It is actually b becomes 3 that's length of 1

S3BAS•3y ago

If you would properly hit the base condition you wouldn’t exceed the call stack Are you calling length on a number? Does that give you the amount of digits?