Checking if a variable has been set is not working

Question

I am trying to set a variable initially only if it has not been set. For some reason the logic I'm using to determine whether the variable has been set seems to run even regardless of whether the variable has been set.

My code is

var l0 = -210;

function add(){
  var lThis;
  if (lThis == null){
     lThis = l0;
     console.log('set lThis to '+ lThis);
  }
  lThis ++;
}

var incrementInterval = setInterval(add, 33);

The console logs "set lThis to -210" every time the interval runs, so the "if (lThis == null)" seems to be doing nothing

See example on codepen

lThis, is a variable name local to the add function and will be initialized and / or re-declared each time the function is invoked. To avoid complications you can also declare lThis globally. - And you never compare literals to null, unless you've nullified its handle manually and deliberately later on the code in order to reset its value. Even then, nullifying a variable that is supposed to carry a literal value is wrong. — Bekim Bacaj
– Bekim Bacaj, Commented Dec 19, 2016 at 23:26

Max Vorobjev · Accepted Answer · 2016-12-19 22:12:07Z

1

With setInterval function add is executed every 33ms from the start, do the whole function scope is recreated, including lThis. I suppose what you want to achieve is to add 1 on every call to lThis within function. One of ways to achieve that is using closure in the following way:

var l0 = -210;

function thunkAdd(){
  var lThis = l0;
  console.log('set lThis to '+ lThis);
  return function inc() {
    console.log(lThis);
    return lThis ++;
  }
}

var incrementInterval = setInterval(thunkAdd(), 33);

Note, this is the old way of writing JavaScript, with new ES6 syntax it can be achieved in much more compact form:

const l0=-210;

function *add() {
  let lThis = l0;
  while (true) yield lThis++;
}

const lAdd = add()
const incrementInterval = setInterval(() => {console.log(lAdd.next())}, 33);

answered Dec 19, 2016 at 22:12

Max Vorobjev

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1 Comment

Mike Eng Over a year ago

That was quite helpful in incrementing the variable. I have a followup I posted at stackoverflow.com/q/41232326/370407 . Maybe you'd have some insight there.

andrgolubev · Accepted Answer · 2016-12-19 21:58:40Z

1

Your variable lThis is a local variable, it exists (and created each time) only in the scope of function add(). If you want to preserve a value of lThis during the setInterval(...) you should consider to move it to the global scope:

var lThis = null;

function add() {
    if (lThis == null) {
        lThis = -210
        console.log('set lThis to '+ lThis);
    }
    lThis++;
}

answered Dec 19, 2016 at 21:58

andrgolubev

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Comments

GG. · Accepted Answer · 2016-12-19 22:16:15Z

1

lThis is a local variable in your example. So lThis will be reinitialized to undefined each time the function add() is called, and you will never reach lThis++. You probably want to move lThis outside of add():

var lThis;

function add() {
  if (!lThis) {
     lThis = l0;
     console.log('set lThis to '+ lThis);
  }
  lThis++;
}

edited Dec 19, 2016 at 22:16

answered Dec 19, 2016 at 21:57

GG.

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Comments

Alex Wiese · Accepted Answer · 2016-12-19 23:35:08Z

1

lThis is a local variable. This variable would be reset to 'null' every 33ms, when add() gets called.

What about another global variable, like this:

var l0 = -210;
var lThis;

function add() {
  if (lThis == null) {
    lThis = l0;
  .......

edited Dec 19, 2016 at 23:35

Alex Wiese

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answered Dec 19, 2016 at 22:06

mzhan017

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Comments

Mulan · Accepted Answer · 2016-12-21 04:27:55Z

0

For what it's worth I think your whole code snippet needs a dramatic rewrite

function add (x) {
  console.log(x)
  setTimeout(add, 33, x + 1)
}

let timeout = add(-210)

If you really want a global variable for some (probably bad) reason

let x = -210

function add () {
  console.log(x)
  x = x + 1
}

let timeout = setInterval(add, 33)

And another

function delay (ms) {
  return new Promise(r => setTimeout(r, ms))
}

async function add (x) {
  console.log(x)
  await delay(33)
  add(x + 1)
}

add(-210)

And... no just kidding. There are no more examples.

edited Dec 21, 2016 at 4:27

answered Dec 21, 2016 at 4:05

Mulan

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