detect "null"

[ulrik]

How can I detect "null"?
I have a variable holding a number, and in some cases it will return as a "null" and I want to detect when that happens, so how can I

if (variable == bla...)
    do something else...

I tried "undefined", "isDefined(), and it seems that "null" is defined, so that will not work.

I tried

Console.print(typeof var);

which results in "number"

[ulrik]

@ulrik this will not work either

if (r === null)

[d.healey]

@ulrik What about preventing it being null?

[ulrik]

@ulrik Ok I solved it in this case, this worked

if (variable == "inf")
    break;

[ustk]

@d-healey I often use undefined or null as intentional state or value for variables (especially for object properties) so I can discriminate if a value is ready to read or not. So when I want to reset it I set it back to undefined or null
Let’s say it my -1

EDIT: I realise that I should more often use null rather than undefined

[d.healey]

@ulrik said in detect "null":

@ulrik this will not work either

if (r === null)

It works for me. Try this.

const t = null;

Console.print(t == void);
Console.print(t == undefined);
Console.print(t == null);

Console.print(t === void);
Console.print(t === undefined);
Console.print(t === null);

[parabuh]

@ulrik you can use shorthand

if (!var) {
}

[d.healey]

@parabuh That won't tell you is the variable is null

[ustk]

@d-healey Hmmm... I just discovered that a value of 0 == undefined Back to school for me!

[ulrik]

@parabuh that will not work

[d.healey]

@ustk I never use var == undefined I always use isDefined(var)

I just noticed this guy commented on my video (I think I might re-record that one)

[ulrik]

@d-healey

if I declare a variable like this

const t = null;

First I'll get

Console.print(t);
API call with undefined parameter 0

Console.print(t == void);
result: 1

Console.print(t == undefined);
result: 1

Console.print(t == null);
result: 1

Console.print(t === void);
result: 0

Console.print(t === undefined);
result: 0

Console.print(t === null);
result: 1

but my variable show up as null, and if I use

Console.print(myvariable === null);
result: 0

Are there different kind of "null"??

[d.healey]

@ulrik Are you sure your variable is null?

[ulrik]

@d-healey that's what it's saying if I Console.print it
my var is inside an array and if trace log it will say something like this:

Interface: [
  3,
  null
]

[d.healey]

@ulrik I think you need to give us an example we can test.

This works for me:

const a = [3, null];

Console.print(a[1] === null);

So you must be doing something different.

[ulrik]

@ulrik Ok, I Console.print each member of the array and got this instead:

Interface: inf

so if trace logged it show up as "null" and if logged independently it show up as "inf"

[ulrik]

@ulrik @d-healey so what does "inf" stand for, infinite?

[d.healey]

@ulrik No idea. I'd recommend you try to always populate your variables yourself so you know what it contains.

[Christoph Hart]

Yes, inf is infinite:

Console.print(1.0 / 0.0)

But the question is: how did you get null in the first place? I honestly didn't know that existed, we've got undefined for a uninitialized variable, void for a invalid array or object field and inf and nan for illegal number values.

var x = 1 / 0.0;
x = 0.0 * x;

var a = [1];
var n;

Console.print(typeof(a[9])); // => void
Console.print(typeof(b)); // => undefined
Console.print(typeof(1.0 / 0.0)); // => number
Console.print(typeof(n)); // => undefined
Console.print(x); // => nan

[ulrik]

@Christoph-Hart ok thanks!

As I said here:

@ulrik Ok, I Console.print each member of the array and got this instead:

Interface: inf

so if trace logged it show up as "null" and if logged independently it show up as "inf"

It seems the

Console.print(trace(array))

is not showing correct information?