Prevent javascript from going out of an array

Question

Is there any way I can prevent javascript from dropping an error if I try to go into a non existing array index?

Example: array[-1] would return error and eventually break all my code. How can I let it just return 'undefined' and let my script go on? I can implement an if statement before checking the array (so that if the index is minor than zero or major than the array size it would skip it) but this would be very tedious!

this is my code:

if (grid[j-1][i])
n++;
if (grid[j+1][i])
n++;
if (grid[j][i+1])
n++;
if (grid[j][i-1])
n++;
if (grid[j-1][i-1])
n++;
if (grid[j+1][i+1])
n++;
if (grid[j-1][i+1])
n++;
if (grid[j+1][i-1])
n++;

It is inside of two loops which both sees J and I starting from zero. I don't want to change them and neither writing another if statement (as you can see, there are already too much of them!). Is there any solution?

Thanks!

You do know you can write all those ifs with one if, right? — gdoron
– gdoron, Commented Apr 22, 2012 at 1:32
@gdoron - I don't think it could (at least not neatly), since more than one rule could apply, and with an if, n would only be incremented once. A loop would work, but then you'd end up with 4 nested loops. — Karl Nicoll
– Karl Nicoll, Commented Apr 22, 2012 at 1:35
@KarlNicoll how bout n+= !!grid[j-1][i] + !!grid[j+1][i] + ... — d_inevitable
– d_inevitable, Commented Apr 22, 2012 at 1:37
@d_inevitable - Nice, but I wouldn't exactly call it "maintainer friendly" :P — Karl Nicoll
– Karl Nicoll, Commented Apr 22, 2012 at 1:42
@Saturnix !!x "casts" a value to a boolean. It's a double negation. For a single negation see e.g. 11heavens.com/falsy-and-truthy-in-javascript — Kijewski
– Kijewski, Commented Apr 22, 2012 at 1:46

Kijewski · Accepted Answer · 2012-04-22 02:45:41Z

1

If you know the measures of your grid, you can put "sentinel cells" around it.

If you add a -1st index to an array x, it does not count to x.length. Putting an additional last element into the list would increment x.length.

I daresay using sentinel cells combined with the arithmetic counting algorithms mentioned by d_inevitable would be the fastest solution, since it would not involve branches. You even can omit the !! because true will evaluate to 1 and false to 0 in an equalization.

Update:

Do not use index -1. Its an awful lot slower that normal array indexes. See http://jsperf.com/index-1.

edited Apr 22, 2012 at 2:45

answered Apr 22, 2012 at 1:59

Kijewski

26.1k14 gold badges108 silver badges149 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

mystery · Accepted Answer · 2012-04-22 01:33:00Z

1

You could use ||, which muffles errors, e.g.:

(grid[j-1] || [])[i] || false

(I haven't tested this, but it should work)

Edit: updated based on am not i am's suggestion

answered Apr 22, 2012 at 1:33

mystery

19.6k4 gold badges45 silver badges65 bronze badges

3 Comments

user1106925 Over a year ago

I think the idea is that grid[-1][i] will throw an error because you can't access [i] on undefined. The || won't help with that.

user1106925 Over a year ago

...you'd need to do something more like this... (grid[-1] || [])[i]

user1106925 Over a year ago

Yep, except that you should be able to get rid of the || false, since when the empty array is used, the [i] will give undefined.

QuantumBlack · Accepted Answer · 2012-04-22 01:36:32Z

0

A less tedious way while still using ifs would be checking the first index if it's defined:

if (typeof grid[j-1] != "undefined" && grid[j-1][i])

answered Apr 22, 2012 at 1:36

QuantumBlack

1,5561 gold badge13 silver badges28 bronze badges

Comments

Niet the Dark Absol · Accepted Answer · 2012-04-22 01:38:39Z

0

You could create a function to do the checks:

function getArrayValue(arr,key) {
    if( key < 0 || key >= arr.length) return null;
    return arr[key];
}

But really you should be avoiding out-of-bounds keys anyway.

answered Apr 22, 2012 at 1:38

Niet the Dark Absol

326k86 gold badges480 silver badges604 bronze badges

Comments

d_inevitable · Accepted Answer · 2012-04-22 04:41:05Z

0

I would do this:

for(m = Math.max(j-1,0) ; m <= Math.min(j+1,grid.length-1) ; m++)
  for (p = Math.max(i-1,0) ; p <= Math.min(i+1, grid[m].length-1) ; p++)
     n += !(m == j && p == i) && !!grid[m][p];

edited Apr 22, 2012 at 4:41

answered Apr 22, 2012 at 1:48

d_inevitable

4,4612 gold badges31 silver badges49 bronze badges

3 Comments

user1106925 Over a year ago

Is it your intention to use n as the grid[m] index as well as the main n += ... counter? I'm not understanding it.

d_inevitable Over a year ago

@amnotiam, ok it should make sense more now. Thanks for pointing it out.

user1106925 Over a year ago

One more thing, I think you have your min and max reversed.

Mark Reed · Accepted Answer · 2012-04-23 17:50:33Z

0

How about this for your solution?

for (dj = -1; dj <= 1; ++dj) {
  for (di = -1; di <= 1; ++di) {
    if ((dj || di) && grid[j+dj] && grid[j+dj][i+di]) {
        n++;
    }
  }
}

If you refactor all those ifs into a single loop like the above, then having to do the extra conditional is not so bad.

edited Apr 23, 2012 at 17:50

answered Apr 22, 2012 at 1:45

Mark Reed

96k17 gold badges149 silver badges189 bronze badges

1 Comment

user1106925 Over a year ago

Why was this -1? I see nothing wrong with it. @downvoter: Please explain why MarkReed's answer is wrong.

Collectives™ on Stack Overflow

Prevent javascript from going out of an array

6 Answers 6

Update:

Comments

3 Comments

Comments

Comments

3 Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

Update:

Comments

3 Comments

Comments

Comments

3 Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related