when does the no pointer variable disappear in javascript?

Question

I found some issue with .length = 0 and now if I change to = [], all things are working well in my javascript project.

var arr = [1,2,3,4,5];
alert(arr);
arr = [];
alert(arr);
arr.length = 0;
alert(arr);

But I have one question if I use arr = [], when the memory of [1,2,3,4,5] disappear? If the array is big and I use [] lots of times it will cause memory lack problems.

You can see my stack problem stackoverflow.com/questions/45924788/… here I changed .length =0 to = [] and solved the problem — artgb
– artgb, Commented Aug 30, 2017 at 15:57

T.J. Crowder · Accepted Answer · 2017-08-30 15:55:10Z

0

But I have one question if I use arr = [], when the memory of [1,2,3,4,5] disappear?

Whenever the JavaScript engine gets around to releasing it, which it will do as and when necessary once nothing has a reference to it anymore. The details vary depending not only on the JavaScript engine involved, but on where that code appears and how frequently it's run (e.g., how aggressively it ends up getting optimized). If it matters (e.g., that's run a lot), it'll get reclaimed really quickly.

answered Aug 30, 2017 at 15:55

T.J. Crowder

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7 Comments

artgb Over a year ago

I want to you to describe why this post is available stackoverflow.com/questions/4804235/… according to your answer.

T.J. Crowder Over a year ago

@artgb: I don't understand the question in your comment...?

artgb Over a year ago

You said if I use arr = [] the memory space will disappear automatically and why .length = 0 is needed?

T.J. Crowder Over a year ago

@artgb: If you've done arr = [], you don't need arr.length = 0.

artgb Over a year ago

Thanks, I understood your idea just now

|

ballenf · Accepted Answer · 2017-08-30 15:58:15Z

0

Your question comes down to how the various Javascript engines perform garbage collection. While they each do it slightly differently, the key to allowing them cleanup allocated memory is to make sure there are no lingering references (including through a closure or async call) remaining to the original array.

Gets trickier if that array has been passed into another function that is still on the stack or itself in a closure.

answered Aug 30, 2017 at 15:58

ballenf

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