CODEWITHSUNDEEP
javascriptarrays
ocomfd
ocomfd

Reputation: 4020

Why is Object.keys([,1,,1]) different from Object.keys([,1,undefined,1])?

I was shocked that the result of

document.writeln(Object.keys([,1,,1]));

is

1,3

which filters the index of an array that doesn't contains anything automatically. But when I try to print out Object.keys([,1,undefined,1]):

document.writeln(Object.keys([,1,undefined,1]));

the result is:

1,2,3

which is different from the previous one, and I try to print the value of index 2 at the first one:

document.writeln([,1,,1][2]);

which is undefined actually. Why would Object.keys([,1,,1]) and Object.keys([,1,undefined,1]) return different results?

Upvotes: 3

Views: 88

Answers (3)

EKW
EKW

Reputation: 2113

You're right to think that it doesn't make much sense. It doesn't. Basically, in javascript there is sometimes a difference between "undefined" and "has never been defined". In the first example, you're not giving the second index any value, so it doesn't print. In the second example, you're giving it a value (even though that value is undefined, it's still getting a value) so it prints it.

Upvotes: 0

Pac0
Pac0

Reputation: 23149

This is one caveat to be careful with Javascript arrays : you can have sparse arrays, arrays with missing values, which is not the same as having undefined values (the array element exists, like a variable can be declared, but it's value is `undefined).

Note that you can get the same behavior using delete on an array element.

As already said, you can test for empty value wiht === undefined, which is even more confusing, because it does not usually cause the same output.

Upvotes: 0

thefourtheye
thefourtheye

Reputation: 239463

In Object.keys([, 1, undefined, 1]) case, you are explicitly defining the third element to be undefined, but in Object.keys([, 1, , 1]), there is no element associated with the index 2. So, 2 is recognized as an array index in the first case, but not in the second case.

Just to represent the holes in the array, when you actually reference it, undefined is returned. That is why [, 1, , 1][2] is undefined.

Upvotes: 3

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