CODEWITHSUNDEEP
javascriptarrays
acostela
acostela

Reputation: 2727

Comparison array number in JS

I'm working with JS and I don't understand certain behavior with arrays. I searched but I can't find a correct answer, so sorry if this is dupe or in relation with another question.

I have the following code using lodash.

    return _.difference(self.list1, self.list2) <= 0;

That is returning an array and I'm comparing it directly with a number, just because I forgot the .length property. I saw that this is "working" unless it is not correct. So I started to do some tests with the JS console and I don't understand what is happening here.

[Object, Object, Object] <= 0 //returns false

[] <= 0 //returns true

[[]] <= 0 //returns true

[[[]]] <= 0 //returns true

[[2]] <= 0 // returns false

[[],[]] <= 0 //returns false

What is JS doing here? Thank you very much.

Upvotes: 0

Views: 119

Answers (1)

T.J. Crowder
T.J. Crowder

Reputation: 1074048

The <= operator will coerce its operands to try to make them comparable. The full details are in the specification's Abstract Relational Comparison operation, but basically:

When you compare an object reference with a primitive, it tries to convert the referenced object to a primitive. In this case, since the primitive is a number, it will try to get a primitive number from the object if the object supports that (arrays don't) and then fall back on getting a string. At which point it has a string and a number, so it coerces the string to number and compares them numerically.

When you convert an array to a string, it does the same thing join does: Joins the string version of all the entries together separated by commas. Of course, if you have only one entry, you'll end up with just that entry (as a string).

Here's a simpler example:

var a = [2];
console.log(a < 10); // true

The steps there are:

  1. Since a contains an object reference, convert that object to a primitive, preferring a number to a string if possible:

    • Arrays don't provide a way of converting to number, so we end up doing (effectively) a.toString()
    • a.toString() calls a.join(), which joins the entries together as strings with a comma in between
    • So we end up with "2".


    Now we have "2" < 10

  2. Since one of those is a number, the operator coerces the other one to number, giving us 2 < 10

  3. The result of 2 < 10 is true

This strategy sometimes, even frequently, means the <= operator ends up comparing NaN to something. For instance, if a were a non-array object:

var a = {};
console.log(a.toString()) // "[object Object]"
console.log(a < 10);      // false

The steps would be:

  1. Since a contains an object reference, convert that object to a primitive, preferring a number to a string if possible:

    • Plain objects don't support conversion to number, so we end up doing (effectively) a.toString().

    • As you can see in the snippet above, that gives us "[object Object]"


    Now we have "[object Object]" < 10

  2. Since one of those is a number, the operator coerces the other one to number. "[object Object]" coerced to a number is NaN.

  3. The result of NaN < 10 is false, because all comparisons involving NaN (including ===) result in false. (Yes, really; NaN === NaN is false.)

(Thanks Rajesh for suggesting the {} < 10 example and providing the initial version of the above.)

Upvotes: 7

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