I found this strange behavior in javascript.
var v1 = [1];
var v2 = [1];
v1 == v2 // false
v1 == 1 //true
[1] == [1] // false
1 == [1] // true
why is it that [1] == [1] returns falseand [1] == 1 returns true?
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Dyrandz Famador
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Arrays are objects, so [1] == [1] is comparing two separate objects. I'm unsure why [1] == 1. – Chris Jul 04 '15 at 03:53
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http://stackoverflow.com/q/359494/720164 – Jonathan Allard Jul 04 '15 at 03:54
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Forget about `==` in javascript. Use only `===` – Kirill Slatin Jul 04 '15 at 04:00
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1`==` is basically evil in Javascript. It is a good practice to avoid it nearly always unless you explicitly want a type conversion and you fully understand what it will do in all cases that your code will encounter. – jfriend00 Jul 04 '15 at 06:16
1 Answers
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The spec says that if the two operands of == have the same type as each other (such as in the [1] == [1] case, where they both are type Object), then == behaves exactly like ===. The two arrays are not the exact same object, so false is returned. Notice that:
var v1 = [1];
var v2 = v1;
v1 == v2; // true
When the operands have different types, they are both coerced. In the case of 1 == [1] Rule 10 from the link above applies first and the array is converted to a primitive, by its toString() which returns '1'. Then rule 6 applies (converting the string '1' to the number 1), and the comparison becomes 1 == 1, and finally they have the same type and are compared with ===. Obviously 1 === 1 evaluates to true.
Paul
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1@DyrandzFamador Yes, you should almost always use `===`. The `==` variant should be used carefully and always with a comment describing the reason for the implicit cast. It's almost always better to use `===` and explicitly convert between types before comparing eg. using `+str` or `parseInt( str, 10 )` to convert a string to a number. – Paul Jul 05 '15 at 04:11