Why does javascript's "in" operator return true when testing if 0 exists in an array that doesn't contain 0?

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Why does the "in" operator in Javascript return true when testing if "0" exists in array, even when the array doesn't appear to contain "0"?

For example, this returns true, and makes sense:

var x = [1,2];
1 in x; // true

This returns false, and makes sense:

var x = [1,2];
3 in x; // false

However this returns true, and I don't understand why:

var x = [1,2];
0 in x;

It refers to the index or key, not the value. 0 and 1 are the valid indices for that array. There are also valid keys, including "length" and "toSource". Try 2 in x. That will be false (since JavaScript arrays are 0-indexed).

See the MDN documentation.

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The in operator doesn't do what you're thinking it does. The in operator returns true if the specified operand is a property of the object. For arrays, it returns true if the operand is a valid index (which makes sense if think of arrays as a special-case object where the properties are simply named 0, 1, 2, ...)

For example, try this:

javascript:var x=[1,4,6]; alert(2 in x);

It'll also return true, because "2" is a valid index into the array. In the same way, "0" is an index into the array, so also returns true.

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Javascript's in operator does not check if a value is contained in an array. It checks if the object has a property or index. So var x = [4,5]; 4 in x; //false 1 in x; //true.

Because length is a property of x, "length" in x; //true

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Modern browsers, except IE, support a couple methods that can find a value in an array.

indexOf and lastIndexOf return the first(or last) index of an exact match of their argument in an array, or -1, if no matching element was found.

if(A.indexOf(0)!= -1){
    // the array contains an element with the value 0.
}

You can add one or both methods to IE and older browsers-

if(![].indexOf){
    Array.prototype.indexOf= function(what, i){
        i= i || 0;
        var L= this.length;
        while(i< L){
            if(this[i]=== what) return i;
            ++i;
        }
        return -1;
    }
    Array.prototype.lastIndexOf= function(what, i){
        var L= this.length;
        i= i || L-1;
        if(isNaN(i) || i>= L) i= L-1;
        else if(i< 0) i += L;
        while(i> -1){
            if(this[i]=== what) return i;
            --i;
        }
        return -1;
    }
}

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I guess you use Python before, in JS, use Array.prototype.includes

let x = [1, 2]
x.includes(1) // true

in operator check the indices of the array not the value

0 in [1, 2] // true
2 in [1, 2] // false

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Comments
  • Thank you providing a workaround to the problem rather than just an explanation as to why it doesn't work.