Determining the presence of a hash key in JavaScript without utilizing hasOwnProperty

Question

Determining the presence of a hash key in JavaScript without utilizing hasOwnProperty

Consider this sample data:

let numbers = [5,8,4,3,9,7]
let tracked = {}

What is the distinction between these two methods of verifying if a key exists? Are there any potential challenges that may arise with the first method?

numbers.filter(function(num) {
    return tracked[num] ? false : (tracked[num] = true);
});

vs.

numbers.filter(function(num) {
    return tracked.hasOwnProperty(num) ? false : (tracked[num] = true);
});

javascript

Answer 1

Answer №1

Firstly, it's important to note that both approaches mentioned in the question are incorrect because they do not utilize the return value of filter which is a new array with filtered elements. To address this issue, consider using forEach or for-of on modern systems. However, it's possible that you are already incorporating this and it just wasn't explicitly stated in your question.

Focusing on your primary inquiry:

The initial method may mistakenly assume an element was not found if it evaluates to a falsy value. Falsy values include 0, "", NaN, null, undefined, and false. On the other hand, the second approach effectively handles these cases.

Another distinction between the two comparisons is that the first one retrieves the property regardless of its placement in the prototype chain, while the second one only considers properties within the object itself (usually preferable). This distinction becomes significant for properties derived from Object.prototype like valueOf.

You could also explore options such as Set or Map.

Answer 2

Firstly, it's important to note that both approaches mentioned in the question are incorrect because they do not utilize the return value of filter which is a new array with filtered elements. To address this issue, consider using forEach or for-of on modern systems. However, it's possible that you are already incorporating this and it just wasn't explicitly stated in your question.

Focusing on your primary inquiry:

The initial method may mistakenly assume an element was not found if it evaluates to a falsy value. Falsy values include 0, "", NaN, null, undefined, and false. On the other hand, the second approach effectively handles these cases.

Another distinction between the two comparisons is that the first one retrieves the property regardless of its placement in the prototype chain, while the second one only considers properties within the object itself (usually preferable). This distinction becomes significant for properties derived from Object.prototype like valueOf.

You could also explore options such as Set or Map.

Answer 3

Answer №2

Here's a brief answer: In this particular case, not using Object.hasOwnProperty() does not pose any issues. You can find more information on this topic on the [wiki].

When could it become problematic?

let seen = {}

console.log(seen["constructor"]) // function Object() { [native code] }

As illustrated above, the use of `seen["constructor"]` will include other strings for inherited properties such as the `constructor` of an Object. However, since you are working with numbers only in this specific case, there are no inherited properties to worry about.

Answer 4

Here's a brief answer: In this particular case, not using Object.hasOwnProperty() does not pose any issues. You can find more information on this topic on the [wiki].

When could it become problematic?

let seen = {}

console.log(seen["constructor"]) // function Object() { [native code] }

As illustrated above, the use of `seen["constructor"]` will include other strings for inherited properties such as the `constructor` of an Object. However, since you are working with numbers only in this specific case, there are no inherited properties to worry about.

Answer 5

Answer №3

Adding on to TJ's point.

The hasOwnProperty method does not check the prototype of the object, whereas bracket or dot notation will. Take a look at this example:

Object.prototype.foo = 1;
var baz = {"bar": 1}

if (baz.hasOwnProperty('foo')) {
    console.log('has foo');
    return true
}

if (baz.bar) {
    console.log('has bar'); // has bar
    return true;
}

Answer 6

Adding on to TJ's point.

The hasOwnProperty method does not check the prototype of the object, whereas bracket or dot notation will. Take a look at this example:

Object.prototype.foo = 1;
var baz = {"bar": 1}

if (baz.hasOwnProperty('foo')) {
    console.log('has foo');
    return true
}

if (baz.bar) {
    console.log('has bar'); // has bar
    return true;
}

Determining the presence of a hash key in JavaScript without utilizing hasOwnProperty

Answer №1

Answer №2

Answer №3

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