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HomeWeb Front-endJS TutorialWhy Does an Empty Array Seem Both True and False in JavaScript?

Why Does an Empty Array Seem Both True and False in JavaScript?

Unveiling the Enigmatic Nature of Empty Arrays in JavaScript

Empty arrays in JavaScript exhibit a peculiar duality, appearing to be simultaneously true and false. This perplexing behavior stems from the implicit conversion performed by equality operators. To delve into the underlying mechanisms, let's explore the code snippet below:

<br>var arr = [];<br>console.log('Array:', arr);<br>if (arr) console.log("It's true!");<br>if (arr == false) console.log("It's false!");<br>if (arr && arr == false) console.log("...what??");<br>

The first if statement verifies the presence of the arr object (Array is an instance of Object), returning true. However, the if statement with arr == false compares the value of the arr object with the primitive value of false. To perform this comparison, arr.toString() is implicitly invoked, returning an empty string "".

This outcome is explained by the behavior of Array.join(). When toString() is called on an array, it returns the concatenated result of invoking join() on that array. Since the array is empty, the result is an empty string, which falls under the category of falsy values in JavaScript. Thus, arr.toString() evaluates to false, resulting in if (arr == false) returning true.

The peculiar case of if (arr && arr == false) highlights the operation of the logical AND (&&) operator. For the entire statement to be true, both conditions must evaluate to truthy values. arr by itself is truthy, but arr == false evaluates to true due to the aforementioned implicit conversion. Therefore, the conjunction evaluates to true, leading to the seemingly contradictory "It's false!" output.

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