Why Does TypeScript Throw an Error When Using `Array.prototype.includes()` with Different Types?

Why Array.prototype.includes() Requires Matching Types for Array Elements

The Array.prototype.includes() method aims to determine if an array contains a specific element. To ensure accurate and reliable results, the searchElement parameter must match the type of the array elements. This type safety prevents erroneous comparisons and maintains the integrity of the array's data integrity.

Consider the following code snippet:

<code class="js">type AllowedChars = 'x' | 'y' | 'z';
const exampleArr: AllowedChars[] = ['x', 'y', 'z'];

function checkKey(e: KeyboardEvent) { 
    if (exampleArr.includes(e.key)) { 
        // ...
    } 
}</code>

The TypeScript compiler flags an error, indicating that the e.key parameter does not match the AllowedChars type of the exampleArr elements. This error is because the includes() method assumes that the searchElement has the same type as the array elements.

It may seem contradictory since Array.prototype.includes() returns a boolean value rather than an assignment, as the error suggests. However, typescript's type system plays a crucial role here.

Including an element in the array implies that the element's type is consistent with the array's elements. However, in the example provided, the e.key parameter's type is string, while the exampleArr elements are of the AllowedChars type. This mismatch requires type assertions or type widening to override the TypeScript type system's expectations.

The simplest and most recommended approach to resolve this issue is to widen the type of the exampleArr to the broader type of string[], or to narrow the searchElement parameter to only include allowed chars. These modifications ensure type compatibility, preventing any potential type-related errors or inconsistencies.

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