Why Doesn't the TypeScript Compiler Detect Missing Properties When Using Object.create(null)?

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Introduction

No Inherited Properties: The object created has no properties from the default JavaScript object.

Creates Pure Objects: These objects are often used to create data structures without any risk of colliding with inherited properties.

Analyzing the TypeScript Compiler Behavior

TypeScript Compilation Process

When examining how TypeScript compiles the code:

Creates an object with a specified prototype (which can be null).

Optionally accepts property descriptors.

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Return Type of any: Both overloads return an any type when called with null. This means that TypeScript cannot infer any specific structure or properties when you create an object this way, resulting in a lack of type-checking.

Why No Errors Occur?

Dynamic Nature of any: Since the returned type is any, TypeScript cannot provide proper validation for the properties you're trying to set later. So when you assign rv.b = 'str';, TypeScript does not flag the absence of rv.a = '...'.

Contrast with Object Literals: In contrast, using an empty object literal (const rv = {};) allows TypeScript to derive a more definite type, leading to errors for missing properties.

Practical Implications

The implications of this behavior can lead to bugs if developers assume that TypeScript is providing full type safety. To ensure type checking:

Use Object Literals: Opt for standard JS objects ({}) when you expect TypeScript to enforce property checks.

Conclusion

By recognizing when to use each approach, you can utilize TypeScript's strengths in type checking, ultimately leading to more robust and error-free code.