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unsafe usage of optional chaining. if it short-circuits with 'undefined' the evaluation will throw typeerror no-unsafe-optional-chaining

Gavin Dennett answered on May 14, 2024 Popularity 2/10 Helpfulness 1/10

answer unsafe usage of optional chaining. if it short-circuits with 'undefined' the evaluation will throw typeerror no-unsafe-optional-chaining

unsafe usage of optional chaining. if it short-circuits with 'undefined' the evaluation will throw typeerror no-unsafe-optional-chaining

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The error message "unsafe usage of optional chaining" indicates that you're using the optional chaining operator (?.) in a way that could potentially lead to a TypeError.

Optional chaining is a feature in JavaScript that allows you to access properties or methods of an object without causing an error if the object is null or undefined. However, if the expression short-circuits with 'undefined', the evaluation will throw a TypeError.

To fix this, youll need to make sure that the object you're trying to access is not null or undefined before using optional chaining. You can do this by adding a null check or using the typeof operator to verify the object's existence.

Here's an example of how you can fix this issue:

Instead of:
const value = obj?.property.subProperty;

Use:
const value = obj !== null && obj !== undefined ? obj.property.subProperty : undefined;

Or:
const value = typeof obj !== 'undefined' && obj !== null ? obj.property.subProperty : undefined;

By adding these checks, you ensure that the object exists before attempting to access its properties, avoiding the unsafe usage of optional chaining and the potential TypeError.

Popularity 2/10 Helpfulness 1/10 Language typescript

Source: Grepper

Tags: evaluation throw typeerror typescript unsafe

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Contributed on May 14 2024

Gavin Dennett

0 Answers Avg Quality 2/10

unsafe usage of optional chaining. if it short-circuits with 'undefined' the evaluation will throw typeerror no-unsafe-optional-chaining

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