Error in Typescript persists even after short-circuit evaluation is used

Kindly review the provided code sample

type type1 = string[] | undefined;

let variable1 : type1 = undefined;

console.log(variable1 && variable1.length);

Upon attempting to run this code in Typescript Playground, an error is generated stating Property 'length' does not exist on type 'never'.

The question arises as to why TypeScript is throwing an error despite handling the possibility of variable1 being falsy using && short-circuit evaluation.

An alternative attempt involving ?. option chaining was also made, however, the outcome remains unchanged from the original code snippet.

The expectation here is for TypeScript to not display any errors. The inclusion of undefined as a potential value for variable1 is crucial, as there are checks based on its type within my application logic.

Answer №1

The reason for this is that TypeScript is already aware of the fact that you intend to assign undefined to the variable variable1.

To prevent TypeScript from automatically detecting the type being assigned to variable1, you can use the following approach:

type customType = string[] | undefined;

let variable1 = undefined as customType;

console.log(variable1 && variable1.length);

Alternatively, you can indirectly reference the customType by using a function like this, making it harder for TypeScript to identify the type immediately:

type customType = string[] | undefined;

function retrieveCustomType(): customType {
    return undefined;
}

let variable1 = retrieveCustomType();

console.log(variable1 && variable1.length);

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