The 'in' operator is unable to find 'colour' within true (function return type)

Here's the TypeScript code I'm working with:

let a: unknown = true;

if(hasColour(a)) {
    console.log(a.colour);  // Using a.colour after confirming a has the colour property
}

I've created a function to check if the color property exists:

function hasColour (obj: any) : boolean {
    return !!obj &&
    "colour" in obj &&
    typeof obj == "object" 

}  

An error is popping up saying:

Cannot use 'in' operator to search for 'colour' in true.

To fix this, instead of specifying boolean as the return type, specify: obj is { colour: string } like so:

function hasColour (obj: any) : obj is { colour: string } {
    return !!obj &&
    "colour" in obj &&
    typeof obj == "object" 

}  

Why does this work? Shouldn't hasColour return a true/false? Why does it work with obj is { colour: string } as a return type?

Answer №1

Explaining the distinction between...

function hasColour (obj: any) : boolean {
    return !!obj &&
    typeof obj == "object" &&
    "colour" in obj
}  

...and...

function hasColour (obj: any) : obj is { colour: string } {
    return !!obj &&
    typeof obj == "object" &&
    "colour" in obj
}

...lies in the fact that the second version functions as a type guard. By employing a type guard, your unknown type will be narrowed down to { colour: string }, and within the if statement where the type guard is used, the compiler will ensure the presence of the colour property.

On the other hand, the first version simply returns a boolean. The compiler cannot infer anything from it, as it lacks information regarding the type check being performed within the function. With a type guard, you explicitly instruct the compiler to make an assertion about the type.

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