When an object in Typescript is clearly a function, it throws a 'cannot invoke' error

Check out this TypeScript code snippet

Take a look here

type mutable<A,B> = {
    mutate: (x : A) => B
}

type maybeMutable<A,B> = {
    mutate? : (x : A) => B; 
}

const myFunction = function<A,B>(config : A extends B ? maybeMutable<A,B> : mutable<A,B>, argument : A){
    let mutate; 
    if ('mutate' in config) {
        mutate = config.mutate; 
    } else {
        mutate = (x : A) => x 
    }
    mutate(argument); 
}

The configuration might have the property mutate or not, depending on the types A and B. If it is present, it must be a function. The code then checks if mutate is in the config, assigns it if so, or sets the default value otherwise. This default value is the identity function once more. How does TypeScript infer that mutate can be undefined and thus throws an error

Cannot invoke an object which is possibly 'undefined'
?

Answer №1

There seems to be an issue with the `if` statement as it is not properly handling the case where the value is `undefined`.

mutate?: (x : A) => A;

This problem includes scenarios such as having `mutate: undefined` and the `in` operator returning `true` even when the object or its prototype chain is `undefined`.

It's important to note that a property in an object can exist but have a value of `undefined`. Hence, using x in obj does not equate to obj.x === undefined.

Visit this link for more information about the 'in' operator in JavaScript.

To address this issue, you can implement the following fix:

type mutable<A,B> = {
    mutate: (x : A) => B
}

type maybeMutable<A,B> = {
    mutate?: (x : A) => A; 
}

const  myFunction = function<A,B>(config : A extends B ? maybeMutable<A,B> : mutable<A,B>, argument : A){
    let mutate; 
    if (typeof config['mutate'] === 'function') {
        mutate = config.mutate; 
    } else {
        mutate = (x : A) => x 
    }
    mutate(argument); 
}

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