Add an option not to use implicit any when I create an empty container (for example [], new Set() or new Map())

[DonaldDuck313]

🔍 Search Terms

implicit any containers

✅ Viability Checklist

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, new syntax sugar for JS, etc.)
• This isn't a request to add a new utility type: https://github.com/microsoft/TypeScript/wiki/No-New-Utility-Types
• This feature would agree with the rest of our Design Goals: https://github.com/Microsoft/TypeScript/wiki/TypeScript-Design-Goals

⭐ Suggestion

Currently, new Map() has type Map<any, any> and new Set() has type Set<any>. I suggest to add an option so that instead new Map() has type Map<never, never> and new Set() has type Set<never>. This should work because Set<never> is assignable to any other set type, and same for maps.

Note that this needs to be an option, otherwise it could break existing code.

📃 Motivating Example

I ran into this issue with code similar to the following:

declare function returnsNullableMap(): Map<number, number> | null;

const myMap = returnsNullableMap() ?? new Map();
myMap.set("", "");    //This compiles but shouldn't

The last line is obviously a mistake since the map contains numbers, not strings. But since new Map() has type Map<any, any>, so does myMap, and the last line compiles.

💻 Use Cases

1. What do you want to use this for?

I would like to completely avoid the any type in my code, and the fact that the code above actually does contain any is not immediately obvious. I use existing options such as noImplicitAny and useUnknownInCatchVariables to avoid any sneaking in at other places, but apparently any can still sneak in in this case. By the way, I would have expected noImplicitAny to cover this case as well, but apparently it doesn't.

2. What workarounds are you using in the meantime?

In the specific code above I can use const myMap = returnsNullableMap() ?? new Map<never, never>(); and it works fine, myMap has the correct type (Map<number, number>). Note that in my actual code my types are much more complicated than just number, so using never saves a lot of typing compared to using the actual type. In this simplified example however new Map<number, number>() would have worked just as well.

3. What shortcomings exist with current approaches?

The main issue is that if I forget the <never, never> (or a similar workaround), my code will contain the any type without making it immediately obvious, allowing mistakes like in the example above. Another minor annoyance is that new Map<never, never>() is more verbose than just new Map(), but I can live with that, what I'm more concerned about is any types sneaking into my code where I don't want them.

[RyanCavanaugh]

This is already an option.

Currently if I do let a = [];, a will be of type any[].

It's an evolving array, not really an any[]. You will get an error under noImplicitAny if you do any access that actually produces an any value, e.g.

function foo(arr: unknown[]) { }

let a = [];
foo(a); // Implicit any error

If you want this behavior in Set or Map, write a new overload in a declaration merge:

export { }
declare global {
    interface SetConstructor {
        new(): Set<never>;
    }
    interface MapConstructor {
        new(): Map<never, never>;
    }
}

const m = new Set(); // m: Set<never>

[DonaldDuck313]

You're right, it already works as it should for arrays. Your workaround for sets and maps is better than mine, but it would still be nice to have a tsconfig.json option for that.

[typescript-bot]

This issue has been marked as "Question" and has seen no recent activity. It has been automatically closed for house-keeping purposes.

[DonaldDuck313]

It's not a question though, it's a feature request.