TypeScript: Validate mapped types and const context

Stefan Baumgartner

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Mapped types are great, as they allow for the flexibility in object structures JavaScript is known for. But they have some crucial implications on the type system. Take this example:

type Messages = 

type ChannelDefinition = {
[key: string]: {
open: Messages,
close: Messages,
fail: Messages

This is from a generic messaging library, that takes a “channel definition” where multiple channel tokens can be defined. The keys from this channel definition object are what the user wants it to be. So this is a valid channel definition:

const impl: ChannelDefinition = {
test: {
message: {

We have a problem when we want to access the keys we defined so flexibly. Let’s say we have a function that opens a channel. We pass the whole channel definition object, as well as the channel we want to open.

declare function openChannel(
def: ChannelDefinition,
channel: keyof ChannelDefinition

So what are the keys of ChannelDefinition? Well, it’s every key: [key: string]. So the moment we assign a specific type, TypeScript treats impl as this specific type, ignoring the actual implementation. The contract is fulfilled. Moving on. This allows for wrong keys to be passed:

// Passes, even though "massages" is no part of impl
openChannel(impl, 'massages')

So we are more interested in the actualy implementation, not the type we assing to our constant. This means we have to get rid of the ChannelDefinition type and make sure we care about the actual type of the object.

First, the openChannel function should take any object that is a subtype of ChannelDefinition, but work with the concrete subtype:

- declare function openChannel(
- def: ChannelDefinition,
- channel: keyof ChannelDefinition
- )
+ declare function openChannel<T extends ChannelDefinition>(
+ def: T,
+ channel: keyof T
+ )

TypeScript now works on two levels:

  1. Checking if T actually extends ChannelDefinition. If so, we work with type T
  2. All our function parameters are typed with the generic T. This also means we get the real keys of T through keyof T.

To benefit from that, we have to get rid of the type definition for impl. The explicit type definition overrides all actual types. From the moment we explicitly specify the type, TypeScript treats it as ChannelDefinition, not the actual underlying subtype. We also have to set const context, so we can convert all strings to their unit type (and thus be compliant with Messages):

- const impl: ChannelDefinition = { ... };
+ const impl: { ... } as const;

Without const context, the inferred type of implis:

/// typeof impl 
test: {
open: string;
close: string;
fail: string;
message: {
open: string;
close: string;
fail: string;

With const context, the actual type of impl is now:

/// typeof impl 
test: {
readonly open: "CHANNEL_OPEN";
readonly close: "CHANNEL_CLOSE";
readonly fail: "CHANNEL_FAIL";
message: {
readonly open: "MESSAGE_CHANNEL_OPEN";
readonly close: "MESSAGE_CHANNEL_CLOSE";
readonly fail: "MESSAGE_CHANNEL_FAIL";

const context allows us to satisfy the contract made by ChannelDefinition. Now, openChannel correctly errors:

openChannel(impl, 'messages') // ✅ satisfies contract
openChannel(impl, 'massages') // 💥 bombs

You might be in a space where you need to work with the concrete type, that satisfies the ChannelDefinition contract, outside of a function. For that, we can mimic the same behaviour using the Validate<T, U> helper type:

type Validate<T, U> = T extends U ? T : never; 

Use this as follows:

const correctImpl = {
test: {
} as const;

const wrongImpl = {
test: {
} as const;

// ✅ returns typeof correctImpl
type ValidatedCorrect
= Validate<typeof correctImpl, ChannelDefinition>;

// 💥 returns never
type ValidatedWrong
= Validate<typeof wrongImpl, ChannelDefinition>;

As always, there’s a pen for you to fiddle around.

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