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Narrowing Function Parameters With Rests And Tuples

· 20 min read

TypeScript's provides several ways to describe the type of a function that can be called in multiple different ways. But the two most common strategies -function overloads and generic functions- don't help much with how the function's internal implementation understands the types of its parameters.

This article will walk through three techniques for describing a parameter type that changes based on a previous parameter. The first two allow using standard JavaScript syntax but aren't quite precise in describing the types inside the function. The third one requires using a ... array spread and [...] tuple type in a funky new way that gets the right types internally.

Dependent Parameter Types? What?

Suppose you wanted to write a function that takes in two parameters:

  • fruit: either "apple" or "banana"
  • info: either...
    • an AppleInfo type if fruit is "apple", or
    • a BananaInfo type if fruit is "banana"

In that function, the second parameter's type is different based on the first parameter.

An initial approach might be to declare the function's two parameters -fruit and info- as union types. In short:

ts
declare function logFruit(
fruit: "apple" | "banana",
info: AppleInfo | BananaInfo
): void;
ts
declare function logFruit(
fruit: "apple" | "banana",
info: AppleInfo | BananaInfo
): void;

That would allow calling the function with the right matched types, such as "apple" and an object matching AppleInfo. That might look something like this:

ts
interface AppleInfo {
color: "green" | "red";
}
 
interface BananaInfo {
curvature: number;
}
 
function logFruit(fruit: "apple" | "banana", info: AppleInfo | BananaInfo) {
switch (fruit) {
case "apple":
console.log(`My apple's color is ${(info as AppleInfo).color}.`);
break;
case "banana":
console.log(
`My banana's curvature is ${(info as BananaInfo).curvature}.`
);
break;
}
}
 
logFruit("apple", { color: "green" }); // Ok
 
logFruit("banana", { color: "green" }); // Should error, but doesn't...
ts
interface AppleInfo {
color: "green" | "red";
}
 
interface BananaInfo {
curvature: number;
}
 
function logFruit(fruit: "apple" | "banana", info: AppleInfo | BananaInfo) {
switch (fruit) {
case "apple":
console.log(`My apple's color is ${(info as AppleInfo).color}.`);
break;
case "banana":
console.log(
`My banana's curvature is ${(info as BananaInfo).curvature}.`
);
break;
}
}
 
logFruit("apple", { color: "green" }); // Ok
 
logFruit("banana", { color: "green" }); // Should error, but doesn't...

Two problems in that code block:

  • It requires manually as asserting info to the right type even when fruit's type has been narrowed, such as within case "apple".
  • Nothing prevents calling logFruit with mismatched types, such as "banana" and an object matching AppleInfo.

The logFruit function needs a way to indicate that info's type is dependent on the type of fruit.

Using Function Overloads

One approach for describing a function as having multiple ways it can be called is with function overloads. To recap function overloads, TypeScript allows code to describe any number of call signatures just before a function's implementation. The function is then allowed to be called in ways matching any of those call signatures.

For example, this eitherWay can be called either with a number input to return a number or a string input to return a string:

ts
function eitherWay(input: number): number;
function eitherWay(input: string): string;
function eitherWay(input: number | string) {
return input;
}
 
eitherWay(123);
function eitherWay(input: number): number (+1 overload)
 
eitherWay("abc");
function eitherWay(input: string): string (+1 overload)
ts
function eitherWay(input: number): number;
function eitherWay(input: string): string;
function eitherWay(input: number | string) {
return input;
}
 
eitherWay(123);
function eitherWay(input: number): number (+1 overload)
 
eitherWay("abc");
function eitherWay(input: string): string (+1 overload)

Describing the logFruit function with overloads -let's call it logFruitOverload- would look something like declaring a signature for "apple" and a signature for "banana":

ts
function logFruitOverload(fruit: "apple", info: AppleInfo): void;
function logFruitOverload(fruit: "banana", info: BananaInfo): void;
function logFruitOverload(
fruit: "apple" | "banana",
info: AppleInfo | BananaInfo
) {
switch (fruit) {
case "apple":
console.log(`My apple's color is ${(info as AppleInfo).color}.`);
break;
case "banana":
console.log(
`My banana's curvature is ${(info as BananaInfo).curvature}.`
);
break;
}
}
 
logFruitOverload("apple", { color: "green" }); // Ok
 
logFruitOverload("banana", { color: "green" }); // Should error
No overload matches this call. Overload 1 of 2, '(fruit: "apple", info: AppleInfo): void', gave the following error. Argument of type '"banana"' is not assignable to parameter of type '"apple"'. Overload 2 of 2, '(fruit: "banana", info: BananaInfo): void', gave the following error. Argument of type '{ color: string; }' is not assignable to parameter of type 'BananaInfo'. Object literal may only specify known properties, and 'color' does not exist in type 'BananaInfo'.2769No overload matches this call. Overload 1 of 2, '(fruit: "apple", info: AppleInfo): void', gave the following error. Argument of type '"banana"' is not assignable to parameter of type '"apple"'. Overload 2 of 2, '(fruit: "banana", info: BananaInfo): void', gave the following error. Argument of type '{ color: string; }' is not assignable to parameter of type 'BananaInfo'. Object literal may only specify known properties, and 'color' does not exist in type 'BananaInfo'.
ts
function logFruitOverload(fruit: "apple", info: AppleInfo): void;
function logFruitOverload(fruit: "banana", info: BananaInfo): void;
function logFruitOverload(
fruit: "apple" | "banana",
info: AppleInfo | BananaInfo
) {
switch (fruit) {
case "apple":
console.log(`My apple's color is ${(info as AppleInfo).color}.`);
break;
case "banana":
console.log(
`My banana's curvature is ${(info as BananaInfo).curvature}.`
);
break;
}
}
 
logFruitOverload("apple", { color: "green" }); // Ok
 
logFruitOverload("banana", { color: "green" }); // Should error
No overload matches this call. Overload 1 of 2, '(fruit: "apple", info: AppleInfo): void', gave the following error. Argument of type '"banana"' is not assignable to parameter of type '"apple"'. Overload 2 of 2, '(fruit: "banana", info: BananaInfo): void', gave the following error. Argument of type '{ color: string; }' is not assignable to parameter of type 'BananaInfo'. Object literal may only specify known properties, and 'color' does not exist in type 'BananaInfo'.2769No overload matches this call. Overload 1 of 2, '(fruit: "apple", info: AppleInfo): void', gave the following error. Argument of type '"banana"' is not assignable to parameter of type '"apple"'. Overload 2 of 2, '(fruit: "banana", info: BananaInfo): void', gave the following error. Argument of type '{ color: string; }' is not assignable to parameter of type 'BananaInfo'. Object literal may only specify known properties, and 'color' does not exist in type 'BananaInfo'.

This function overloads approach does improve on the first one around union types:

  • Calling logFruitOverload with "apple" permits passing an object matching AppleInfo as info.
  • Calling logFruitOverload with "banana" and an AppleInfo object is a type error.

Progress!

But, the type of info inside the case "apple": is still AppleInfo | BananaInfo. You can see this by hovering your cursor or mouse over the info in info as AppleInfo. Explicit as assertions are still needed to access fruit-specific properties.

TypeScript isn't able to narrow the type of info even though the two overloads explicitly stated that each fruit string matched up with a specific interface. The function's implementation signature -which is what its internal implementation respects- didn't understand the relationship.

Using Generic Functions

Another attempt at type safety for this function might be to turn it generic. The type of info depends on the type of fruit; therefore, using a type parameter for fruit would allow for narrowing the type of info based on the type of fruit.

This implementation declares a Fruit type parameter that must be one of the keys of an InfoForFruit interface ("apple" | "banana"), then declares that info must the corresponding InfoForFruit value under that Fruit key:

ts
interface InfoForFruit {
apple: AppleInfo;
banana: BananaInfo;
}
 
function logFruitGeneric<Fruit extends keyof InfoForFruit>(
fruit: Fruit,
info: InfoForFruit[Fruit]
) {
switch (fruit) {
case "apple":
console.log(`My apple's color is ${(info as AppleInfo).color}.`);
break;
case "banana":
console.log(
`My banana's curvature is ${(info as BananaInfo).curvature}.`
);
break;
}
}
 
logFruitGeneric("apple", { color: "green" }); // Ok
 
logFruitGeneric("banana", { color: "green" }); // Should error
Argument of type '{ color: string; }' is not assignable to parameter of type 'BananaInfo'. Object literal may only specify known properties, and 'color' does not exist in type 'BananaInfo'.2345Argument of type '{ color: string; }' is not assignable to parameter of type 'BananaInfo'. Object literal may only specify known properties, and 'color' does not exist in type 'BananaInfo'.
ts
interface InfoForFruit {
apple: AppleInfo;
banana: BananaInfo;
}
 
function logFruitGeneric<Fruit extends keyof InfoForFruit>(
fruit: Fruit,
info: InfoForFruit[Fruit]
) {
switch (fruit) {
case "apple":
console.log(`My apple's color is ${(info as AppleInfo).color}.`);
break;
case "banana":
console.log(
`My banana's curvature is ${(info as BananaInfo).curvature}.`
);
break;
}
}
 
logFruitGeneric("apple", { color: "green" }); // Ok
 
logFruitGeneric("banana", { color: "green" }); // Should error
Argument of type '{ color: string; }' is not assignable to parameter of type 'BananaInfo'. Object literal may only specify known properties, and 'color' does not exist in type 'BananaInfo'.2345Argument of type '{ color: string; }' is not assignable to parameter of type 'BananaInfo'. Object literal may only specify known properties, and 'color' does not exist in type 'BananaInfo'.

This generic version gives a friendlier error message in the case of calling logFruitGeneric with "banana" and the wrong shape of info. But TypeScript again unfortunately isn't able to narrow down the type of info inside case "apple".

Using Rest Parameters

For a third and final attempt, let's combine three pieces of syntax:

  1. Rest parameters: using a ... to allow any number of parameters as an array
  2. Tuple types: restricting the type of an array to a fixed size, with an explicit type for element
  3. Array destructuring assignment: taking the elements in an array and assigning them to local variables

This logFruitTuple version uses those three techniques to allow passing in fruit and info parameters adhering to the FruitAndInfo tuple type:

ts
type FruitAndInfo = ["apple", AppleInfo] | ["banana", BananaInfo];
 
function logFruitTuple(...[fruit, info]: FruitAndInfo) {
switch (fruit) {
case "apple":
console.log(`My apple's color is ${info.color}.`);
break;
case "banana":
console.log(`My banana's curvature is ${info.curvature}.`);
break;
}
}
 
logFruitTuple("apple", { color: "green" }); // Ok
 
logFruitTuple("banana", { color: "green" }); // Should error
Argument of type '["banana", { color: "green"; }]' is not assignable to parameter of type 'FruitAndInfo'. Type '["banana", { color: "green"; }]' is not assignable to type '["banana", BananaInfo]'. Type at position 1 in source is not compatible with type at position 1 in target. Type '{ color: "green"; }' is not assignable to type 'BananaInfo'. Object literal may only specify known properties, and 'color' does not exist in type 'BananaInfo'.2345Argument of type '["banana", { color: "green"; }]' is not assignable to parameter of type 'FruitAndInfo'. Type '["banana", { color: "green"; }]' is not assignable to type '["banana", BananaInfo]'. Type at position 1 in source is not compatible with type at position 1 in target. Type '{ color: "green"; }' is not assignable to type 'BananaInfo'. Object literal may only specify known properties, and 'color' does not exist in type 'BananaInfo'.
ts
type FruitAndInfo = ["apple", AppleInfo] | ["banana", BananaInfo];
 
function logFruitTuple(...[fruit, info]: FruitAndInfo) {
switch (fruit) {
case "apple":
console.log(`My apple's color is ${info.color}.`);
break;
case "banana":
console.log(`My banana's curvature is ${info.curvature}.`);
break;
}
}
 
logFruitTuple("apple", { color: "green" }); // Ok
 
logFruitTuple("banana", { color: "green" }); // Should error
Argument of type '["banana", { color: "green"; }]' is not assignable to parameter of type 'FruitAndInfo'. Type '["banana", { color: "green"; }]' is not assignable to type '["banana", BananaInfo]'. Type at position 1 in source is not compatible with type at position 1 in target. Type '{ color: "green"; }' is not assignable to type 'BananaInfo'. Object literal may only specify known properties, and 'color' does not exist in type 'BananaInfo'.2345Argument of type '["banana", { color: "green"; }]' is not assignable to parameter of type 'FruitAndInfo'. Type '["banana", { color: "green"; }]' is not assignable to type '["banana", BananaInfo]'. Type at position 1 in source is not compatible with type at position 1 in target. Type '{ color: "green"; }' is not assignable to type 'BananaInfo'. Object literal may only specify known properties, and 'color' does not exist in type 'BananaInfo'.

The function properly narrows info's type based on fruit, and properly flags the incorrect call with "banana" and an AppleInfo-shaped object. Hooray!

Step-by-step, here's how that parameter works:

  1. ... is a rest parameter, collecing any arguments may be passed to the function into an array
  2. [fruit, info] collects the first two elements of that array, storing them in fruit and info variables, respectively
  3. : FruitAndInfo annotates the type of the parameters to be FruitAndInfo, which is a union allowing either of the two tuples:
    • ["apple", AppleInfo]
    • ["banana", BananaInfo]

Putting it all together: the function accepts two parameters, and assigns them the type FruitAndInfo together. That's exactly what we wanted!

How It Works

TypeScript is smart enough that when elements of a tuple are narrowed, other elements of that same tuple are narrowed too. That's why TypeScript knows that if fruit is an "apple" inside logFruitTuple, it can narrow the type of info to AppleInfo.

You can isolate that smartness separately from function parameters. In the following snippet, when the destructured fruit variable is narrowed to "apple", info is narrowed to AppleInfo:

ts
type FruitAndInfo = ["apple", AppleInfo] | ["banana", BananaInfo];
 
const [fruit, info]: FruitAndInfo = Math.random()
? ["apple", { color: "green" }]
: ["banana", { curvature: 35 }];
 
if (fruit === "apple") {
info;
const info: AppleInfo
}
ts
type FruitAndInfo = ["apple", AppleInfo] | ["banana", BananaInfo];
 
const [fruit, info]: FruitAndInfo = Math.random()
? ["apple", { color: "green" }]
: ["banana", { curvature: 35 }];
 
if (fruit === "apple") {
info;
const info: AppleInfo
}

logFruitTuple's [fruit, info] parameter is typed as FruitAndInfo, so it benefits from the tuple types narrowing. Smart.

Named Tuples

Thanks to Emil van Galen on Twitter for noting the usage of named tuples!

One downside of the tuples approach is that, by default, arguments do not have names. The logFruitTuple function from earlier uses auto-generated __0_0 and __0_ when suggesting parameter names in editors:

ts
logFruitTuple(__0_0: "apple", __0_1: AppleInfo): void
ts
logFruitTuple(__0_0: "apple", __0_1: AppleInfo): void

You can work around this naming behavior by assigning explicit names to the tuple type's elements.

ts
type FruitAndInfo =
| [fruit: "apple", info: AppleInfo]
| [fruit: "banana", info: BananaInfo];
 
function logFruitTupleNamed(...[fruit, info]: FruitAndInfo) {
switch (fruit) {
case "apple":
console.log(`My apple's color is ${info.color}.`);
break;
case "banana":
console.log(`My banana's curvature is ${info.curvature}.`);
break;
}
}
ts
type FruitAndInfo =
| [fruit: "apple", info: AppleInfo]
| [fruit: "banana", info: BananaInfo];
 
function logFruitTupleNamed(...[fruit, info]: FruitAndInfo) {
switch (fruit) {
case "apple":
console.log(`My apple's color is ${info.color}.`);
break;
case "banana":
console.log(`My banana's curvature is ${info.curvature}.`);
break;
}
}

Putting fruit: and info: before each of the elements in the tuples will let TypeScript know to suggest those names in the function:

ts
logFruitTuple(fruit: "apple", info: AppleInfo): void
ts
logFruitTuple(fruit: "apple", info: AppleInfo): void

See Default rest parameter names to destructuring names when of an anonymous tuple on the TypeScript issue tracker for more details.


Which Should You Use?

It depends.

Most of the time, these kinds of complex function signatures are a sign of overly complicated code. Types that are difficult to represent in the type system can be difficult to work with because they can be conceptually complex. When possible, try to avoid needing this kind of function signature.

Technique Tradeoffs

If you must use one of these techniques, consider which one will cause your project's developer(s) the least pain in the future. Many TypeScript developers prefer to avoid function overloads because of how wacky they appear at first glance. Generic functions can also be intimidating to many TypeScript developers, though generally less so in cases such as this one that only have one type parameter.

... rest tuple parameters are a nifty trick that also run the risk of confusing developers. But the increased type safety internally may be worth that confusion. My recommendation would be to wait for a time that they're very well suited towards, try them out once, and see how they feel. You may find that you like them or you may not.

Wrapping arguments in an array and then immediately destructuring that array also runs a small risk of hurting runtime performance if used in a very commonly run path. Though, this performance hit is likely optimized to be zero or near-zero in most instances of the function. Always diagnose runtime performance problems before jumping to conclusions.

Alternative: Objects

Instead of wrestling with multiple function parameters, consider using a single object with multiple properties. Doing so would allow using techniques such as the following discriminated union to represent different allowed shapes.

ts
interface AppleInfo {
color: "green" | "red";
}
 
interface AppleAndInfo {
fruit: "apple";
info: AppleInfo;
}
 
interface BananaInfo {
curvature: number;
}
 
interface BananaAndInfo {
fruit: "banana";
info: BananaInfo;
}
 
function logFruitTuple({ fruit, info }: AppleAndInfo | BananaAndInfo) {
switch (fruit) {
case "apple":
console.log(`My apple's color is ${info.color}.`);
break;
case "banana":
console.log(`My banana's curvature is ${info.curvature}.`);
break;
}
}
 
logFruitTuple({ fruit: "apple", info: { color: "green" } }); // Ok
 
logFruitTuple({ fruit: "banana", info: { color: "green" } }); // Should error
Type 'string' is not assignable to type '"green" | "red"'.2322Type 'string' is not assignable to type '"green" | "red"'.
ts
interface AppleInfo {
color: "green" | "red";
}
 
interface AppleAndInfo {
fruit: "apple";
info: AppleInfo;
}
 
interface BananaInfo {
curvature: number;
}
 
interface BananaAndInfo {
fruit: "banana";
info: BananaInfo;
}
 
function logFruitTuple({ fruit, info }: AppleAndInfo | BananaAndInfo) {
switch (fruit) {
case "apple":
console.log(`My apple's color is ${info.color}.`);
break;
case "banana":
console.log(`My banana's curvature is ${info.curvature}.`);
break;
}
}
 
logFruitTuple({ fruit: "apple", info: { color: "green" } }); // Ok
 
logFruitTuple({ fruit: "banana", info: { color: "green" } }); // Should error
Type 'string' is not assignable to type '"green" | "red"'.2322Type 'string' is not assignable to type '"green" | "red"'.

Whichever strategy you choose, remember to keep the code as readable and straightforward to understand as possible. This quote from Brian Kerninghan applies to the type system as well as runtime code:

Debugging is twice as hard as writing the code in the first place. Therefore, if you write the code as cleverly as possible, you are, by definition, not smart enough to debug it.

Happy typing!

Further Reading


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