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content/20201001110806-typescript_functions.md
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content/20201001110806-typescript_functions.md
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---
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id: 6df4db64-8ad2-4190-9d4e-3e6f77021a31
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title: TypeScript Functions
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---
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# TypeScript 4.0
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## Variadic Tuple Types
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### Pre 4.0 situation
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Consider the following
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[function](20201006111125-javascript_function_declerations) using
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[Tuples](20200929163624-typescript_tuple_type):
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``` javascript
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function concat(arr1, arr2) {
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return [...arr1, ...arr2];
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}
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```
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The only way to type this in [TypeScript](20200929161126-typescript)
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used to be:
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``` typescript
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function concat(arr1: [], arr2: []): [];
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function concat<A>(arr1: [A], arr2: []): [A];
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function concat<A, B>(arr1: [A, B], arr2: []): [A, B];
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function concat<A, B, C>(arr1: [A, B, C], arr2: []): [A, B, C];
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function concat<A, B, C, D>(arr1: [A, B, C, D], arr2: []): [A, B, C, D];
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function concat<A, B, C, D, E>(arr1: [A, B, C, D, E], arr2: []): [A, B, C, D, E];
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function concat<A, B, C, D, E, F>(arr1: [A, B, C, D, E, F], arr2: []): [A, B, C, D, E, F];)
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function concat<A2>(arr1: [], arr2: [A2]): [A2];
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function concat<A1, A2>(arr1: [A1], arr2: [A2]): [A1, A2];
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function concat<A1, B1, A2>(arr1: [A1, B1], arr2: [A2]): [A1, B1, A2];
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function concat<A1, B1, C1, A2>(arr1: [A1, B1, C1], arr2: [A2]): [A1, B1, C1, A2];
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function concat<A1, B1, C1, D1, A2>(arr1: [A1, B1, C1, D1], arr2: [A2]): [A1, B1, C1, D1, A2];
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function concat<A1, B1, C1, D1, E1, A2>(arr1: [A1, B1, C1, D1, E1], arr2: [A2]): [A1, B1, C1, D1, E1, A2];
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function concat<A1, B1, C1, D1, E1, F1, A2>(arr1: [A1, B1, C1, D1, E1, F1], arr2: [A2]): [A1, B1, C1, D1, E1, F1, A2];
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```
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### Post 4.0 situation
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1. Tail example
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Consider the following example:
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``` javascript
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function tail(arg) {
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const [_, ...result] = arg;
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return result
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}
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```
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TypeScript 4.0 bring 2 changes here.
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1. Changes
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1. Generic spreads in tuple types
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The [spreads](20201014094144-spread) in [tuple
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type](20200929163624-typescript_tuple_type) syntax to be
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[generic](20200929163051-typescript_generics):
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``` typescript
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function tail<T extends any[]>(arr: readonly [any, ...T]) {
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const [_ignored, ...rest] = arr;
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return rest;
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}
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const myTuple = [1, 2, 3, 4] as const;
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const myArray = ["hello", "world"];
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// type [2, 3, 4]
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const r1 = tail(myTuple);
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// type [2, 3, 4, ...string[]]
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const r2 = tail([...myTuple, ...myArray] as const);
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```
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2. Rest elements can occur anywhre in a tuple
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[Rest](20200922162500-rest_parameters) elements can occur
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anywhere in a [tuple](20200929163624-typescript_tuple_type),
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not just at the end:
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``` typescript
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type Strings = [string, string];
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type Numbers = [number, number];
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// [string, string, number, number, boolean]
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type StrStrNumNumBool = [...Strings, ...Numbers, boolean];
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```
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2. Concat example
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Note that in cases when we [spread](20201014094144-spread) in a type
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without a known length, the resulting type becomes unbounded as
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well, and all the following elements factor into the resulting rest
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element type.
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``` typescript
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type Strings = [string, string];
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type Numbers = number[]
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// [string, string, ...Array<number | boolean>]
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type Unbounded = [...Strings, ...Numbers, boolean];
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```
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Therefore the `concat()` example can be typed with:
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``` typescript
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type Arr = readonly any[];
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function concat<T extends Arr, U extends Arr>(arr1: T, arr2: U): [...T, ...U] {
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return [...arr1, ...arr2];
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}
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```
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# Examples
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## Parameter annotations
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``` typescript
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function tralala(tra: number, la: number) {}
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```
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## Return type annotation
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``` typescript
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interface Foo {
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bar: string
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}
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function tralala(): Foo {
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return {bar: 'tralala'}
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}
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```
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## Optional Parameters
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``` typescript
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function foo(bar: number, optional?: string) {}
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```
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``` typescript
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function foo(bar: number, optional: string = 'This is an optional string') {}
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```
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## Overloading
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This is for documentation purposes but imho just makes things more
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confusing.
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``` typescript
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// Overloads
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function padding(all: number);
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function padding(topAndBottom: number, leftAndRight: number);
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function padding(top: number, right: number, bottom: number, left: number);
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// Actual implementation that is a true representation of all the cases the function body needs to handle
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function padding(a: number, b?: number, c?: number, d?: number) {
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if (b === undefined && c === undefined && d === undefined) {
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b = c = d = a;
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}
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else if (c === undefined && d === undefined) {
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c = a;
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d = b;
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}
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return {
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top: a,
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right: b,
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bottom: c,
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left: d
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};
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}
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padding(1); // Okay: all
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padding(1,1); // Okay: topAndBottom, leftAndRight
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padding(1,1,1,1); // Okay: top, right, bottom, left
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padding(1,1,1); // Error: Not a part of the available overloads
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```
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## Declaring Functions
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Use this to declare function type without providing implementation:
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``` typescript
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type LongHand = {
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(a: number): number;
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};
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type ShortHand = (a: number) => number;
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```
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Overloading is supported as well:
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``` typescript
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type LongHandAllowsOverloadDeclarations = {
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(a: number): number;
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(a: string): string;
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};
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```
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