CODEWITHSUNDEEP
arraystypestypescriptdestructuring
thr0w
thr0w

Reputation: 1544

Types when destructuring arrays

function f([a,b,c]) {
  // this works but a,b and c are any
}

it's possible write something like that?

function f([a: number,b: number,c: number]) {
  // being a, b and c typed as number 
}

Upvotes: 139

Views: 66758

Answers (5)

Saurabh Thakur
Saurabh Thakur

Reputation: 149

As a simple answer I would like to add that you can do this:

function f([a,b,c]: number[]) {}

Upvotes: 5

henriquehbr
henriquehbr

Reputation: 1122

With TypeScript 4.0, tuple types can now provide labels

type Range = [start: number, end: number]

Upvotes: 13

Rajnikant
Rajnikant

Reputation: 2236

my code was something like below 

type Node = { 
    start: string;
    end: string;
    level: number;
};

const getNodesAndCounts = () => { 
    const nodes : Node[]; 
    const counts: number[];
    // ... code here

return [nodes, counts];
}

const [nodes, counts] = getNodesAndCounts(); // problematic line needed type

typescript was giving me error in line below TS2349: Cannot invoke an expression whose type lacks a call signature; 

nodes.map(x => { 
//some mapping; 
    return x;
);

Changing line to below resolved my problem;

const [nodes, counts] = <Node[], number[]>getNodesAndCounts();

Upvotes: 7

Ryan Cavanaugh
Ryan Cavanaugh

Reputation: 220944

This is the proper syntax for destructuring an array inside an argument list:

function f([a,b,c]: [number, number, number]) {

}

Upvotes: 205

Marcelo Camargo
Marcelo Camargo

Reputation: 2298

Yes, it is. In TypeScript, you do it with types of array in a simple way, creating tuples.

type StringKeyValuePair = [string, string];

You can do what you want by naming the array:

function f(xs: [number, number, number]) {}

But you wouldn't name the interal parameter. Another possibility is use destructuring by pairs:

function f([a,b,c]: [number, number, number]) {}

Upvotes: 17

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