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
}
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这是在参数列表中解构数组的正确语法:

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


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

type StringKeyValuePair = [string, string];

您可以通过命名数组来完成您想要的操作:

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

But you wouldn't name the interal parameter. 另一种可能性是成对使用解构:

function f([a,b,c]: [number, number, number]) {}
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我的代码如下

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

在 TS2349下面的一行输入错误: 不能调用类型缺少调用签名的表达式;

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

把线改到下面解决了我的问题;

const [nodes, counts] = <Node[], number[]>getNodesAndCounts();
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使用 TypeScript 4.0，元组类型现在可以提供标签

type Range = [start: number, end: number]
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As a simple answer I would like to add that you can do this:

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

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