Issue with TypeScript when utilizing the Spread syntax?

https://i.sstatic.net/b9Yz6.png Every time I implement the spread operator like demonstrated below

public outputTestData(p1: number, p2: number, p3: number):void {
    console.log(p1, p2, p3);
}
let data = [2, 2, 5];
this.outputTestData( ... data );

An error message pops up in my editor saying:

[ts] Expected # arguments, but got a minimum of 0.

Why does TypeScript flag an error when utilizing the spread operator to pass arguments?

Oddly enough, there's no issue when the code is executed – the spread operator smoothly allows me to use the array as function arguments without any hitches. However, on VSCode, it highlights an error that seems unnecessary.

Answer №2

More recent versions of TypeScript are designed to automatically handle this through flow analysis. However, in case you encounter any issues, you can manually specify the array types like this to ensure a minimum length:

function displayNumbers(num1: number, num2: number, num3: number): void {
    console.log(num1, num2, num3);
}
let numbersArray: [number, number, number] = [2, 2, 5];
this.displayNumbers(...numbersArray);

Answer №3

Make sure to verify the typescript version being used by Vscode, as it could potentially be outdated and causing issues with code evaluation. Check the version number located at the bottom right corner of your IDE window.

If my assumption is incorrect, please provide the definition of your drawInfo object for further analysis. It's possible that optional properties within drawInfo are causing the evaluator to anticipate 0 parameters when using the spread operator.

UPDATE: Upon closer inspection, it seems that drawInfo is actually an array. However, since arrays can have variable lengths and may not always contain three properties, the typescript evaluator may flag this as an error.

To ensure that drawInfo consistently includes three values, consider defining it as a specific type instead of an array and refrain from using the spread operator.

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