Compiling TypeScript: Using the `as` operator to convert and then destructure an array results in a compilation error, requiring an

I am currently utilizing TypeScript version 2.9.2.

There is a static method in a third-party library called URI.js, which is declared as follows:

joinPaths(...paths: (string | URI)[]): URI;

I have a variable named urlPaths that is defined as urlPaths: string | string[]. The code below is causing an error [ts] Expression expected. at the spread operator:

URI.joinPaths(typeof urlPaths === 'string' ? urlPaths as string : ...(urlPaths as string[]))

However, if I separate the ternary operator expression into its own variable, it works without errors:

const paths = typeof urlPaths === 'string' ? [urlPaths as string] : urlPaths as string[];
URI.joinPaths(...paths);

What could be incorrect with my syntax in this scenario?

typescript

Answer №1

The spread syntax is a useful feature in JavaScript that can be used on function arguments. It's important to ensure that the ... is placed at the outermost level of your code:

URI.joinPaths(... (typeof urlPaths === 'string' ? [urlPaths as string] : (urlPaths as string[])));

It's worth mentioning that the type assertions in this code snippet are redundant. TypeScript is capable of inferring the types without them, especially since typeof urlPaths === 'string' functions as a type guard and defines urlPaths: string | string[].

URI.joinPaths(... (typeof urlPaths === 'string' ? [urlPaths] : urlPaths));

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