How to declare a function that throws an error in Typescript

Question

In Java I would declare a function like this:

public boolean Test(boolean test) throws Exception {   if (test == true)     return false;   throw new Exception(); } 

And I can use this function without handling the exception.

If it is possible, how to do the same in Typescript? The compiler will tell me that I can't use the function without a try/catch.

Answer 1

There is no such feature in TypeScript. It's possible to specify error type only if a function returns an error, not throws it (this rarely happens and is prone to be antipattern).

The only relevant type is never. It is applicable only if a function definitely throws an error, it cannot be more specific than that. It's a type as any other, it won't cause type error as long as it doesn't cause type problems:

function Test(): never => {   throw new Error(); }  Test(); // won't cause type error let test: boolean = Test(); // will cause type error 

When there is a possibility for a function to return a value, never is absorbed by return type.

It's possible to specify it in function signature, but for reference only:

function Test(test: boolean): boolean | never {   if (test === true)     return false;    throw new Error(); } 

It can give a hint to a developer that unhandled error is possible (in case when this is unclear from function body), but this doesn't affect type checks and cannot force try..catch; the type of this function is considered (test: boolean) => boolean by typing system.

Answer 2

You can mark the function with @throws jsdoc at least. Even though it does not provide static analysis errors in typescript compiler, some good IDE or linter may still report a warning if you try to disregard the function that throws...

/**   * @throws {Error}  */ function someFunc() {     if (Math.random() < 0.5) throw Error(); } someFunc();