Typescript infer type without using explicit type variable for callback

Question

I want to wrap rxjs subscribe's next callback with my function:

type Handler<T> = (value: T) => void;

export function withTryCatch<T>(callback?: Handler<T>): Handler<T> {

  return (value: T) => {
    try {
      callback?.(value);
    } catch(err) {
      // error handling
    }
  };
}

Problem with this example bellow is, that it does not automatically infer type from subscribe's next function. In this example, user type is stated as unknown. Only way how to make user desired type, is to explicitly set withTryCatch type variable T (see commented code below - withTryCatch<UserModel>).

 store$
    .pipe(
      map(userSelector)
    )
    // .subscribe(withTryCatch<UserModel>((user) => {
    .subscribe(withTryCatch((user) => {
      // possible error code
    }));

Is there any way how to avoid using withTryCatch<UserModel>?

Marek Szkudelski · Accepted Answer · 2020-03-30 11:09:29Z

2

This issue is separated from place where you use withTryCatch function, in that case it's rxjs subscribe method. When you just invoke it the generic T type parameter is unknown. When you call it with some type parameter then T is of course known. You could use Typescript inferring by typing user argument like here:

withTryCatch((user: UserModel) => {

});

You need to pass callback directly to subscribe to use Typescript inferring. Unfortunately it is impossible with that kind of wrapper function

answered Mar 30, 2020 at 11:09

Marek Szkudelski

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