Sonata · HackerTrans

2 năm trướcdiscuss

Great to see this moving forward.

It does seem a bit of a mixed message to enable this by default while still saying the functionality is experimental. I fear many people will take this as a signal to start using the feature in production, then find a later Node.js update breaks things if the behaviour changes.

Sonata

2 năm trướcdiscuss

The approach taken to this problem by Cats Effect (a Scala concurrency library) is interesting. It allows cancellation of a fiber from outside, but let's blocks of code be marked as uncancelable. If a fiber is cancelled while executing one of these blocks, it will complete the block before cancelling. This protects against cancellation in between two operations which leaves the program in a broken state.

The drawback of this approach is that the onus is on anybody writing code which might be cancelled to correctly mark the uncancelable regions.

Sonata

2 năm trướcdiscuss

As long as they don't make them mutable!

Sonata

3 năm trướcdiscuss

Nice to see some Scala content on HN.

Hope he finishes the book.

Sonata

3 năm trướcdiscuss

Here a common example:

Given the following tables:

  CREATE TABLE person (
    id INT PRIMARY KEY,
    name VARCHAR NOT NULL
  );

  CREATE TABLE pet (
    id INT PRIMARY KEY,
    name VARCHAR NOT NULL,
    owner_id INT NOT NULL
  );

It is common to want to join them, like so:

  SELECT owner.id, owner.name, COUNT(pet.id) AS numberOfPets 
  FROM person AS owner
  LEFT OUTER JOIN pet ON owner.id = pet.owner_id
  GROUP BY owner.id

This doesn't work in standard SQL, because all columns in the SELECT list have to either be aggregated or included in the GROUP BY. owner.name is neither. That is a bit silly though because we know each result row will only have one unambiguous value for the owner name, since the GROUP BY is on a unique column from the same table as the owner name.

We can solve this with ANY_VALUE:

  SELECT owner.id, ANY_VALUE(owner.name) AS name, COUNT(pet.id) AS numberOfPets 
  FROM person AS owner
  LEFT OUTER JOIN pet ON owner.id = pet.owner_id
  GROUP BY owner.id

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