Why is HealthCheck memoized?

[parthik19]

The status field in HealthCheck[F] is val and not def on line 46 of Healthcheck.scala. Doesn't this mean that whenever status is called, healthCheck.status will return whatever is cached from the first call of status? Shouldn't it recompute the status since a successful status at one point doesn't mean the service will be healthy at every point afterwards?

[gvolpe]

Hi @parthik19, it is not memoized, that is the power of referential transparency :)

This is a key property of functional programming. It allows us to treat program as values; we can inline a value such as one returning F[RedisStatus] without changing the meaning of the resulting program.

Have a look at those links, and I also recommend to play around with similar code that uses Console[F].println instead. You'll see that the different calls to status are evaluated per every call. There's no memoization whatsoever :)

[parthik19]

Thank you for the explanation! Allow me to ask another question -

Hypothetically, suppose HealthCheck[F] defined its status field as Sync[F].pure(/* compute AppStatus here */)

Then, it would memoized, correct?

[BalmungSan]

Hi @parthik19 yeah you are correct, in that case the result would be "memoized", since it would even be evaluated at the moment of creation of the program / value. That is why pure only make sense for plain values which could be inlined without affecting the RT of the whole expression.

[parthik19]

Okay makes sense, thanks!