Comment by StavrosK

7 years ago

Yes, in mathematics, not programming. And a function that doubles a number isn't idempotent even by that definition.

6 comments

StavrosK

Reply
Y_Y  7 years ago

Of course a doubling function is not idempotent!

I think the confusion arises because side-effectful functions can be considered as having type

    f :: (RealWorld, OtherArgs) -> (RealWorld, OtherOutputs)

and so for a garage door toggle you have something like

    t :: RealWorld -> RealWorld

where the new state is the old one with the door opened/closed as appropriate.

Now the idempotence condition becomes:

    t(t(world)) == t(world)

but clearly

       t(t(doorOpenWorld)
    =  t(doorClosedWorld)
    =  doorOpenWorld
    != t(doorOpenWorld)
    =  doorclosedWorld

so this is where the notion comes from. If you abuse notation and just say a function of no arguments can be idempotent then you'll get confusion like this.

  • smaddox  7 years ago

    But `GET(GET(x))` doesn't make sense, in general (and if it did, then you would not expect it to be idempotent), so clearly idempotency in this context is meant to mean side-effect free. They should probably just say side-effect free, though, to avoid the confusion.

    • Ajedi32  7 years ago

      It's possible to be idempotent without being side-effect free. If you PUT some record, for example, then that operation _will_ have side effects (modifying the record). If you then PUT that same data again, the result will be the same (it's idempotent).

      3 replies →