Haskell - Is effect order deterministic in case of Applicative?

Question

When executing the IO action defined by someFun <$> (a :: IO ()) <$> (b :: IO ()), is the execution of the a and b actions

Answer 1

It's certainly deterministic, yes. It will always do the same thing for any specific instance. However, there's no inherent reason to choose left-to-right over right-to-left for the order of effects.

However, from the documentation for Applicative:

If f is also a Monad, it should satisfy pure = return and (<*>) = ap (which implies that pure and <*> satisfy the applicative functor laws).

The definition of ap is this, from Control.Monad:

ap :: (Monad m) => m (a -> b) -> m a -> m b
ap =  liftM2 id

And liftM2 is defined in the obvious way:

liftM2 f m1 m2 = do { x1 <- m1; x2 <- m2; return (f x1 x2) }

What this means is that, for any functor that is a Monad as well as an Applicative, it is expected (by specification, since this can't be enforced in the code), that Applicative will work left-to-right, so that the do block in liftM2 does the same thing as liftA2 f x y = f <$> x <*> y.

Because of the above, even for Applicative instances without a corresponding Monad, by convention the effects are usually ordered left-to-right as well.

More broadly, because the structure of an Applicative computation is necessarily independent of the "effects", you can usually analyze the meaning of a program independently of how Applicative effects are sequenced. For example, if the instance for [] were changed to sequence right-to-left, any code using it would give the same results, just with the list elements in a different order.