I'm trying to understand the following output of PAKCS:
test> ((1+1)+) _impl#+#Prelude.Num#Prelude.Int (1 _impl#+#Prelude.Num#Prelude.Int 1) test> :t (+) (1+1) (+) (1+1) :: Num a => a -> a
It surprises me that the subexpression 1+1 is not reduced. The full expression has no normal form because the first occurrence of + cannot be eliminated. Yet PAKCS seems to treat it as if it was a value. This makes a certain practical sense in the context of the REPL, but if this is like a value then why is the subexpression not reduced?
If I go one step further and apply $!! to force normalization of the partial application, I see that $!! ignores the subexpression 1+1 (trace shown):
test> id $!! ((1+1)+) Call: id $!! (_impl#+#Prelude.Num#Prelude.Int (1 _impl#+#Prelude.Num#Prelude.Int 1)) Call: id (_impl#+#Prelude.Num#Prelude.Int (1 _impl#+#Prelude.Num#Prelude.Int 1)) Exit: id (_impl#+#Prelude.Num#Prelude.Int (1 _impl#+#Prelude.Num#Prelude.Int 1)) (HNF: _impl#+#Prelude.Num#Prelude.Int (1 _impl#+#Prelude.Num#Prelude.Int 1)) Exit: id $!! (_impl#+#Prelude.Num#Prelude.Int (1 _impl#+#Prelude.Num#Prelude.Int 1)) (HNF: _impl#+#Prelude.Num#Prelude.Int (1 _impl#+#Prelude.Num#Prelude.Int 1)) _impl#+#Prelude.Num#Prelude.Int (1 _impl#+#Prelude.Num#Prelude.Int 1)
This seems to imply that (1+1)+ is a normal form, which it clearly is not.
Is this perhaps some corner that is not well defined, or are there good reasons to NOT reduce an expression such as ((1+1)+) to (2+)? I can imagine that attaining correct semantics might require this delay if the subexpression is non-deterministic but I'm not entirely convinced. Certainly, it is not obvious to me.
And is there some theoretical basis upon which I can understand these seeming quirks?
-Andy