How does the Haskell compiler emit code for `(==) (fromInteger 0) (fromInteger 0)` when the intermediary values are never concretely typed?

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My question is related to type class instance deduction in the presence of ambiguously typed intermediary values

Prelude> :t fromInteger 0
fromInteger 0 :: Num a => a
Prelude> :t (==)
(==) :: Eq a => a -> a -> Bool
Prelude> :t (==) (fromInteger 0)
(==) (fromInteger 0) :: (Eq a, Num a) => a -> Bool
Prelude> :t (==) (fromInteger 0) (fromInteger 1)
(==) (fromInteger 0) (fromInteger 1) :: Bool
Prelude> (==) (fromInteger 0) (fromInteger 1)
False

Magic! It's unclear how or whether a was made concrete, yet the code ran successfully!

According to the type inference rules, the type variables denoted by a above unify successfully with each other because they have compatible Num a constraints across the different terms. However, a never binds to a concrete type. My question is, at runtime, which instance dictionary (or specialization, whatever) is used for the (==) function?

Is this a case where Haskell relies on a simple binary memcmp style comparison? Or perhaps it just picks the first instance in its list of Num instances since in theory it shouldn't matter (as long as the algebraic properties of that instance are implemented correctly...)

instance haskell
1 Answers

Yes it is made concrete by defaulting, which is pretty much your second theory.

The expression [..] is ambiguous because the literal 4 is of Num a => a type in Haskell. 4 can be an Int, a Float or any other type that is an instance of Num, so the compiler can’t choose any particular type for the same reason above. But the Haskell Committee thought that this is too much restriction. After much debates, they compromised and added an ad-hoc rule for choosing a particular default type.

(Although it does matter which instance is chosen. For example 2^64 == 0 is True if Int is chosen, but False if Integer is chosen. Integer is first in the default list for good reason.)

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