How can we deduce the type of a Haskell expression?
To get the general idea, let's look at a few examples.
Note: The discussion here is correct for monomorphic functions, but it is a bit simplistic for polymorphic functions. However, it should be of assistance in understanding how types are assigned to Haskell expressions.
Expressed in prefix form, functional composition can be defined with the equation:
(.) f g x = f (g x)
We begin the process of type inference by assigning types to the parameter names and to the function's defining express (i.e., its result). We introduce new type names t1, t2, t3 and t4 for the components of f as follows:
f :: t1 -- parameter of f
g :: t2 -- parameter of f
x :: t3 -- parameter of f
f (g x) :: t4 -- defining expression for f
The type of (.) is therefore given by:
(.) :: t1 -> t2 -> t3 -> t4
We are not finished because there are certain relationships among the new types that must be taken into account. To see what these relationships are, we use the following inference rules.
Using the application rule on f (g x) :: t4, we introduce a new type t5 such that:
g x :: t5
f :: t5 -> t4
Using the application rule for g x :: t5, we introduce another new type t6 such that:
x :: t6
g :: t6 -> t5
Using the equality rule on the two types deduced for each of f, g, and x, respectively, we get the following identities:
t1 = (t5 -> t4) -- f
t2 = (t6 -> t5) -- g
t3 = t6 -- x
For function (.), we thus deduce type signature:
(.) :: (t5 -> t4) -> (t6 -> t5) -> t6 -> t4
If we replace the type names by Haskell generic type variables that follow the usual naming convention, we get:
(.) :: (b -> c) -> (a -> b) -> a -> c
Now let's consider the function definition:
f x y = fst x + fst y
Note that the names (+) and fst occur on the right side of the definition, but do not occur on the left.
From the Haskell standard prelude, we can see that:
(+) :: Num a => a -> a -> a
fst :: (a, b) -> a
We must be careful. The two occurrences of the polymorphic function fst in the definition for f need not bind the type variables a and b to the same concrete types. For example, consider the expression:
fst (2, True) + fst (1, "hello")
This expression is well-typed despite the fact that the first occurrence of fst has the type
Num a => (a,Bool) -> a
and the second occurrence has type
Num a => (a, [Char]) -> a
Furthermore, the two occurrences of the type variable a are not, in general, required to bind to the same type. (However, as we will see, they do in this expression because of the addition operation.)
To handle the situation with the multiple applications of fst, we use the following rule.
Following this rule, we rewrite the definition of f in the form
f x y = fst1 x + fst2 y
and assume two different instantiations of the generic type of fst:
fst1 :: (u1, u2) -> u1
fst2 :: (v1, v2) -> v1
After making the above transformation, we proceed by assigning types to the parameters and definition of f, introducing three new types:
x :: t1 -- parameter of f
y :: t2 -- parameter of f
fst1 x + fst2 y :: t3 -- defining expression for f
Thus we have the following type for f:
f :: t1 -> t2 -> t3
Now we can rewrite the defining expression for f fully in prefix form to get:
(+) (fst1 x) (fst2 y)
Then, using the application rule on the above expression, we deduce:
(fst2 y) :: t4
(+) (fst1 x) :: t4 -> t3
Using the application rule on (fst2 y) :: t4, we get:
y :: t5
fst2 :: t5 -> t4
Similarly, using the application rule on (+) (fst1 x) :: t4 -> t3, we get:
(fst1 x) :: t6
(+) :: t6 -> t4 -> t3
Going further and applying the application rule to (fst1 x) :: t6, we deduce:
x :: t7
fst1 :: t7 -> t6
Now we have introduced types for all the symbols appearing in the definition of function f. We begin simplification by using the equality rule for x, y, fst1, fst2, and (+), respectively. We thus deduce the type equations:
t1 = t7 -- x
t2 = t5 -- y
((u1, u2) -> u1) = (t7 -> t6) -- fst1
((v1, v2) -> v1) = (t5 -> t4) -- fst2
(Num a => a -> a -> a) = (t6 -> t4 -> t3) -- (+)
Now, using the function rule on the last three equations above, we derive:
t7 = (u1, u2)
t6 = u1
t5 = (v1, v2)
t4 = v1
t3 = t4 = t6 = v1 = u1 = (Num a => a)
We had assigned type f :: t1 -> t2 -> t3 originally. Substituting from the above, we deduce the following type:
f :: Num a => (a, u2) -> (a, v2) -> a
Finally, we can replace the type names u2 and v2 by Haskell generic type variables that follow the usual naming convention. We get the following inferred type for function f:
f :: Num a => (a, b) -> (a, c) -> a
For this example, consider the definition:
fix f = f (fix f)
To deduce a type for fix, we proceed as before and introduce types for the parameters and defining expression of f:
f :: t1 -- parameter of fix
f (fix f) :: t2 -- defining expression for fix
Thus, fix has the type:
fix :: t1 -> t2
Using the application rule on the expression f (fix f), we obtain:
(fix f) :: t3
f :: t3 -> t2
Then using the application rule on the expression fix f, we get:
f :: t4
fix :: t4 -> t3
Using the equality rule on f and fix, we deduce:
t1 = t4 = (t3 -> t2) -- f
(t1 -> t2) = (t4 -> t3) -- fix
Then, using the function rule on the second equation, we obtain the identities:
t1 = t4
t2 = t3
Since fix :: t1 -> t3, we derive the type:
fix :: (t3 -> t3) -> t3
If we replace t3 by a Haskell generic type variable that follows the usual naming convention, we get the following inferred type for fix:
fix :: (a -> a) -> a
Finally, let us consider an example in which the typing is wrong. Let us define selfapply as follows:
selfapply f = f f
Proceeding as in the previous examples, we introduce new types for the parameters and defining expression of f:
f :: t1 -- parameter of selfapply
f f :: t2 -- defining expression for selfapply
Thus we have the type:
selfapply :: t1 -> t2
Using the application rule on f f, we get:
f :: t3
f :: t3 -> t2
But the equality rule for f tells us that:
t1 = t3 = (t3 -> t2)
or just
t1 = (t1 -> t2)
However, the equation t1 = (t1 -> t2) does not possess a solution for t1 and the definition of selfapply is thus rejected by the type checker.
Haskell function definitions must also conform to the following rules.
These notes are based closely on:
Some of the material was also based on:
I thank Hong Mei Gao for assisting me in preparation of this set of notes.
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