CODEWITHSUNDEEP
genericshaskellboilerplate
Pradeep
Pradeep

Reputation: 4127

understanding syb boilerplate elimination

In the example given in http://web.archive.org/web/20080622204226/http://www.cs.vu.nl/boilerplate/ 

-- Increase salary by percentage
increase :: Float -> Company -> Company
increase k = everywhere (mkT (incS k))

-- "interesting" code for increase
incS :: Float -> Salary -> Salary
incS k (S s) = S (s * (1+k))

how come increase function compiles without binding anything for the first Company mentioned in its type signature.

Is it something like assigning to a partial function? Why is it done like that?

Upvotes: 3

Views: 180

Answers (1)

OndraSej
OndraSej

Reputation: 692

Yes, it's the same concept as partial application. The line is a shorter (but arguably less clear) equivalent of

increase k c = everywhere (mkT (incS k)) c

As everywhere takes two parameters but is only given one, the type of everywhere (mkT (incS k)) is Company -> Company. Because this is exactly what increase k returns for each Float k, the resulting type of increase is Float -> Company -> Company.

Upvotes: 3

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