BSD-3-Clause licensed by Matt Noonan
Maintained by [email protected]
This version can be pinned in stack with:lawful-0.1.0.0@sha256:d531511cdeeeb157ec4c8907dc529f546f28b8ab0a303b6c893b397ea7fa5f65,815

Module documentation for 0.1.0.0

Depends on 1 package(full list with versions):
Used by 1 package in nightly-2019-04-24(full list with versions):

lawful: Assert that your typeclass instances are lawful

What is this package for?

This small library provides a single two-parameter typeclass Lawful c t, where c is a typeclass and t is a type. Declaring an instance of Lawful C T is an assertion that “the instance for C T obeys the laws of the class C (whatever that means!)”

For example, a lawful instance of Eq T should satisfy the reflexive law x == x for all x :: T. This is certainly true for most types, such as Int or [a] when Eq a is lawful, so we can define

Lawful Eq Int
Lawful Eq a => Lawful Eq [a]

But it isn’t true for Double:

λ> nan = 0 / 0 :: Double
λ> nan == nan
False

Why is there a c t constraint on Lawful c t?

This constraint lets you use a Lawful c t wherever a c t would be expected, as in:

same :: Lawful Eq a => a -> a -> Bool
same x y = x == y

How do I know what laws are expected from a typeclass?

If everybody more-or-less agrees on what the right laws are, hopefully they bothered to write them down somewhere. If they didn’t, then sorry! You’re on your own!

Shouldn’t all typeclass instances be lawful anyway, making this package useless?

Wouldn’t that be nice?