Lost in Technopolis

by John Wiegley

Getting Started with Lenses

Posted by John Wiegley on November 21, 2012 with labels:

The following is the first in a series of articles I hope to write as a gentle introduction to Edward Kmett’s excellent lens library.

Control.Lens provides a composable way to access and modify sub-parts of data structures (where by modify I mean: return a new copy with that part changed). In this introduction I won’t be talking about the theory or laws behind Lens – both of which are worthy of study – but rather how you can use them to write simpler, more expressive code.

The mighty tuple

You’ve probably used tuples quite often by now, and may have discovered that (,) e is a Functor, letting you do things like:

ghci> fmap (+1) (1,2)
  (1,3)

After which a very common question is: “How do I do that for the first element?” One way is with the Arrow library:

ghci> first (+1) (1,2)
  (2,2)

But this method does not generalize for n-tuples. What about the 3rd element of a 3-tuple? Enter lenses. Here’s the lens equivalent of fmap and first:

ghci> over _1 (+1) (1,2)
  (2,2)

ghci> over _2 (+1) (1,2)
  (1,3)

ghci> over _3 (+1) (1,2,3)
  (1,2,4)

In this example, _1 is a lens, which means it represents both a getter and setter focused on a single element of a data structure. In this case we are using it in both senses, since we are first getting the value from the tuple, applying (+1), and then setting the result back to create a new tuple.

You can also apply your function to both elements of a pair at the same time:

ghci> over both (+1) (3,4)
  (4,5)

There is also an infix operator notation for this:

ghci> both %~ (+1) $ (3,4)
  (4,5)

For the simple case of integer addition, you can use +~ n instead of %~ (+n):

ghci> both +~ 1 $ (3,4)
  (4,5)

Most of the math operators are available in this form: -~, *~, //~, ^~, etc. Or you can use .~ to force both members to a specific value:

ghci> both .~ 9 $ (3,4)
  (9,9)

In the operator form it’s fairly easy to chain applications, which makes it easier to apply the same operator to the odd members of, say, a 5-tuple. I’ll use a new function for this, set, which sets the element selected by the lens. Since the result is the new value, we can compose these:

ghci> set _1 9 . set _3 9 $ (1,1,1,1,1)
  (9,1,9,1,1)

Consider what that would look like without lenses:

ghci> let (a,b,c,d,e) = (1,1,1,1,1) in (9, b, 9, d, e)
  (9,1,9,1,1)

The main difference is that you have to capture and pass through all the members, just to modify the few you’re interested in. But even more, the lens version above will work on any tuple with 3 or more elements, while the non-lens version is fixed to working with 5-tuples.

There are lots of these modifier operators for lenses, here’s a selection:

Operator Meaning
.~ Replace the focused element with some x
?~ Replace the focused element with Just x
%~ Apply a function to the element
+~ Add to it
-~ Subtract
*~ Multiple
//~ Divide
^~ Take the integral power
^^~ Take the fractional power
**~ Take an arbitrary power
||~ Logically or with a boolean
&&~ Logically and with a boolean

In the next post, we will look at how lenses interact with some of the other basic functors: Maybe, Either and List.


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