In managing projects at FP Complete, I get to see both the
software development and devops sides of our engineering practice.
Over the years, I've been struck by the recurrence of a single word
appearing repeatedly in both worlds: immutability.
On the software side, one of the strongest tenets of functional
programming is immutable data structures. These are values which -
once created - can never be changed again through the course of
running the application. These reduce coupling between components,
simplify concurrency and parallelism, and decrease the total number
of moving pieces in a system, making it easier to maintain and
develop over time.
On the devops side, immutable infrastructure is relatively a
more recent discovery. By creating machine images and replacing
rather than modifying existing components, we have a more reliable
hosting setup to target, minimize the differences between test and
production systems, and reduce the amount of error-prone, manual
fiddling that leads to 3am coffee-fueled emergency recovery
sessions.
It's no secret that containerization in general, and Docker in
particular, has become very popular in the devops space. I've
noticed that there's a strong parallel between how Docker images
are built, and a technique from functional programming - the ST
(State Thread) type. This blog post will explain both sides of the
puzzle, and then explain how they match up.
Dockerfile:
mutable steps, immutable outcome
A Docker image is a complete Linux filesystem, providing all of
the tools, libraries, and data files needed for its task. As a
simple example, I recently created
a simple Docker image containing the Stack build tool (more on
that later) and Apache FOP for generating some PDFs. In the Docker
world, the formula you use for creating a Docker image is a
Dockerfile. Let's look at the (very simple) Dockerfile I wrote:
FROM fpco/pid1:16.04
RUN DEBIAN_FRONTEND=noninteractive apt-get update && \
DEBIAN_FRONTEND=noninteractive apt-get install -y wget default-jre && \
wget -qO- https://get.haskellstack.org/ | sh
RUN wget -q https://github.com/fpco/docker-fop/releases/download/fop-2.1/fop-2.1-bin.tar.gz && \
tar zxf fop-2.1-bin.tar.gz && \
rm -f fop-2.1-bin.tar.gz && \
mv fop-2.1 /usr/local/share
In this file, I'm starting off from the fpco/pid1
base image, which provides us with a filesystem to start off
with (it would obviously be pretty difficult to create a complete
filesystem each time we wanted to create a new image). Then we
provide a series of actions to take to modify that image.
Looking at the example above, we:
- Update the list of APT packages available
- Install
wget
and the default Java Runtime
Environment
- Install the Stack build tool by running a script
- Download the FOP binary bundle
- Unpack the bundle and move it to /usr/local/share
Look at that list of steps. In no world could those actions be
called "immutable." Every single one of them mutates the
filesystem, either modifying files, adding files, or removing
files. The end result of this mutation process is a new filesystem,
captured in a Docker image.
And here's the important bit: this new image is totally
immutable. You cannot in any way modify the image. You can
create a new image based on it, but the original will remain
unchanged. For all of history, this image will remain
identical.*
In other words: a Dockerfile is a series of mutations which
generates an immutable data structure.
* You may argue that you can delete the image, or you could
create a new image with the same name. That's true, but as long as
you're working with the image in question, it does not change. By
contrast, each time you access the /tmp/foobar
file,
it may have different contents.
The ST type
In a purely functional programming language like Haskell, data
is immutable by default. This means that, if you have a variable
holding an Int
, you cannot change it. Consider this
example code, playing around with a Map
structure
(also known as a dictionary or lookup table):
myMap <- makeSomeMap
print myMap
useMap myMap
print myMap
We make our initial Map
using the
makeSomeMap
function, print its contents, pass it to
some other function (useMap
), and then print it again.
Pop quiz: is there any way that the two print
operations will print different values?
If you're accustomed to mutable languages like Java or Python,
you'd probably say yes: myMap
is (presumably) an
object with mutable state, and the useMap
function
might modify it. In Haskell, that can't happen: you've passed a
reference to myMap
to your useMap
function, but useMap
is not allowed to modify
it.
Of course, we would like to be able to create different values,
so saying "you can't ever change anything" is a little daunting.
The primary way of working with Haskell's immutable data structures
is to have functions which create new values based on old ones. In
this process, we create a new value by giving it some instructions
for the change. For example, if in our example above, the
myMap
value had a mapping from names to ages, we could
insert an extra value:
myMap <- makeSomeMap
let myModifiedMap = insert "Alice" 35 myMap
print myModifiedMap
useMap myModifiedMap
print myModifiedMap
However, this isn't real mutation: the original
myMap
remains the same. There are cases in which
creating a completely new version of the data each time would be
highly inefficient. Most sorting algorithms fall into this
category, as they involve a large number of intermediate swaps. If
each of those swaps generated a brand new array, it would be very
slow with huge amounts of memory allocation.
Instead, Haskell provides the ST
type, which allows
for local mutations. While within an ST
block,
you can directly mutate local variables, such as mutable vectors.
But none of those mutated values can escape the ST
block, only immutable variants. To see how this works, look at this
Haskell code (save it to Main.hs
and run with
stack Main.hs
using the Stack build
tool):
#!/usr/bin/env stack
import Data.Vector (Vector, fromList, modify, freeze, thaw)
import Data.Vector.Algorithms.Insertion (sort)
import Control.Monad.ST
immutableSort :: Vector Int -> Vector Int
immutableSort original = runST $ do
mutableVector <- thaw original
sort mutableVector
freeze mutableVector
main = do
let unsortedVector = fromList [1, 4, 2, 0, 8, 9, 5]
sortedVector = immutableSort unsortedVector
print unsortedVector
print sortedVector
The immutableSort
function takes an immutable
vector of integers, and returns a new immutable vector of integers.
Internally, though, it runs everything inside an ST
block. First we thaw the immutable vector into a mutable
copy of the original. Now that we have a fresh copy, we're free to
- within the ST
block - modify it to our heart's
content, without impacting the original at all. To do this, we use
the mutating sort
function. When we're done, we
freeze that mutable vector into a new immutable vector,
which can be passed outside of the ST
block.
(I've also included a shorter version of the function which uses
the modify
function to automate the freezing and
thawing. Under the surface, it's doing almost exactly the same
thing... see extra credit at the bottom for more details.)
Using this technique, we get to have our cake and eat it too: an
efficient sorting algorithm (insertion sort) based on mutations to
a random-access vector, while maintaining the invariant that our
original vector remains unchanged.
Parallels between Docker and functional programming
After analyzing both Dockerfiles and the ST type, I think we can
draw some interesting parallels. Both techniques accept that there
are some things which are either easier or more efficient to do
with direct mutation. But instead of throwing out the baby with the
bathwater, they both value immutability as a goal. To achieve this,
both of them have the concept of constrained mutation: you
can only mutate in some specific places.
There's another interesting parallel to be observed: both Docker
and functional programming hide some mutation from the user.
For example, when you code 2 + 3
, under the surface
your compiler is generating something like:
- Write the value
2
to a machine register
- Write the value
3
to another machine register
- Perform the ADD machine instruction
- Copy the result in the output machine register to some location
in memory
All four of these steps are inherently mutating the state of
your machine, but you probably never think about that. (This
applies to all common programming languages, not just
functional languages.) While mutation is happening all the time,
we'd often rather not think about it, and instead focus on the
higher level goal (in this case: add two numbers together).
When you launch a Docker container, Docker is making a lot of
mutating changes. When you execute docker run busybox echo
Hello World!
, Docker creates a new control group (c-group),
creates some temporary files, forks processes, and so on. Again,
each of these actions is inherently a state mutation, but taken as
a whole, we can view the sum total as an immutable action that uses
a non-changing file system to run a command in an isolated
environment that generates some output on the command line.
Of course, you can also use Docker to run mutating commands,
such as bind-mounting the host file system and modifying files.
Similarly, from within a functional programming language you can
cause mutations of similar magnitude. But that's up to you; the
system itself tries to hide away a bunch of intermediate mutations
as a single, immutable action.
Further insights
I always enjoy finding a nexus between two seemingly unrelated
fields. While the line of reasoning that brought them there are
quite distinct, I'm very intrigued that both the devops and
functional programming worlds seem to be thriving today on
immutability. I'd be interested to hear others' experiences with
similar intersections between these worlds, or other worlds.
FP Complete is regularly in the business of combining modern devops practices with cutting edge functional programming. If you'd like
to learn more, check out our consulting
offerings or reach out for a free consultation.
If you're interested in learning more about Haskell, check out
our Haskell syllabus.
I made a comment above about "almost the same thing" with the
two versions of immutable sort. The primary difference is in
safe versus unsafe freezing. In our longer version,
we're using the safe variants of both freeze and thaw, which
operate by making a new copy of the original buffer. In the case of
thaw
, this ensures that the original, immutable
version of the vector is never modified. In the case of
freeze
, this ensures that we don't create a
falsely-immutable vector, which can have its values changed when
the original, mutable vector is tweaked.
Based on this, our long version of the function does the
following operations:
- Create a new memory buffer the same size as the original. Let's
call this buffer A.
- Copy the values into A from the original.
- Sort the values inside A using mutation.
- Create a new memory buffer of the same size. Let's call this
buffer B.
- Copy the values from A into B.
- Make B immutable and return it.
But if you pay close attention, that intermediate memory buffer
A can never be modified after the end of our ST block, and
therefore making that extra B buffer and copying into it is
unnecessary. Therefore, the modify
helper function
does an unsafe freeze on the A memory buffer, avoiding the
unneeded allocation and copy. While this operation may be unsafe in
general, we know in our usage it's perfect safe. This is another
great tenet of functional programming: wrapping up operations which
may be dangerous on their own into helper functions that guarantee
safety.
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