How to make a programming language

What actually happens in an interpreter when it is evaluating your code? Presumably some very impressive voodoo understandable only to doctors of computer science. The kind of people who are fluent in C++ and can blast through the Euler Project in Haskell before their morning Weetabix.

Right? Wrong. It’s actually relatively straightforward and an interpreter can be built by mere mortals such as you and I. Let’s explore some basic language interpreter concepts in Elixir, a nice friendly high level language. No pointer arithmetic or segfaults here!

The first thing to know is that all code syntax can be expressed as a tree. Take the expression below, in which + is called with the numbers 1 and 2.

This can be expressed as a tree with + as the root nodes, and the numbers as its child nodes.

1 + 2
call: +
├── number: 1
└── number: 2

We can do the same for more complex expression.

1 + 2 == 3
call: ==
├── call: +
│    ├── number: 1
|    └── number: 2
└── number: 3

Here the == is the root node and it has two children nodes, a call to + and the number 3. The call to + has two child nodes, the number 1 and the number 2.

All source code can be expressed as a tree data structure (an abstract syntax tree to be precise). Let’s implement this in Elixir using a struct for each type of node we’ve seen so far, as well as a boolean node, which would be the result of ==.

defmodule Number do
defstruct :value
def new(value), do: %Number{value: value}
end

defstruct :left, :right
def new(left, right), do: %Add{left: left, right: right}
end

defmodule Equals do
defstruct :left, :right
def new(left, right), do: %Equals{left: left, right: right}
end

defmodule Boolean do
defstruct :value
def new(value), do: %Boolean{value: value}
end

Now we can model 1 + 1 like this:

Number.new(1))

And we can model 1 + 2 == 4 like this:

Number.new(2)),
Number.new(4))

Now we can express programs in our code, providing those programs do nothing other than add numbers and check for equality. How do we go one step further and execute this tree? First we need to think about what happens when we step through code.

One way to think about this is to model it as the tree going through a series of reduction steps. A node tries to reduce its first child, and if it can then that’s one reduction step. If not it tries to reduce its next child, and so on until it has no more children to reduce, after that the only step is to reduce itself.

This all seems a bit abstract, so let’s look at 1 + 2 == 4 again.

First == tried tries to reduce its first child, which is the left hand side expression 1 + 2.

+ can be reduced, but first it needs to try and reduce its children. 1 and 2 are numbers, which can’t be reduced, so + reduces itself it a number by summing the value of its two children, making 3.

That’s one reduction step. The expression is now 3 == 4.

For the next step == tries to reduce its first child, 3, but cannot as numbers cannot be reduced. Then it tries to reduce its second child, 4, but again it cannot. Now there are no more children to reduce, it reduces itself to false by checking if the children are equal.

# Initial expression
1 + 2 == 4

# Reduce step 1
3 == 4

# Reduce step 2
false

Or in the Elixir AST:

# Initial expression
Number.new(2)),
Number.new(4))

# Reduce step 1
Equals.new(Number.new(3),
Number.new(4))

# Reduce step 2
Boolean.new(false)

To teach the Elixir AST how to reduce I’m doing to define an Elixir protocol and then implement it for each of the structs (it’s a bit like an interface in an OO language).

This protocol will have a function reduce, which takes a node and returns either a reduced node, or the atom :noop, signifying that it cannot be reduced.

defprotocol Node do
def reduce(node)
end

Numbers cannot be reduced, so they return :noop.

defimpl Node, for: Number do
def reduce(_number) do
:noop
end
end
result = Node.reduce(Number.new(1))
assert result == :noop

And the same for Boolean.

defimpl Node, for: Boolean do
def reduce(_boolean) do
:noop
end
end
result = Node.reduce(Boolean.new(true))
assert result == :noop

Add is more complex. It tries to reduce each of its children, and then reduces itself if they both :noop.

{:ok, reduced} ->
:noop ->
{:ok, reduced} ->
:noop ->
end
end
end
end
assert result == {:ok, Number.new(2)}

This code’s pretty grim, so I’ll use Elixir’s monadic with to tidy it up a bit. This code does exactly the same thing, except the happy path is written first, and then the other paths are in the else block.

with {:left, :noop} <- {:left, Node.reduce(add.left)},
{:right, :noop} <- {:right, Node.reduce(add.right)} do
else
{:left, {:ok, reduced}} ->

{:right, {:ok, reduced}} ->
end
end
end

And then lastly there’s Equals, which is pretty much the same as Add.

defimpl Node, for: Equals do
with {:left, :noop} <- {:left, Node.reduce(add.left)},
{:right, :noop} <- {:right, Node.reduce(add.right)} do
else
{:left, {:ok, reduced}} ->

{:right, {:ok, reduced}} ->
end
end
end

Can you spot the difference? It’s just this line in the middle.

Maybe there’s an abstraction to be extracted here. For now I’ll leave it as it is- duplication is better than the wrong abstraction.

Right, so if we can reduce expressions by a single step the only thing left to do is perform this step repeatedly until the expression can be reduced no more.

defmodule Program do
def run(expression) do
case Node.reduce(expression) do
{:ok, reduced} ->
run(reduced)
:noop ->
expression
end
end
end

And if we try it out…