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| title | date | draft | tags | |||
|---|---|---|---|---|---|---|
| Compiling a Functional Language Using C++, Part 3 - Operations On Trees | 2019-08-06T14:26:38-07:00 | true |
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I called tokenizing and parsing boring, but I think I failed to articulate the real reason that I feel this way. The thing is, looking at syntax is a pretty shallow measure of how interesting a language is. It's like the cover of a book. Every language has one, and it so happens that to make our "book", we need to start with making the cover. But the content of the book is what matters, and that's where we've arrived now. We must make decisions about our language, and give meaning to programs written in it. But before we can give our programs meaning, we need to make sense of the current domain of programs that we receive from our parser. Let's consider a few wonderful examples.
defn main = { plus 320 6 }
defn plus x y = { x + y }
This is a valid program, as far as we're concerned. But are all programs we get from the parser valid? See for yourself:
data Bool = { True, False }
defn main { 3 + True }
Obviously, that's not right. The parser accepts it - this matches our grammar. But giving meaning to this program is not easy, since we have no clear way of adding 3 and some data type. Similarly:
defn main { 1 2 3 4 5 }
What is this? It's a sequence of applications, starting with 1 2. Numbers
are not functions. Their type is Int, not Int -> a. I can't even think of a type 1
would need to have for this program to be valid.
Before we give meaning to programs in our language, we'll need to toss away the ones that don't make sense. To do so, we will use type checking. During the process of type checking, we will collect information about various parts of our abstract syntax trees, classifying them by the types of values they create. Using this information, we'll be able to throw away blatantly incorrect programs.
Basic Type Checking
You may have noticed in the very first post that I have chosen to avoid polymorphism. This will significantly simplify our type checking algorithm. If a more robust algorithm is desired, take a look at the Hindley-Milner type system. Personally, I enjoyed this section of Write You a Haskell.
Let's start with the types of constants - those are pretty obvious. The constant 3 is an integer,
and we shall mark it as such: 3 :: Int. Variables seem like the natural
next step, but they're fairly different. Without outside knowledge, all we can do is say
that a variable has some type. If we stick with Haskell's notation (used in polymorphic types),
we can say a variable has a type a, where a can be replaced with any other type. If we
know more information, like the fact that x was declared to be an integer, we can instead
say that. This tells us that throughout type checking we'll have to keep some kind
of record of names and their associated types.
Next, let's take a look at functions, which are admittedly more interesting
than the previous two examples. I'm not talking about the case of seeing something
like a function name f. This is the same as the variable case - we don't even know
it's a function unless there is context, and if there is context, then that
context is probably the most useful information we have. I'm talking about
something like the application of a function to another value, in the form
f x. In this case, we know that f :: a -> b, a function from something
to something. However, we know even more. For this program to be correct,
the a in f :: a -> b, and the type of x (let's call it c), must
be compatible. In order to do that, we will use what can be considered
simplified unification.
Conceptually, what this means is that we will attempt to perform substitutions
in various equations in search of a solution.
Basic Examples
Let's try an example. We'll try to determine the type of the following expression, and extract any other information from this expression that we might use later.
foo 320 6
In out parse tree, this will be represented as (foo 320) 6, meaning
that the outermost application will be at the top. Let's assume
we know nothing about foo.
To figure out the type of the application, we will need to know the types
of the thing being applied, and the thing that serves as the argument.
The latter is easy: the type of 6 is Int. Before we proceed
into the left child of the application, there's one more
piece of information we can deduce: since foo 320 is applied to
an argument, it has to be of type a -> b.
Let's proceed to the left child. It's another application, this time
of foo to 320. Again, the right child is simple: the type of
320 is Int. Again, we know that foo has to have a type
c -> d (we're using different variable names to avoid ambiguity).
Now, we need to combine the pieces of information that we have. Since
foo :: c -> d, we know that its first parameter must be of
type c. We also know that its first parameter is of type Int.
The only way for both of these to be tree is for c = Int.
This also tells us that foo :: Int -> d. Finally,
since foo has now been applied to its first argument,
we know that the foo 320 :: d.
We've done what we can from this innermost application; it's time to return
to the outermost one. We now know that the left child is of type d, and
that it also has to be of type a -> b. The only way for this to be true
is for d = a -> b. So, foo 320 is a function from a to b.
Again, we can conclude the first parameter has to be of type
a. We also know that the first parameter is of type Int. Clearly,
this means that a = Int. After the application, we know
that the whole expression has some type d.
Let's revisit what we know about foo. Last time we checked in on it,
foo was of type Int -> d. But since we know that d = a -> b,
and that a = Int, we can now say that foo :: Int -> Int -> b.
We haven't found any issues with the expression, and we learned
some new information about the type of foo. Awesome!
Let's apply this to a simplified example from the beginning of this post. Let's check the type of:
1 2
Once again, the application is what we see first. The right child
of the application, just like in the previous example, is Int.
We also kno that since 1 is being applied as a function,
its type must be a -> b. However, we also know that the left
child, being a number, is also of type Int! There's no
way to combine a -> b with Int, and thus, there is no solution
we can find for the type of 1 2. This means our program is invalid.
We can toss it away, give an error, and exit.
Some Notation
If you go to the Wikipedia page on the Hindley-Milner type system, you will see quite a lot of symbols and greek letters. This is a good thing, because the way that I presented to you the rules for figuring out types of expressions is very verbose. You have to read several paragraphs of text, and that's only for the first three logical rules! If you're anything like me, you want to be able to read just the important parts, and with some notation, I'll be able to show you these important parts concisely, while continuing to explain the rules in detail in paragraphs of text.
Let's start with inference rules. An inference rule is an expression in the form:
\\frac{A\_1 \\ldots A\_n} {B\_1 \\ldots B\_m}
This reads, "given that the premises \(A_1\) through \(A_n\) are true, it holds that the conclusions \(B_1\) through \(B_n\) are true".
For example, we can have the following inference rule:
\\frac
{\\text{if it's cold, I wear a jacket} \\quad \\text{it's cold}}
{\\text{I wear a jacket}}
Since you wear a jacket when it's cold, and it's cold, we can conclude that you will wear a jacket.
When talking about type systems, it's common to represent a type with \(\tau\). The letter, which is the greek character "tau", is used as a placeholder for some concrete type. It's kind of like a template, to be filled in with an actual value. When we plug in an actual value into a rule containing \(\tau\), we say we are instantiating it. Similarly, we will use \(e\) to serve as a placeholder for an expression (matched by our \(A_{add}\) grammar rule from part 2). Next, we have the typing relation, written as \(e:\tau\). This says that "expression \(e\) has the type \(\tau\)".
Alright, this is enough to get us started with some typing rules. Let's start with one for numbers. If we define \(n\) to mean "any expression that is a just a number, like 3, 2, 6, etc.", we can write the typing rule as follows:
\\frac{}{n : \\text{Int}}
There's nothing above the line -
there are no premises that are needed for a number to
have the type Int.
Now, let's move on to the rule for function application:
\\frac
{e_1 : \\tau\_1 \\rightarrow \\tau\_2 \\quad e_2 : \\tau_1}
{e_1 \\; e_2 : \\tau\_2}
It's the variable rule that forces us to adjust our notation. Our rules don't take into account the context that we've already discussed. Let's fix that! It's convention to use the symbol \(\Gamma\) for the context. We then add notation to say, "using the context \(\Gamma\), we can deduce that \(e\) has type \(\tau\)". We will write this as \(\Gamma \vdash e : \tau\).
But what is our context? It's just a set of pairs in the form \(x : \tau\), where \(x\) represents a variable name. Each pair tells us that the variable \(x\) is known to have a type \(\tau\).
Since \(\Gamma\) is just a regular set, we can write \(x : \tau \in \Gamma\), meaning that in the current context, it is known that \(x\) has the type \(\tau\).
Let's update our rules with this new addition.
The integer rule just needs a slight adjustment:
\frac{}{\\Gamma \\vdash n : \\text{Int}}
The same is true for the application rule:
\frac
{\\Gamma \\vdash e_1 : \\tau\_1 \\rightarrow \\tau\_2 \\quad \\Gamma \\vdash e_2 : \\tau_1}
{\\Gamma \\vdash e_1 \\; e_2 : \\tau\_2}
And finally, we can represent the variable rule:
\\frac{x : \\tau \\in \\Gamma}{\\Gamma \\vdash x : \\tau}
In these three expressions, we've captured all of the rules that we've seen so far. It's important to know that These rules leave out the process of unification altogether: we use unification to find types that satisfy the rules.
Checking More Expressions
So far, we've only checked types of numbers, applications, and variables. Our language has more than that, though!