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There are more types than classes

As programmers, we are very incautious with our use of the word “type”. The concept of “type” is sufficiently abstract and specific that we are tempted to understand it by analogy, so much that we begin to confuse analogy with sameness.

The colloquial “runtime type”, a fair approximation of “class”, makes it tempting to equate types with “classes, interfaces, traits, that sort of thing”, which I will name classes for the rest of this article. But they aren’t the same. The type system is much richer and more interesting than the class system, even in Java.

To appreciate this richness, we must stop thinking of types as classes and stop drawing conclusions from that weak analogy. Luckily, the compiler will readily reveal how unlike classes types are, if we ask it some simple questions.

One value with class, many variables with type

val greeting: String = "hi there!"

Here I have constructed a String and assigned it to a variable. (I have also constructed the char array in the String and various other details, but immediately handed those off to the String and forgotten about them.) This value has class String. It has several classes, really.

  1. String
  2. java.io.Serializable
  3. CharSequence
  4. Comparable[String]
  5. Object/AnyRef

That seems like a lot of classes for one value. And they are genuine classes of greeting, though 2-5 are all implied by #1.

greeting also has all five of these types. We can ask the compiler to verify that this type truth holds, entirely separately from the class truth.

scala> (greeting: String, greeting: java.io.Serializable,
        greeting: CharSequence, greeting: Comparable[String],
        greeting: AnyRef)
res3: (String, java.io.Serializable, CharSequence,
       Comparable[String], AnyRef) =
  (hi there!,hi there!,hi there!,hi there!,hi there!)

So we have exhausted the classes, but aren’t quite done with types.

scala> greeting: greeting.type
res4: greeting.type = hi there!

greeting.type is not like the other five types we just tested. It is a strict subtype of String, and has no class with the same name.

// If and only if call compiles, A is a subtype of B.
def conformance[A, B >: A]: Unit = ()

scala> conformance[greeting.type, String]

scala> conformance[String, greeting.type]
<console>:14: error: type arguments [String,greeting.type] do not conform
              to method conformance's type parameter bounds [A,B >: A]

Fine, we can accept that object identity is represented at the type level without our universe imploding, by inventing the theory that this is about object identity; hold on, though:

scala> val salutation = greeting
salutation: String = hi there!

Fine, salutation is just another name for greeting, right?

scala> conformance[salutation.type, String]

scala> implicitly[greeting.type =:= salutation.type]
<console>:14: error: Cannot prove that greeting.type =:= salutation.type.

Now we have seven. I’ll spare you spelling out the induction: each new variable defined like salutation will yield a new alias with a distinct type. This is not about objects; this is about variables!

// find a type for the literal "hi there!"
scala> val literalHiThere = shapeless.Witness("hi there!")
literalHiThere: shapeless.Witness.Aux[String("hi there!")] = shapeless.Witness$$anon$1@1d1537bb

scala> conformance[greeting.type, literalHiThere.T]
<console>:15: error: type arguments [greeting.type,literalHiThere.T] do not conform
              to method conformance's type parameter bounds [A,B >: A]

scala> conformance[literalHiThere.T, greeting.type]
<console>:15: error: type arguments [literalHiThere.T,greeting.type] do not conform
              to method conformance's type parameter bounds [A,B >: A]

As local variables are a strictly compile-time abstraction, and we have anyway seen that the numbers don’t match up, that should be the end of the “types are classes” confusion for you. But maybe this is just some Scala oddity! And anyhow I haven’t even begun to demonstrate the overwhelming richness of the type model as it blindingly outshines the paucity of the class model. Let’s go further.

No values, infinite types: method type parameters

To our small program of a greeting, we can add a small method.

def pickGreeting[G](grt: G, rand: Int) = grt

scala> pickGreeting(greeting, 42)
res9: String = hi there!

It seems like G must be String, because the argument passed to pickGreeting is a string, and in that case so must its return value be, according to the implementation. And from the perspective of this call, outside pickGreeting’s implementation, it is String indeed.

But that implementation’s perspective matters, too; it is also part of our program. And it sees things quite differently. We can ask its thoughts on the matter by adding to its body

def pickGreeting[G](grt: G, rand: Int) = {
  implicitly[G =:= String]
  grt
}

<console>:12: error: Cannot prove that G =:= String.
         implicitly[G =:= String]
                   ^

In fact, G bears no direct relationship to String at all.

// replace implicitly with
conformance[G, String]

<console>:13: error: type arguments [G,String] do not conform
              to method conformance's type parameter bounds [A,B >: A]
         conformance[G, String]
                    ^

// or with
conformance[String, G]

<console>:13: error: type arguments [String,G] do not conform
              to method conformance's type parameter bounds [A,B >: A]
         conformance[String, G]
                    ^

Let’s apply the pigeonhole principle. Imagine that you had a list of every class that ever was or ever will be. Imagine that, somehow, all of these classes, from String to AbstractFactoryMethodProxyBuilder, were on your classpath, available to your program.

Next, imagine that you had the time and inclination to try the =:= test with every last one of these classes.

implicitly[G =:= javax.swing.JFrame]
implicitly[G =:= AbstractFactoryMethodProxyBuilder]
// ad [in]finitum

Your search will be futile; every class on your list-of-every-class will give the same compiler error we got with String.

So, since G is not equal to anything on this list, it must be something else that doesn’t appear on the list. Because this list contains all classes, G must be something other than a class.

It must not necessarily be anything

It seems like it might be convenient to say “well, in this program G is only ever String by substitution, so therefore it is, even if the compiler doesn’t see that.” However, thinking like this misses out on the second key advantage of type parameterization, the one not based on multiplicity of substitution, or the type-safety of callers: blindness.

The implementation of type-parameterized classes and methods are required to treat each type parameter uniquely, uniformly, and without prejudice. The compiler enforces this by making the implementation blind to what that parameter, like G, could be. It can only use what the caller, the “outside”, has told it about G—arguments whose types contain G, like List[G], (G, G) => G, or G itself, like the argument to pickGreeting. This is information-hiding at the type level; if you find information-hiding a useful tool for implementing correct programs, you will find the same of the fresh, unique, and mysterious types induced by each introduction of a type parameter.

Each operation a language permits by default, not via an argument, on values of a type parameter is a leak in this abstraction. This includes testing the value’s class, converting to string, and comparing to other values of supposedly utter mystery for equality. The ability to create a “default” value is also a leak. A function is always permitted to ask only that of these that it needs from the caller; make them default, and this design choice is taken away. That is why Object#equals is little better for type-safety than reflection-based calls, and why total type erasure is a desirable feature rather than a design flaw—plugging these leaks gives the programmer as much freedom to abstract by information-hiding as she wishes.

How many calls are there?

Put another way, when implementing the code in the scope of a type parameter, your implementation must be equally valid for all possible G substitutions, including the ones that haven’t been invented yet. This is why we call it universal quantification.

But it is not merely each declaration of a type parameter that yields a distinct type—each call does! Consider two consecutive calls to pickGreeting.

pickGreeting(greeting, 42)
pickGreeting(33, 84)

Externally, there are two G types. However, the possibility of writing this demands another level of uniqueness treatment when typechecking pickGreeting’s definition: whatever G is now, like String, it might be something else in the next call, like Int in the above example. With recursion, it might even be two different things at the same time. There’s nothing to hold this at two, either: there may be an unbounded number of substitutions for a given type parameter within a single program, at a single point in time.

While G may be the same between two invocations of pickGreeting, it might not. So we have no choice but to treat the G types of each call as separate types. There may be infinitely many calls, so there are so many types.

Incidentally, the same happens for singleton types. Each time val greeting comes into scope, it induces a separate singleton type. It is easy enough to arrange for an unbounded number of scope entries in a particular program. This isn’t so practical as the type parameter phenomenon, though.

More types from variable copies

Suppose we’d like to wait a while to compute our greeting. We can define a type-and-class to represent that conveniently.

// like Coyoneda Id, if that helps
sealed abstract class Later[A] {
  type I
  val i: I
  val f: I => A
}

def later[In, A](now: In)(later: In => A)
  : Later[A] = new Later[A] {
    type I = In
    val i = now
    val f = later
}

val greeting3 = later(3){
  n => List.fill(n)("hi").mkString(" ")
}

How many classes are involved here, in the type of greeting3?

  1. Later, obviously;
  2. Function1, the greeting3.f overall class;
  3. String, the output type of greeting3.f;
  4. Int, the I type.

How many types?

The first difference is that greeting3.I is not Int.

scala> implicitly[greeting3.I =:= Int]
<console>:13: error: Cannot prove that greeting3.I =:= Int.

They are unrelated for much the same reason as G was unrelated to String in the previous example: the only things code following val greeting3 may know are those embodied in the greeting3.i and greeting3.f members. You can almost think of them as “arguments”.

But that’s not all.

val salut3 = greeting3

scala> greeting3.f(greeting3.i)
res11: String = hi hi hi

scala> salut3.f(salut3.i)
res12: String = hi hi hi

scala> greeting3.f(salut3.i)
<console>:14: error: type mismatch;
 found   : salut3.i.type (with underlying type salut3.I)
 required: greeting3.I
       greeting3.f(salut3.i)
                          ^

scala> implicitly[greeting3.I =:= salut3.I]
<console>:14: error: Cannot prove that greeting3.I =:= salut3.I.

Just like every call to pickGreeting induces a new G type, each simple val copy of greeting3 will induce a new, unique I type. It doesn’t matter that they’re all the same value; this is a matter of variables, not values, just as with singleton types.

But that’s still not all.

One value with class, many variable references with types

The preceding is more delicate than it seems.

var allo = greeting3

scala> allo.f(allo.i)
<console>:13: error: type mismatch;
 found   : Later[String]#I
 required: _1.I where val _1: Later[String]
       allo.f(allo.i)
                   ^

All we have done differently is use a mutable var instead of an immutable val. Why is this enough to throw a wrench in the works?

Suppose you had another value of the Later[String] type.

val bhello = later("olleh")(_.reverse)

The I substitution here is String. So the f takes a String argument, and the I is a String.

bhello is of a compatible type with the allo var. So this assignment will work.

allo = bhello

In a sense, when this mutation occurs, the I type also mutates, from Int to String. But that isn’t quite right; types cannot mutate.

Suppose that this assignment happened in the middle of that line of code that could not compile. We could imagine the sequence of events, were it permitted.

  1. allo.f (which is greeting3.f) evaluates. It is the function (n: Int) => List.fill(n)("hi").mkString(" ").
  2. The allo = bhello assignment occurs.
  3. allo.i (which is bhello.i) evaluates. It is the string "olleh".
  4. We attempt to pass "olleh" as the (n: Int) argument to complete the evaluation, and get stuck.

Just as it makes no difference what concrete substitutions you make for G, it makes no difference whether such an assignment could ever happen in your specific program; the compiler takes it as a possibility because you declared a var. (def allo = greeting3 gets the same treatment, lest you think non-functional programs get to have all the fun here.) Each reference to allo gets a new I type member. That failing line of code had two allo references, so was working with two incompatible I types.

Since the number of references to a variable in a program is also unbounded…you get the picture.

This also occurs with existential type parameters, which are equally expressive to type members. Accordingly, Java also generates new types from occurrences of expressions of existential type.

How do we tell the two apart?

All of this is simply to say that we must be working with two separate concepts here.

  1. The runtime shape and properties of the values that end up flying around when a program actually runs. This we call class.
  2. The compile-time, statically-discoverable shape and properties of the expressions that fly around when a program is written. This we call type.

The case with var is revealing. Maybe the I type will always be the same for a given mutable variable. But demonstrating that this holds true for one run of the program (#1, class) isn’t nearly good enough to prove that it will be true for all runs of the program (#2, type).

We refuse to apply the term “type” to the #1, ‘class’ concept because it does not live up to the name. The statement “these two types are the same” is another level of power entirely; “these two values have the same class” is extraordinarily weak by comparison.

It is tempting to use the term “runtime type” to refer to classes. However, in the case of Scala, as with all type systems featuring parametric polymorphism, classes are so dissimilar to types that the similar-sounding term leads to false intuition, not helpful analogy. It is a detriment to learning, not an aid.

Types are compile-time, and classes are runtime.

When are types real?

The phase separation—compile-time versus runtime—is the key to the strength of types in Scala and similar type systems. The static nature of types means that the truths they represent must be universally quantified—true in all possible cases, not just some test cases.

We need this strength because the phase separation forbids us from taking into account anything that cannot be known about the program without running it. We need to think in terms of “could happen”, not “pretty sure it doesn’t”.

How do classes give rise to types?

There appears to be some overlap between the classes of greeting and its types. While greeting has the class String, it also has the type String.

We want types to represent static truths about the expressions in a program. That’s why it makes sense to include a “model of the classes” in the type system. When we define a class, we also define an associated type or family of types.

When we use a class to construct a value, as in new Blob, we would like to assign as much specific meaning to that expression as we can at compile time. So, because we know right now that this expression will make a value of class Blob, we assign it the type Blob too.

How do the types disappear?

There’s a common way to throw away type information in Scala, especially popular in object-oriented style.

val absGreeting: CharSequence = greeting

absGreeting has the same value as greeting, so it has the same five classes. However, it only has two of those five types, because we threw away the other three statically. It has lost some other types, too, namely greeting.type, and acquired some new ones, namely absGreeting.type.

Once a value is constructed, the expression will naturally cast off the types specifying its precise identity, as it moves into more abstract contexts. Ironically, the best way to preserve that information as it passes through abstract contexts is to take advantage of purely abstract types—type parameters and type members.

scala> pickGreeting[greeting.type](greeting, 100)
res16: greeting.type = hi there!

While the implementation must treat its argument as being of the abstract type G, the caller knows that the more specific greeting.type must come out of that process.

How do the types come back?

There is a feature in Scala that lets you use class to get back some type information via a dynamic, runtime test.

absGreeting match {
  case hiAgain: String =>
    conformance[hiAgain.type, String] // will compile
}

The name “type test” for this feature is poorly chosen. The conclusion affects the type level—hiAgain is, indeed, proven statically to be of type String—but the test occurs only at the class level.

The compiler will tell you about this limitation sometimes.

def pickGreeting2[G](grt: G, rand: Int): G =
  ("magic": Any) match {
    case ok: G => ok
    case _ => sys error "failed!"
  }

<console>:13: warning: abstract type pattern G is unchecked
              since it is eliminated by erasure
           case ok: G => ok
                    ^

But reflecting the runtime classes back to compile-time types is a subtle art, and the compiler often can’t explain exactly what you got wrong.

def pickGreeting3[G](grt: G, rand: Int): G =
  grt match {
    case _: String =>
      "Surely type G is String, right?"
    case _ => grt
  }

<console>:14: error: type mismatch;
 found   : String("Surely type G is String, right?")
 required: G
             "Surely type G is String, right?"
             ^

I’ve touched upon this mistake in previous articles, but it’s worth taking at least one more look. Let’s examine how tempting this mistake is.

String is a final class. So it is true that G can contain no more specific class than String, if the first case matches. For example, given trait MyAwesomeMixin, G cannot be String with MyAwesomeMixin if this case succeeds, because that can’t be instantiated; you would need to create a subclass of String that implemented MyAwesomeMixin.

This pattern match isn’t enough evidence to say that G is exactly String. There are still other class-based types it could be, like Serializable.

pickGreeting3[java.io.Serializable](greeting, 4055)

Instead, it feels like this pattern match confirms Serializable as a possibility, instead of denying it.

But we don’t need G = String for this code to compile; we only need G >: String. If that was true, then "Surely type G is String, right?", a String, could simply upcast to G.

However, even G >: String is unproven. There are no subclasses of String, but there are infinitely many subtypes of String. Including the G created by each entry into pickGreeting3, every abstract and existential type bounded by String, and every singleton type of String variable definitions.

This mistake is, once again, confusing a demonstration of one case with a proof. Pattern matching tells us a great deal about one value, the grt argument, but very little about the type G. All we know for sure is that “grt is of type G, and also of type String, so these types overlap by at least one value.” In the type system, if you don’t know something for sure, you don’t know it at all.

Classes are a concrete source of values

In the parlance of functional Scala, concrete classes are often called “data constructors”.

When you are creating a value, you must ultimately be concrete about its class, at the bottom of all the abstractions and indirections used to hide this potentially messy detail.

def pickGreeting4[G]: G = new G

<console>:11: error: class type required but G found
       def pickGreeting4[G]: G = new G
                                     ^

You’ll have to do something else here, like take an argument () => G, to let pickGreeting4 construct Gs.

The truly essential role that classes play is that they encapsulate instructions for constructing concrete values of various types. In a safe program, this is the only feature of classes you’ll use.

In Scala, classes leave fingerprints on the values that they construct, without fail. This is merely an auxiliary part of their primary function as value factories, like a “Made in class Blah” sticker on the back. Pattern matching’s “type tests” work by checking this fingerprint of construction.

Most runtime “type test” mechanisms do not work for types

These fingerprints only come from classes, not types. So “type tests” only work for “classy” types, like String and MyAwesomeMixin. They also work for specific singleton types because construction also leaves an “object identity” fingerprint that the test can use.

The ClassTag typeclass does not change this restriction. When you add a ClassTag or TypeTag context bound, you also prevent that type parameter from working with most types.

scala> implicitly[reflect.ClassTag[greeting3.I]]
<console>:13: error: No ClassTag available for greeting3.I

As such, judicious use of ClassTag is not a great solution to excessive use of type tests in abstract contexts. There are so many more types than classes that this is to confine the expressivity of your types to a very small, class-reflective box. Set them free!

“But doesn’t Python/JavaScript/&c have both types and classes at runtime?”

In JavaScript, there’s a very general runtime classification of values called “type”, meant to classify built-in categories like string, number, and the like.

>> typeof "hi"
"string"
>> typeof 42
"number"
>>> typeof [1, 'a']
"object"

Defining a class with the new class keyword doesn’t extend this partition with new “types”; instead, it further subdivides one of those with a separate classification.

>> class Foo() {}
>> class Bar() {}
>> typeof (new Foo)
"object"
>> typeof (new Bar)
"object"
>> new Foo().constructor
function Foo()
>> new Bar().constructor
function Bar()

So, if you treat JavaScript’s definition of the word “type” as analogous to the usage in this article, then yes, JavaScript has “runtime types”.

But JavaScript can only conveniently get away with this because its static types are uninteresting. It has one type—the type of all values—and no opportunities to do interesting type-level modeling, at least not as part of the standard language.

Hence, JavaScript is free to repurpose the word “type” for a flavor of its classes, because our “types” aren’t a tool you make much use of in JavaScript. But when you come back to Scala, Haskell, the ML family, et al, you need a word for the static concept once again.

Thinking about types as just classes leads to incorrect conclusions

Setting aside the goal of principled definition of terms, this separation is the one that makes the most sense for a practitioner of Scala. Consider the practicalities:

Types and classes have different behavior, are equal and unequal according to different rules, and there are a lot more types than classes. So we need different words to distinguish them.

Saying “compile-time type” or “runtime type” is not a practical solution—no one wants to speak such an unwieldy qualifier every time they refer to such a commonly-used concept.

While I’ve given a sampling of the richness of the type system in this article, it’s not necessary to know that full richness to appreciate or remember the difference between the two: types are static and compile-time; classes are dynamic and runtime.

This article was tested with Scala 2.12.1, Shapeless 2.3.2, and Firefox 53.0a2.

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