Wikipedia says:

fun factorial n = 
    if n = 0 then 1 else n * factorial (n-1) 

A Standard ML compiler is required to infer the static type int -> int of this function without user-supplied type annotations. I.e., it has to deduce that n is only used with integer expressions, and must therefore itself be an integer, and that all value-producing expressions within the function return integers.

I don't understand how a compiler could infer this. It sounds like SML is essentially solving the halting problem for the factorial function, and showing that it only halts on positive integer inputs.

Am I missing something?


The classic Hindley-Milner type inference algorithm requires each literal value to be unambiguously inferable to a type: 0 is an int, while 0.0 is a real. Haskell's type system has been enhanced to include a certain kind of bounded polymorphism, so that 0 can have any type that is a number (more precisely, of any type that is an instance of the Num type class: 0 is of type Num a => a).

See the wikipedia article on Hindley-Milner for further details on the algorithm (the article explanations are much better than anything I could write on that topic).

  • $\begingroup$ thinking about it, I wonder if I actually answer your question. $\endgroup$ – didierc Feb 2 '13 at 3:01
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    $\begingroup$ if ... then ... else ... is more accurately seen as an expression that should take three (sub)expressions, one having bool type and the other two having matching but otherwise unrestricted types. Otherwise, yeah, I think you gave a sufficient answer. $\endgroup$ – BlueBomber Feb 2 '13 at 3:43
  • $\begingroup$ I agree, I tried to turn the if expression in a function, to avoid adding a special case to handle in the algorithm. That might be not necessary for the purpose of this answer, so you have a point. $\endgroup$ – didierc Feb 2 '13 at 4:01
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    $\begingroup$ My question is: how does SML know the function isn't of type Real -> Real? Since 0 and 1 are both real numbers, it can't just look at the constants' types. $\endgroup$ – Xodarap Feb 2 '13 at 13:57
  • $\begingroup$ Now I understand better your question! $\endgroup$ – didierc Feb 2 '13 at 15:35

I think it is important to realise what the type $\mathtt{int \rightarrow int}$ means and does not mean.

It does not mean that a program P having that type must return an integer, contrary to what the original poster said. Instead it means if P terminates on a given input (of type int), P terminates by returning an integer. This way the halting problem is not an issue.

In addition, well-typing guarantees that programs don't suffer run-time type errors even when not terminating.

The original Milner, Damas paper Principal type-schemes for functional programs is very readable. With 6 pages it's also short.

Finally, it is highly instructive, and not much effort, to implement a type-inferencer for a toy language. I recommend doing this to really understand how types work in programming languages.

  • $\begingroup$ Then why doesn't the type read Real -> Real? Surely the equivalent statement is true: "If P terminates on an input of type Real, P terminates by returning a Real" $\endgroup$ – Xodarap Feb 2 '13 at 13:58
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    $\begingroup$ @Xodarap that's because subterms like 0 and 1 are not of type real. SML has no subtyping (IIRC). $\endgroup$ – Martin Berger Feb 2 '13 at 14:47
  • $\begingroup$ @Xodarap, generally speaking, it depends on the type system of the language. Some other languages' type systems might do just that: subsume $int$ to $real$ if necessary to do that, but it would require a type system that recognizes that $int$s are $real$s, which generally implies that the language supports subtyping. $\endgroup$ – BlueBomber Feb 4 '13 at 17:06

Working with types is not as "hard" as working with values with respect to computability. You can know statically that a function will return a bool without knowing statically that it will return true, for example. Moreover, it's entirely possible and sometimes trivially easy to prove that a specific given function halts with a certain input, but this is a different problem than writing a general algorithm that determines whether or not any other given function will halt on an input.

With respect to how type inference works, let me recommend some sources which may help:

  1. Unification on Wikipedia
  2. Type inference on Wikipedia
  3. Types and Programming Languages by Benjamin C. Pierce (chapter 22, especially)
  4. Hindley-Milner method on Wikipedia.

The type inference algorithm happens essentially in two phases: constraint generation, and unification. During constraint generation, the AST of the program is traversed and a set of type constraints is generated.

For example, if somewhere in the program there is the application expression myfun (3), the inference algorithm could infer that the type of myfun is $\tt{int} \rightarrow \alpha$. If in another part of the program the expression 4 + myfun(y) is encountered, the constraint that myfun has type $\beta \rightarrow \tt{int}$ is generated. During the unification step, such constraints will be (successfully or unsuccessfully) unified. During this phase the set of constraints might include

  • $ \tt{myfun} : \tt{int} \rightarrow \alpha$
  • $ \tt{myfun} : \beta \rightarrow \tt{int}$
  • $ \tt{y} : \tt{int}$
  • $ \tt{y} : \beta$

From this set (really system) of constraints, we could infer that $\alpha = \tt{int}$ and $\beta = \tt{int}$, and everything would be consistent and the program would type check. It's possible for unification to fail, though, depending on what else is in the set of constraints.

  • $\begingroup$ The sample code has the following feature: no type errors will be generated if it is called with say 1.5, as you can compare 1.5 to 1 easily. However, the function will fail to terminate if it is called with 1.5. How does SML know this? $\endgroup$ – Xodarap Feb 2 '13 at 14:01
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    $\begingroup$ @Xodarap, in your sample code, SML doesn't infer that the type is int because it is what will make calls to factorial terminate; it infers that because you used the constant value expression 1 instead of 1.0. $\endgroup$ – BlueBomber Feb 6 '13 at 16:14

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