If Statements, Loops and Recursion

If statements (actually, these are if expressions)

OCaml has an if statement with two variations, and the obvious meaning:

if boolean-condition then expression
if boolean-condition then expression else other-expression

Unlike in the conventional languages you'll be used to, if statements are really expressions. In other words, they're much more like boolean-condition ? expression : other-expression than like the if statements you may be used to.

Here's a simple example of an if statement:

# let max a b =
    if a > b then a else b;;
val max : 'a -> 'a -> 'a = <fun>

As a short aside, if you type this into the OCaml interactive toplevel (as above), you'll notice that OCaml decides that this function is polymorphic, with the following type:

max : 'a -> 'a -> 'a

And indeed OCaml lets you use max on any type:

# max 3 5;;
- : int = 5 # max 3.5 13.0;;
- : float = 13. # max "a" "b";;
- : string = "b"

This is because > is in fact polymorphic. It works on any type, even objects (it does a binary comparison).

[Note that the Pervasives module defines min and max for you.]

Let's look a bit more closely at the if expression. Here's the range function which I showed you earlier without much explanation. You should be able to combine your knowledge of recursive functions, lists and if expressions to see what it does:

# let rec range a b =
    if a > b then []
    else a :: range (a+1) b;;
val range : int -> int -> int list = <fun>

Let's examine some typical calls to this function. Let's start with the easy case of a > b. A call to range 11 10 returns [] (the empty list) and that's it.

What about calling range 10 10? Since 10 > 10 is false, the else-clause is evaluated, which is: 10 :: (range 11 10) (I've added the brackets to make the order of evaluation more clear). We've just worked out that range 11 10 = [], so this is: 10 :: []. Remember our formal description of lists and the :: (cons) operator? 10 :: [] is just the same as [ 10 ].

Let's try range 9 10:

range 9 10
→ 9 :: (range 10 10)
→ 9 :: [ 10 ]
→ [9; 10]

It should be fairly clear that range 1 10 evaluates to [ 1; 2; 3; 4; 5; 6; 7; 8; 9; 10 ].

What we've got here is a simple case of recursion. Functional programming can be said to prefer recursion over loops, but I'm jumping ahead of myself. We'll discuss recursion more at the end of this chapter.

Back, temporarily, to if statements. What does this function do?

# let f x y =
    x + if y > 0 then y else 0;;
val f : int -> int -> int = <fun>

Clue: add brackets around the whole of the if expression. It clips y like an electronic diode.

The abs (absolute value) function is defined in Pervasives as:

# let abs x =
    if x >= 0 then x else -x;;
val abs : int -> int = <fun>

Also in Pervasives, the string_of_float function contains a complex pair of nested if expressions:

let string_of_float f =
  let s = format_float "%.12g" f in
  let l = string_length s in
  let rec loop i =
    if i >= l then s ^ "."
    else if s.[i] = '.' || s.[i] = 'e' then s
    else loop (i+1) in
  loop 0

Let's examine this function. Suppose the function is called with f = 12.34. Then s = "12.34", and l = 5. We call loop the first time with i = 0.

i is not greater than or equal to l, and s.[i] (the ith character in s) is not a period or 'e'. So loop (i+1) is called, ie. loop 1.

We go through the same dance for i = 1, and end up calling loop 2.

For i = 2, however, s.[i] is a period (refer to the original string, s = "12.34"). So this immediately returns s, and the function string_of_float returns "12.34".

What is loop doing? In fact it's checking whether the string returned from format_float contains a period (or 'e'). Suppose that we called string_of_float with 12.0. format_float would return the string "12", but string_of_float must return "12." or "12.0" (because floating point constants in OCaml must contain a period to differentiate them from integer constants). Hence the check.

The strange use of recursion in this function is almost certainly for efficiency. OCaml supports for loops, so why didn't the authors use for loops? We'll see in the next section that OCaml's for loops are limited in a way which prevents them from being used in string_of_float. Here, however, is a more straightforward, but approximately twice as slow, way of writing string_of_float:

let string_of_float f =
  let s = format_float "%.12g" f in
  if String.contains s '.' || String.contains s 'e'
  then s
  else s ^ "."

Using begin ... end

Here is some code from lablgtk:

if GtkBase.Object.is_a obj cls then
  fun _ -> f obj
else begin
  eprintf "Glade-warning: %s expects a %s argument.\n" name cls;
  raise Not_found

begin and end are what is known as syntactic sugar for open and close parentheses. In the example above, all they do is group the two statements in the else-clause together. Suppose the author had written this instead:

if GtkBase.Object.is_a obj cls then
  fun _ -> f obj
  eprintf "Glade-warning: %s expects a %s argument.\n" name cls;
  raise Not_found

Fully bracketing and properly indenting the above expression gives:

(if GtkBase.Object.is_a obj cls then
   fun _ -> f obj
   eprintf "Glade-warning: %s expects a %s argument.\n" name cls
raise Not_found

Not what was intended at all. So the begin and end are necessary to group together multiple statements in a then or else clause of an if expression. You can also use plain ordinary parentheses ( ... ) if you prefer (and I do prefer, because I loathe Pascal :-). Here are two simple examples:

# if 1 = 0 then
      print_endline "THEN"
    else begin
      print_endline "ELSE";
      failwith "else clause"
ELSE Exception: Failure "else clause". # if 1 = 0 then print_endline "THEN" else ( print_endline "ELSE"; failwith "else clause" );;
ELSE Exception: Failure "else clause".

For loops and while loops

OCaml supports a rather limited form of the familiar for loop:

for variable = start_value to end_value do
for variable = start_value downto end_value do

A simple but real example from lablgtk:

for i = 1 to n_jobs () do
  do_next_job ()

In OCaml, for loops are just shorthand for writing:

let i = 1 in
do_next_job ();
let i = 2 in
do_next_job ();
let i = 3 in
do_next_job ();
let i = n_jobs () in
do_next_job ();

OCaml doesn't support the concept of breaking out of a for loop early i.e. it has no break, continue or last statements. (You could throw an exception and catch it outside, and this would run fast but often looks clumsy.)

The expression inside an OCaml for loop should evaluate to unit (otherwise you'll get a warning), and the for loop expression as a whole returns unit:

# for i = 1 to 10 do i done;;
Warning 10: this expression should have type unit. - : unit = ()

Functional programmers tend to use recursion instead of explicit loops, and regard for loops with suspicion since it can't return anything, hence OCaml's relatively powerless for loop. We talk about recursion below.

While loops in OCaml are written:

while boolean-condition do

As with for loops, there is no way provided by the language to break out of a while loop, except by throwing an exception, and this means that while loops have fairly limited use. Again, remember that functional programmers like recursion, and so while loops are second-class citizens in the language.

If you stop to consider while loops, you may see that they aren't really any use at all, except in conjunction with our old friend references. Let's imagine that OCaml didn't have references for a moment:

let quit_loop = false in
while not quit_loop do
  print_string "Have you had enough yet? (y/n) ";
  let str = read_line () in
  if str.[0] = 'y' then
    (* how do I set quit_loop to true ?!? *)

Remember that quit_loop is not a real "variable" - the let-binding just makes quit_loop a shorthand for false. This means the while loop condition (shown in red) is always true, and the loop runs on forever!

Luckily OCaml does have references, so we can write the code above if we want. Don't get confused and think that the ! (exclamation mark) means "not" as in C/Java. It's used here to mean "dereference the pointer", similar in fact to Forth. You're better off reading ! as "get" or "deref".

let quit_loop = ref false in
while not !quit_loop do
  print_string "Have you had enough yet? (y/n) ";
  let str = read_line () in
  if str.[0] = 'y' then
    quit_loop := true

Looping over lists

If you want to loop over a list, don't be an imperative programmer and reach for your trusty six-shooter Mr. For Loop! OCaml has some better and faster ways to loop over lists, and they are all located in the List module. There are in fact dozens of good functions in List, but I'll only talk about the most useful ones here.

First off, let's define a list for us to use:

# let my_list = [1; 2; 3; 4; 5; 6; 7; 8; 9; 10];;
val my_list : int list = [1; 2; 3; 4; 5; 6; 7; 8; 9; 10]

If you want to call a function once on every element of the list, use List.iter, like this:

# let f elem =
    Printf.printf "I'm looking at element %d now\n" elem in
  List.iter f my_list;;
I'm looking at element 1 now I'm looking at element 2 now I'm looking at element 3 now I'm looking at element 4 now I'm looking at element 5 now I'm looking at element 6 now I'm looking at element 7 now I'm looking at element 8 now I'm looking at element 9 now I'm looking at element 10 now - : unit = ()

List.iter is in fact what you should think about using first every time your cerebellum suggests you use a for loop.

If you want to transform each element separately in the list - for example, doubling each element in the list - then use List.map. This function will be familiar to people who've programmed in Perl before.

# List.map (( * ) 2) my_list;;
- : int list = [2; 4; 6; 8; 10; 12; 14; 16; 18; 20]

Perl has the useful function "grep" for filtering only elements of a list which satisfy some condition - eg. returning all even numbers in a list. In OCaml this function is called List.filter:

# let is_even i =
    i mod 2 = 0 in
  List.filter is_even my_list;;
- : int list = [2; 4; 6; 8; 10]

To find out if a list contains some element, use List.mem (short for member):

# List.mem 12 my_list;;
- : bool = false

List.for_all and List.exists are the same as the "forall" and "exist" operators in predicate logic.

For operating over two lists at the same time, there are "-2" variants of some of these functions, namely iter2, map2, for_all2, exists2.

The map and filter functions operate on individual list elements in isolation. Fold is a more unusual operation that is best thought about as "inserting an operator between each element of the list". Suppose I wanted to add all the numbers in my list together. In hand-waving terms what I want to do is insert a plus sign between the elements in my list:

# 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10;;
- : int = 55

The fold operation does this, although the exact details are a little bit more tricky. First of all, what happens if I try to fold an empty list? In the case of summing the list it would be nice if the answer was zero, instead of error. However if I was trying to find the product of the list, I'd like the answer to be one instead. So I obviously have to provide some sort of "default" argument to my fold. The second issue doesn't arise with simple operators like + and *: what happens if the operator I'm using isn't associative, ie. (a op b) op c not equal to a op (b op c)? In that case it would matter if I started from the left hand end of the list and worked right, versus if I started from the right and worked left. For this reason there are two versions of fold, called List.fold_left (works left to right) and List.fold_right (works right to left, and is also less efficient).

Let's use List.fold_left to define sum and product functions for integer lists:

# let sum = List.fold_left ( + ) 0;;
val sum : int list -> int = <fun> # let product = List.fold_left ( * ) 1;;
val product : int list -> int = <fun> # sum my_list;;
- : int = 55 # product my_list;;
- : int = 3628800

That was easy! Notice that I've accidentally come up with a way to do mathematical factorials:

# let fact n = product (range 1 n);;
val fact : int -> int = <fun> # fact 10;;
- : int = 3628800

(Notice that this factorial function isn't very useful because it overflows the integers and gives wrong answers even for quite small values of n. A real factorial function would use the Big_int module.)

Looping over strings

The String module also contains many dozens of useful string-related functions, and some of them are concerned with looping over strings.

String.fill and String.blit are the equivalents to C memset and strcpy respectively. String.copy copies a string, like strdup.

There is also a String.iter function which works like List.iter, except over the characters of the string.


Now we come to a hard topic - recursion. Functional programmers are defined by their love of recursive functions, and in many ways recursive functions in f.p. are the equivalent of loops in imperative programming. In functional languages loops are second-class citizens, whilst recursive functions get all the best support.

Writing recursive functions requires a change in mindset from writing for loops and while loops. So what I'll give you in this section will be just an introduction and examples.

In the first example we're going to read the whole of a file into memory (into a long string). There are essentially three possible approaches to this:

Approach 1

Get the length of the file, and read it all in one go using the really_input method. This is the simplest, but it might not work on channels which are not really files (eg. reading keyboard input) which is why we look at the other two approaches.

Approach 2

The imperative approach, using a while loop which is broken out of using an exception.

Approach 3

A recursive loop, breaking out of the recursion again using an exception.

We're going to introduce a few new concepts here. Our second two approaches will use the Buffer module - an expandable buffer which you can think of like a string onto which you can efficiently append more text at the end. We're also going to be catching the End_of_file exception which the input functions throw when they reach the end of the input. Also we're going to use Sys.argv.(1) to get the first command line parameter.

(* Read whole file: Approach 1 *)
open Printf
let read_whole_chan chan =
  let len = in_channel_length chan in
  let result = String.create len in
  really_input chan result 0 len;
let read_whole_file filename =
  let chan = open_in filename in
  read_whole_chan chan
let () =
  let filename = Sys.argv.(1) in
  let str = read_whole_file filename in
  printf "I read %d characters from %s\n" (String.length str) filename

Approach 1 works but is not very satisfactory because read_whole_chan won't work on non-file channels like keyboard input or sockets. Approach 2 involves a while loop:

(* Read whole file: Approach 2 *)
open Printf
let read_whole_chan chan =
  let buf = Buffer.create 4096 in
    while true do
      let line = input_line chan in
      Buffer.add_string buf line;
      Buffer.add_char buf '\n'
    assert false (* This is never executed
                    (always raise Assert_failure). *)
    End_of_file -> Buffer.contents buf
let read_whole_file filename =
  let chan = open_in filename in
  read_whole_chan chan
let () =
  let filename = Sys.argv.(1) in
  let str = read_whole_file filename in
  printf "I read %d characters from %s\n" (String.length str) filename

The key to approach 2 is to look at the central while loop. Remember that I said the only way to break out of a while loop early was with an exception? This is exactly what we're doing here. Although I haven't covered exceptions yet, you probably won't have any trouble understanding the End_of_file exception thrown in the code above by input_line when it hits the end of the file. The buffer buf accumulates the contents of the file, and when we hit the end of the file we return it (Buffer.contents buf).

One curious point about this is the apparently superfluous statement (assert false) just after the while loop. What is it for? Remember that while loops, like for loops, are just expressions, and they return the unit object (()). However OCaml demands that the return type inside a try matches the return type of each caught exception. In this case because End_of_file results in a string, the main body of the try must also "return" a string — even though because of the infinite while loop the string could never actually be returned. assert false has a polymorphic type, so will unify with whatever value is returned by the with branch.

Here's our recursive version. Notice that it's shorter than approach 2, but not so easy to understand for imperative programmers at least:

(* Read whole file: Approach 3 *)
open Printf
let read_whole_chan chan =
  let buf = Buffer.create 4096 in
  let rec loop () =
    let line = input_line chan in
    Buffer.add_string buf line;
    Buffer.add_char buf '\n';
    loop () in
    loop ()
    End_of_file -> Buffer.contents buf
let read_whole_file filename =
  let chan = open_in filename in
  read_whole_chan chan
let () =
  let filename = Sys.argv.(1) in
  let str = read_whole_file filename in
  printf "I read %d characters from %s\n" (String.length str) filename

Again we have an infinite loop - but in this case done using recursion. loop calls itself at the end of the function. The infinite recursion is broken when input_line throws an End_of_file exception.

It looks like approach 3 might overflow the stack if you gave it a particularly large file, but this is in fact not the case. Because of tail recursion (discussed below) the compiler will turn the recursive loop function into a real while loop (!) which runs in constant stack space.

In the next example we will show how recursion is great for constructing or examining certain types of data structures, particularly trees. Let's have a recursive type to represent files in a filesystem:

# type filesystem = File of string | Directory of filesystem list;;
type filesystem = File of string | Directory of filesystem list

The opendir and readdir functions are used to open a directory and read elements from the directory. I'm going to define a handy readdir_no_ex function which hides the annoying End_of_file exception that readdir throws when it reaches the end of the directory:

# open Unix  (*  You may need to #load "Unix.cma" *)
  let readdir_no_ex dirh =
      Some (readdir dirh)
      End_of_file -> None;;
val readdir_no_ex : Unix.dir_handle -> string option = <fun>

The type of readdir_no_ex is this. Recall our earlier discussion about null pointers.

readdir_no_ex : dir_handle -> string option

I'm also going to define a simple recursive function which I can use to convert the filesystem type into a string for (eg) printing:

# let rec string_of_filesystem fs =
    match fs with
    | File filename -> filename ^ "\n"
    | Directory fs_list ->
        List.fold_left (^) "" (List.map string_of_filesystem fs_list);;
val string_of_filesystem : filesystem -> string = <fun>

Note the use of fold_left and map. The map is used to (recursively) convert each filesystem in the list into a string. Then the fold_left (^) "" concatenates the list together into one big string. Notice also the use of pattern matching. (The library defines a function called String.concat which is essentially equivalent to fold_left (^) , but implemented more efficiently).

Now let's define a function to read a directory structure, recursively, and return a recursive filesystem data structure. I'm going to show this function in steps, but I'll print out the entire function at the end of this section. First the outline of the function:

let rec read_directory path =
  let dirh = opendir path in
  let rec loop () =
    (* ..... *) in
  Directory (loop ())

The call to opendir opens up the given path and returns a dir_handle from which we will be able to read the names using readdir_no_ex later. The return value of the function is going to be a Directory fs_list, so all we need to do to complete the function is to write our function loop which returns a list of filesystems. The type of loop will be:

loop : unit -> filesystem list

How do we define loop? Let's take it in steps again.

let rec loop () =
  let filename = readdir_no_ex dirh in
  (* ..... *)

First we read the next filename from the directory handle. filename has type string option, in other words it could be None or Some "foo" where foo is the name of the next filename in the directory. We also need to ignore the "." and ".." files (ie. the current directory and the parent directory). We can do all this with a nice pattern match:

let rec loop () =
  let filename = readdir_no_ex dirh in
  match filename with
  | None -> []
  | Some "." -> loop ()
  | Some ".." -> loop ()
  | Some filename ->
     (* ..... *)

The None case is easy. Thinking recursively (!) if loop is called and we've reached the end of the directory, loop needs to return a list of entries - and there's no entries - so it returns the empty list ([]).

For "." and ".." we just ignore the file and call loop again.

What do we do when loop reads a real filename (the Some filename match below)? Let pathname be the full path to the file. We 'stat' the file to see if it's really a directory. If it is a directory, we set this by recursively calling read_directory which will return Directory something. Notice that the overall result of read_directory is Directory (loop ()). If the file is really a file (not a directory) then we let this be File pathname. Then we do something clever: we return this :: loop (). This is the recursive call to loop () to calculate the remaining directory members (a list), to which we prepend this.

# let rec read_directory path =
    let dirh = opendir path in
    let rec loop () =
      let filename = readdir_no_ex dirh in
      match filename with
      | None -> []
      | Some "." -> loop ()
      | Some ".." -> loop ()
      | Some filename ->
          let pathname = path ^ "/" ^ filename in
          let stat = lstat pathname in
          let this = if stat.st_kind = S_DIR then
                       read_directory pathname
                       File pathname in
          this :: loop () in
    Directory (loop ());;
val read_directory : string -> filesystem = <fun>

That's quite a complex bit of recursion, but although this is a made-up example, it's fairly typical of the complex patterns of recursion found in real-world functional programs. The two important lessons to take away from this are:

  • The use of recursion to build a list:

    let rec loop () =
      a match or if statement
      | base case -> []
      | recursive case -> element :: loop ()

    Compare this to our previous range function. The pattern of recursion is exactly the same:

    let rec range a b =
      if a > b then []            (* Base case *)
      else a :: range (a+1) b     (* Recursive case *)
  • The use of recursion to build up trees:

    let rec read_directory path =
      (* blah blah *)
      if file-is-a-directory then
        read_directory path-to-file
        Leaf file

    All that remains now to make this a working program is a little bit of code to call read_directory and display the result:

    let path = Sys.argv.(1) in
    let fs = read_directory path in
    print_endline (string_of_filesystem fs)


After I posted this example to the caml-list mailing list I received many follow-ups. (You can read the full thread.)

First of all there is a basic fault with read_directory directory which will cause it to fail if you try to run it on very large directory structures. I don't explicitly call closedir to close the directory handle. The garbage collector doesn't help, because in OCaml the garbage collector does not collect and close open file and directory handles.

The fix is pretty trivial: add a call to closedir at the end of the read_directory function:

let rec read_directory path =
  let dirh = opendir path in
    (* ... *)
  let result = Directory (loop ()) in
  closedir dirh;

Next up is the issue of readdir throwing an exception in a rather "unexceptional" situation, namely End_of_file. I don't agree that throwing an exception in this case is the right thing to do. In typical programs I want to never have to write a try ... with clause, because exceptions are supposed to mean "something really bad happened which I'm not prepared to deal with", like running out of disk space or memory. Throwing an exception as part of the routine running of a function (every program which uses readdir has to be prepared to handle End_of_file), that's not an exceptional situation.

However, Eric Cooper came up with a way to write the function and check the exception, and this highlights another frequent technique used by functional programmers. Namely, passing an accumulator which contains the result of the function call, but passed as an argument to the function (!) Really the accumulator means "this is the result so far", so in the exceptional case (End_of_file) we give up and return the result so far. Here is his code with all the references to the accumulator argument in red:

# let rec read_filesystem path =
    if (lstat path).st_kind = S_DIR then
      Directory (read_directory path)
      File path
  and read_directory path =
    let dirh = opendir path in
    let rec loop entries =
        match readdir dirh with
        | "." | ".." -> loop entries
        | filename ->
           loop (read_filesystem (path ^ "/" ^ filename) :: entries)
      with End_of_file -> entries in
    let list = loop [] in
    closedir dirh;
val read_filesystem : string -> filesystem = <fun> val read_directory : string -> filesystem list = <fun>

Notice End_of_file -> entries which means "when we get the exception, give up and return the result so far".

Next up - it was pointed out that the easiest way to do this is actually not to use recursion at all, but instead to do the loop imperatively (a while loop is probably best) and have a reference to a list to store the ongoing accumulated result. Because readdir throws an exception, we have a simple way to break out of the while loop, and in the with-clause we can just return !list (dereference the reference to the list of entries we've built up).

It all works nicely, and proves that writing code imperatively is often best, particularly as in this case where opendir ... readdir ... closedir is essentially an imperative API, designed for C programmers. It's no surprise, therefore, that the imperative solution is easier. OCaml, not being a bondage and discipline language, doesn't mind you using the imperative approach when it makes sense.

Here is the outline of the imperative approach by Fabrice Le Fessant:

let list = ref [] in
let dir = opendir "......." in
  while true do
    match readdir dir with
    | "." | ".." -> ()
    | filename -> list := filename :: !list
  assert false
with End_of_file -> !list

Recursion example: maximum element in a list

Remember the basic recursion pattern for lists:

let rec loop () =
  a match or if statement
  | base case -> []
  | recursive case -> element :: loop ()

The key here is actually the use of the match / base case / recursive case pattern. In this example - finding the maximum element in a list - we're going to have two base cases and one recursive case. But before I jump ahead to the code, let's just step back and think about the problem. By thinking about the problem, the solution will appear "as if by magic" (I promise you :-)

First of all, let's be clear that the maximum element of a list is just the biggest one, e.g. the maximum element of the list [1; 2; 3; 4; 1] is 4.

Are there any special cases? Yes, there are. What's the maximum element of the empty list []? There isn't one. If we are passed an empty list, we should throw an error.

What's the maximum element of a single element list such as [4]? That's easy: it's just the element itself. So list_max [4] should return 4, or in the general case, list_max [x] should return x.

What's the maximum element of the general list x :: remainder (this is the "cons" notation for the list, so remainder is the tail - also a list)?

Think about this for a while. Suppose you know the maximum element of remainder, which is, say, y. What's the maximum element of x :: remainder? It depends on whether x > y or x <= y. If x is bigger than y, then the overall maximum is x, whereas conversely if x is less than y, then the overall maximum is y.

Does this really work? Consider [1; 2; 3; 4; 1] again. This is 1 :: [2; 3; 4; 1]. Now the maximum element of the remainder, [2; 3; 4; 1], is 4. So now we're interested in x = 1 and y = 4. That head element x = 1 doesn't matter because y = 4 is bigger, so the overall maximum of the whole list is y = 4.

Let's now code those rules above up, to get a working function:

# let rec list_max xs =
      match xs with
      | [] ->   (* empty list: fail *)
         failwith "list_max called on empty list"
      | [x] -> (* single element list: return the element *)
      | x :: remainder -> (* multiple element list: recursive case *)
         max x (list_max remainder);;
val list_max : 'a list -> 'a = <fun>

I've added comments so you can see how the rules / special cases we decided upon above really correspond to lines of code.

Does it work?

# list_max [1; 2; 3; 4; 1];;
- : int = 4 # list_max [];;
Exception: Failure "list_max called on empty list". # list_max [5; 4; 3; 2; 1];;
- : int = 5 # list_max [5; 4; 3; 2; 1; 100];;
- : int = 100

Notice how the solution proposed is both (a) very different from the imperative for-loop solution, and (b) much more closely tied to the problem specification. Functional programmers will tell you that this is because the functional style is much higher level than the imperative style, and therefore better and simpler. Whether you believe them is up to you. It's certainly true that it's much simpler to reason logically about the functional version, which is useful if you wanted to formally prove that list_max is correct ("correct" being the mathematical way to say that a program is provably bug-free, useful for space shuttles, nuclear power plants and higher quality software in general).

Tail recursion

Let's look at the range function again for about the twentieth time:

# let rec range a b =
    if a > b then []
    else a :: range (a+1) b;;
val range : int -> int -> int list = <fun>

I'm going to rewrite it slightly to make something about the structure of the program clearer (still the same function however):

# let rec range a b =
    if a > b then []
      let result = range (a+1) b in
      a :: result;;
val range : int -> int -> int list = <fun>

Let's call it:

# List.length (range 1 10);;
- : int = 10 # List.length (range 1 1000000);;
Stack overflow during evaluation (looping recursion?).

Hmmm ... at first sight this looks like a problem with recursive programming, and hence with the whole of functional programming! If you write your code recursively instead of iteratively then you necessarily run out of stack space on large inputs, right?

In fact, wrong. Compilers can perform a simple optimisation on certain types of recursive functions to turn them into while loops. These certain types of recursive functions therefore run in constant stack space, and with the equivalent efficiency of imperative while loops. These functions are called tail-recursive functions.

In tail-recursive functions, the recursive call happens last of all. Remember our loop () functions above? They all had the form:

let rec loop () =
  (* do something *)
  loop ()

Because the recursive call to loop () happens as the very last thing, loop is tail-recursive and the compiler will turn the whole thing into a while loop.

Unfortunately range is not tail-recursive, and the longer version above shows why. The recursive call to range doesn't happen as the very last thing. In fact the last thing to happen is the :: (cons) operation. As a result, the compiler doesn't turn the recursion into a while loop, and the function is not efficient in its use of stack space.

Now recall we discussed in the addendum above Eric Cooper's "accumulator" technique, and I said that this technique was widely used in functional programming. We will now see why this is the case. It's because it allows you to write functions such as range above in a tail-recursive manner, which means they will be efficient and work properly on large inputs.

Let's plan our rewritten range function which will use an accumulator argument to store the "result so far":

let rec range2 a b accum =
  (* ... *)
let range a b =
  range2 a b []

(I could and probably should have used a nested function here.)

The accum argument is going to accumulate the result. It's the "result so far". We pass in the empty list ("no result so far"). The easy case is when a > b:

let rec range2 a b accum =
  if a > b then accum
    (* ... *)

If a > b (i.e. if we've reached the end of the recursion), then stop and return the result (accum).

Now the trick is to write the else-clause and make sure that the call to range2 is the very last thing that we do, so the function is tail-recursive:

# let rec range2 a b accum =
    if a > b then accum
    else range2 (a+1) b (a :: accum);;
val range2 : int -> int -> int list -> int list = <fun>

There's only one slight problem with this function: it constructs the list backwards! However, this is easy to rectify by redefining range as:

# let range a b = List.rev (range2 a b []);;
val range : int -> int -> int list = <fun>

It works this time, although it's a bit slow to run because it really does have to construct a list with a million elements in it:

# List.length (range 1 1000000);;
- : int = 1000000

The following implementation is twice as fast as the previous one, because it does not need to reverse a list:

# let rec range2 a b accum =
    if b < a then accum
    else range2 a (b-1) (b :: accum);;
val range2 : int -> int -> int list -> int list = <fun> # let range a b = range2 a b [];;
val range : int -> int -> int list = <fun>

That was a brief overview of tail recursion, but in real world situations determining if a function is tail recursive can be quite hard. What did we really learn here? One thing is that recursive functions have a dangerous trap for inexperienced programmers. Your function can appear to work for small inputs (during testing), but fail catastrophically in the field when exposed to large inputs. This is one argument against using recursive functions, and for using explicit while loops when possible.

Mutable records, references (again!) and arrays

Previously we mentioned records in passing. These are like C structs:

# type pair_of_ints = { a : int; b : int };;
type pair_of_ints = { a : int; b : int; } # {a=3; b=5};;
- : pair_of_ints = {a = 3; b = 5} # {a=3};;
Error: Some record fields are undefined: b

One feature which I didn't cover: OCaml records can have mutable fields. Normally an expression like {a = 3; b = 5} is an immutable, constant object. However if the record has mutable fields, then there is a way to change those fields in the record. This is an imperative feature of OCaml, because functional languages don't normally allow mutable objects (or references or mutable arrays, which we'll look at in a moment).

Here is an object defined with a mutable field. This field is used to count how many times the object has been accessed. You could imagine this being used in a caching scheme to decide which objects you'd evict from memory.

# type name = { name : string; mutable access_count : int };;
type name = { name : string; mutable access_count : int; }

Here is a function defined on names which prints the name field and increments the mutable access_count field:

# let print_name name =
    print_endline ("The name is " ^ name.name);
    name.access_count <- name.access_count + 1;;
val print_name : name -> unit = <fun>

Notice a strange, and very non-functional feature of print_name: it modifies its access_count parameter. If you read chapter 5 closely, you'll see that this function is not "pure". OCaml is a functional language, but not to the extent that it forces functional programming down your throat.

Anyway, let's see print_name in action:

# let n = { name = "Richard Jones"; access_count = 0 };;
val n : name = {name = "Richard Jones"; access_count = 0} # n;;
- : name = {name = "Richard Jones"; access_count = 0} # print_name n;;
The name is Richard Jones - : unit = () # n;;
- : name = {name = "Richard Jones"; access_count = 1} # print_name n;;
The name is Richard Jones - : unit = () # n;;
- : name = {name = "Richard Jones"; access_count = 2}

Only fields explicitly marked as mutable can be assigned to using the <- operator. If you try to assign to a non-mutable field, OCaml won't let you:

# n.name <- "John Smith";;
Error: The record field name is not mutable

References, with which we should be familiar by now, are implemented using records with a mutable contents field. Check out the definition in Pervasives:

# type 'a ref = { mutable contents : 'a };;
type 'a ref = { mutable contents : 'a; }

And look closely at what the OCaml toplevel prints out for the value of a reference:

# let r = ref 100;;
val r : int Stdlib.ref = {Stdlib.contents = 100}

Arrays are another sort of mutable structure provided by OCaml. In OCaml, plain lists are implemented as linked lists, and linked lists are slow for some types of operation. For example, getting the head of a list, or iterating over a list to perform some operation on each element is reasonably fast. However, jumping to the nth element of a list, or trying to randomly access a list - both are slow operations. The OCaml Array type is a real array, so random access is fast, but insertion and deletion of elements is slow. Arrays are also mutable so you can randomly change elements too.

The basics of arrays are simple:

# let a = Array.create 10 0;;
Warning 3: deprecated: Stdlib.Array.create Use Array.make instead. val a : int array = [|0; 0; 0; 0; 0; 0; 0; 0; 0; 0|] # for i = 0 to Array.length a - 1 do a.(i) <- i done;;
- : unit = () # a;;
- : int array = [|0; 1; 2; 3; 4; 5; 6; 7; 8; 9|]

Notice the syntax for writing arrays: [| element; element; ... |]

The OCaml compiler was designed with heavy numerical processing in mind (the sort of thing that FORTRAN is traditionally used for), and so it contains various optimisations specifically for arrays of numbers, vectors and matrices. Here is some benchmark code for doing dense matrix multiplication. Notice that it uses for-loops and is generally very imperative in style:

# let size = 30
  let mkmatrix rows cols =
    let count = ref 1
    and last_col = cols - 1
    and m = Array.make_matrix rows cols 0 in
    for i = 0 to rows - 1 do
      let mi = m.(i) in
      for j = 0 to last_col do
        mi.(j) <- !count;
        incr count
  let rec inner_loop k v m1i m2 j =
    if k < 0 then v
    else inner_loop (k - 1) (v + m1i.(k) * m2.(k).(j)) m1i m2 j
  let mmult rows cols m1 m2 m3 =
    let last_col = cols - 1
    and last_row = rows - 1 in
    for i = 0 to last_row do
      let m1i = m1.(i) and m3i = m3.(i) in
      for j = 0 to last_col do
        m3i.(j) <- inner_loop last_row 0 m1i m2 j
  let () =
    let n =
      try int_of_string Sys.argv.(1)
      with Invalid_argument _ -> 1
    and m1 = mkmatrix size size
    and m2 = mkmatrix size size
    and m3 = Array.make_matrix size size 0 in
    for i = 1 to n - 1 do
      mmult size size m1 m2 m3
    mmult size size m1 m2 m3;
    Printf.printf "%d %d %d %d\n" m3.(0).(0) m3.(2).(3) m3.(3).(2) m3.(4).(4);;
270165 1061760 1453695 1856025 val size : int = 30 val mkmatrix : int -> int -> int array array = <fun> val inner_loop : int -> int -> int array -> int array array -> int -> int = <fun> val mmult : int -> int -> int array array -> int array array -> int array array -> unit = <fun>