OCaml

Implementation of programming language OCaml

The official OCaml implementation.

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OCaml Inria logo

Links:

Examples:

Hello, World!:

Example for versions OCaml 3.11

print_endline is a built-in function defined with the following type: 

string -> unit = <func>

This means that it takes 1 string as a parameter, and returns the unit type, (). 

let () = print_endline "Hello World";;

Factorial:

Example for versions OCaml 3.11

This example uses an auxiliary function fact, so that tail recursion is possible.

let rec fact n accum =
    if n <= 1 then 
        accum
    else
        fact (n-1) (accum*n);;

let factorial n =
    fact n 1;;

let () =
    for n = 0 to 16 do
        Printf.printf "%d! = %d\n" n (factorial n)
    done;

Factorial:

Example for versions OCaml 3.11

This example shows the naive way to implement the factorial function. However, it is not tail recursive, since the recursive function call is not the only statement on the line. 

let rec factorial n =
    if n <= 1 then
      1
    else
      factorial (n-1) * n;;

let () =
  for n = 0 to 16 do
    Printf.printf "%d! = %d\n" n (factorial n)
  done;

Fibonacci numbers:

Example for versions OCaml 3.11

This example uses straightforward recursive solution. Printf.printf does formatted output. 

let rec fibonacci n =
  if n < 3 then
    1
  else
    fibonacci (n-1) + fibonacci (n-2)

let () =
  for n = 1 to 16 do
    Printf.printf "%d, " (fibonacci n)
  done;
  print_endline "..."

Quadratic equation:

Example for versions OCaml 3.11

ocaml quadratic_equation.ml 3. 5. 2. 

let square x = x *. x;;

let delta  a b c = ( square b-. (4. *. a *. c ));;

let   solve a b c   =
    if a=0. 
        then Printf.printf "Not a quadratic equation\n"
            else
                if   delta a b c  >= 0.
                    then
                        let   x1=( -.b +. sqrt(delta a b c )) /. (2. *. a)
                            and x2=  (-.b -.  sqrt(delta a b c ))  /. (2. *. a)  
                        in 
                            Printf.printf "x1 =%.5f x2=%.5f \n"  x1  x2
                    else 
                        let   x= (-.b   /. (2. *. a))
                            and i=     sqrt(4. *. a *. c -. square b)  /. (2. *. a) 
                        in Printf.printf "x+ =%.5f+i%.5f  x-=%.5f-i%.5f \n"  x  i  x i
    ;;



let () =
solve (float_of_string Sys.argv.(1))( float_of_string Sys.argv.(2) )(float_of_string Sys.argv.(3))