Function composition

From Rosetta Code
Task
Function composition
You are encouraged to solve this task according to the task description, using any language you may know.

Create a function, compose, whose two arguments f and g, are both functions with one argument. The result of compose is to be a function of one argument, (lets call the argument x), which works like applying function f to the result of applying function g to x.

I.e:

 compose(f, g)(x) == f( g(x) )

Reference: Function composition

Hint: Implementing compose correctly requires creating a closure. If your language does not support closures directly, you will need to implement it yourself.

Ada

The interface of a generic functions package. The package can be instantiated with any type that has value semantics. Functions are composed using the operation '*'. The same operation applied to an argument evaluates it there: f * x. Functions can be composed with pointers to Ada functions. (In Ada functions are not first-class): <lang ada>generic

  type Argument is private;      

package Functions is

  type Primitive_Operation is not null
     access function (Value : Argument) return Argument;
  type Func (<>) is private;
  function "*" (Left : Func; Right : Argument) return Argument;
  function "*" (Left : Func; Right : Primitive_Operation) return Func;
  function "*" (Left, Right : Primitive_Operation) return Func;
  function "*" (Left, Right : Func) return Func;

private

  type Func is array (Positive range <>) of Primitive_Operation;

end Functions;</lang> Here is an implementation; <lang ada>package body Functions is

  function "*" (Left : Func; Right : Argument) return Argument is
  Result : Argument := Right;
  begin
     for I in reverse Left'Range loop
        Result := Left (I) (Result);
     end loop;
     return Result;
  end "*";
  function "*" (Left, Right : Func) return Func is
  begin
     return Left & Right;
  end "*";
  function "*" (Left : Func; Right : Primitive_Operation) return Func is
  begin
     return Left & (1 => Right);
  end "*";
  
  function "*" (Left, Right : Primitive_Operation) return Func is
  begin
     return (Left, Right);
  end "*";

end Functions;</lang> The following is an example of use: <lang ada>with Ada.Numerics.Elementary_Functions; use Ada.Numerics.Elementary_Functions; with Ada.Text_IO; use Ada.Text_IO; with Functions;

procedure Test_Compose is

  package Float_Functions is new Functions (Float);
  use Float_Functions;
  Sin_Arcsin : Func := Sin'Access * Arcsin'Access;

begin

  Put_Line (Float'Image (Sin_Arcsin * 0.5));

end Test_Compose;</lang> Sample output:

 5.00000E-01

ALGOL 68

Translation of: Python
Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8.8d.fc9.i386

Note: Returning PROC (REAL x)REAL: f1(f2(x)) from a function apparently violates standard ALGOL 68's scoping rules. ALGOL 68G warns about this during parsing, and then rejects during runtime. <lang algol68>MODE F = PROC(REAL)REAL; # ALGOL 68 is strong typed #

  1. As a procedure for real to real functions #

PROC compose = (F f, g)F: (REAL x)REAL: f(g(x));

OP (F,F)F O = compose; # or an OPerator that can be overloaded #

  1. Example use: #

F sin arc sin = compose(sin, arc sin); print((sin arc sin(0.5), (sin O arc sin)(0.5), new line))</lang> Output:

+.500000000000000e +0 +.500000000000000e +0

ALGOL 68 is a stack based language, and the following apparently does not violate it's scoping rules.

Works with: ALGOL 68 version Standard - Jan 1975 Boston SC allowed Partial Parametrization.
Works with: ALGOL 68G version Any - tested with release mk15-0.8b.fc9.i386

<lang algol68>MODE F = PROC(REAL)REAL; # ALGOL 68 is strong typed #

  1. As a procedure for real to real functions #

PROC compose = (F f, g)F: ((F f2, g2, REAL x)REAL: f2(g2(x)))(f, g, ); # Curry #

PRIO O = 7; OP (F,F)F O = compose; # or an OPerator that can be overloaded #

  1. Example use: #

F sin arc sin = compose(sin, arc sin); print((sin arc sin(0.5), (sin O arc sin)(0.5), new line))</lang>

AutoHotkey

contributed by Laszlo on the ahk forum <lang AutoHotkey>MsgBox % compose("sin","cos",1.5)

compose(f,g,x) { ; function composition

  Return %f%(%g%(x))

}</lang>

C

Only works for functions taking a double and returning a double: <lang c>#include <stdlib.h>

/* generic interface for functors from double to double */ typedef struct double_to_double {

 double (*fn)(struct double_to_double *, double);

} double_to_double;

  1. define CALL(f, x) f->fn(f, x)


/* functor returned by compose */ typedef struct compose_functor {

 double (*fn)(struct compose_functor *, double);
 double_to_double *f;
 double_to_double *g;

} compose_functor; /* function to be used in "fn" in preceding functor */ double compose_call(compose_functor *this, double x) {

 return CALL(this->f, CALL(this->g, x));

} /* returns functor that is the composition of functors

  f & g. caller is responsible for deallocating memory */

double_to_double *compose(double_to_double *f,

                         double_to_double *g) {
 compose_functor *result = malloc(sizeof(compose_functor));
 result->fn = &compose_call;
 result->f = f;
 result->g = g;
 return (double_to_double *)result;

}


  1. include <math.h>

/* we can make functors for sin and asin by using

  the following as "fn" in a functor */

double sin_call(double_to_double *this, double x) {

 return sin(x);

} double asin_call(double_to_double *this, double x) {

 return asin(x);

}


  1. include <stdio.h>

int main() {

 double_to_double *my_sin = malloc(sizeof(double_to_double));
 my_sin->fn = &sin_call;
 double_to_double *my_asin = malloc(sizeof(double_to_double));
 my_asin->fn = &asin_call;
 double_to_double *sin_asin = compose(my_sin, my_asin);
 printf("%f\n", CALL(sin_asin, 0.5)); /* prints "0.500000" */
 free(sin_asin);
 free(my_sin);
 free(my_asin);
 return 0;

}</lang>

C++

Note: this is already implemented as __gnu_cxx::compose1() <lang cpp>#include <functional>

  1. include <cmath>
  2. include <iostream>

// functor class to be returned by compose function template <class Fun1, class Fun2> class compose_functor :

 public std::unary_function<typename Fun2::argument_type,
                            typename Fun1::result_type>

{ protected:

 Fun1 f;
 Fun2 g;

public:

 compose_functor(const Fun1& _f, const Fun2& _g)
   : f(_f), g(_g) { }
 typename Fun1::result_type
 operator()(const typename Fun2::argument_type& x) const
 { return f(g(x)); }

};

// we wrap it in a function so the compiler infers the template arguments // whereas if we used the class directly we would have to specify them explicitly template <class Fun1, class Fun2> inline compose_functor<Fun1, Fun2> compose(const Fun1& f, const Fun2& g) { return compose_functor<Fun1,Fun2>(f, g); }

int main() {

 std::cout << compose(std::ptr_fun(::sin), std::ptr_fun(::asin))(0.5) << std::endl;
 return 0;

}</lang>

C#

<lang csharp>using System; class Program {

   static void Main(string[] args)
   {
       Func<int, int> outfunc = Composer<int, int, int>.Compose(functA, functB);
       Console.WriteLine(outfunc(5)); //Prints 100
   }
   static int functA(int i) { return i * 10; }
   static int functB(int i) { return i + 5; }
   class Composer<A, B, C>
   {
       public static Func<C, A> Compose(Func<B, A> a, Func<C, B> b)
       {
           return delegate(C i) { return a(b(i)); };
       }
   }

}</lang>

Common Lisp

<lang lisp>(defun compose (f g) (lambda (x) (funcall f (funcall g x))))</lang> Example use: <lang lisp>>(defun compose (f g) (lambda (x) (funcall f (funcall g x)))) COMPOSE >(let ((sin-asin (compose #'sin #'asin))))

  (funcall sin-asin 0.5))

0.5</lang>

D

D 2.0 version of compose function (template). <lang D>import std.stdio; import std.math;

T delegate(S) compose(T, U, S)(T delegate(U) f, U delegate(S) g) {

   return (S s) { return f(g(s)); };

}</lang>

Compose working both in D 1.0 and 2.0: <lang D>T delegate(S) compose(T, U, S)(T delegate(U) f, U delegate(S) g) {

   struct Wrapper {
       typeof(f) fcp;
       typeof(g) gcp;
       T foobar(S s) { return fcp(gcp(s)); }
   }
   Wrapper* hold = new Wrapper;
   hold.fcp = f;
   hold.gcp = g;
   return &hold.foobar;

}</lang>

E

<lang e>def compose(f, g) {

 return fn x { return f(g(x)) }

}</lang>

Erlang

<lang erlang>-module(fn). -export([compose/1, multicompose/2]).

compose(F,G) -> fun(X) -> F(G(X)) end.

multicompose(Fs) ->

   lists:foldl(fun compose/2, fun(X) -> X end, Fs).</lang>

Using them: <lang erlang>1> (fn:compose(fun math:sin/1, fun math:asin/1))(0.5). 0.5 2> Sin_asin_plus1 = fn:multicompose([fun math:sin/1, fun math:asin/1, fun(X) -> X + 1 end]).

  1. Fun<tests.0.59446746>

82> Sin_asin_plus1(0.5). 1.5</lang>

F#

The composition operator in F# is >> for forward composition and << for reverse. <lang fsharp>let compose f g = f << g ;;</lang> Usage: <lang fsharp>> let sin_asin = compose sin asin ;; val sin_asin : (float -> float) > sin_asin 0.5 ;; val it : float = 0.5</lang> To see that this can be a closure we can use arithmetic functions: <lang fsharp>> let add_then_halve = compose (fun n -> n / 2) (fun n -> n + 10) ;; val add_then_halve : (int -> int) > add_then_halve 6 ;; val it : int = 8</lang>

Forth

<lang forth>: compose ( xt1 xt2 -- xt3 )

 >r >r :noname
    r> compile,
    r> compile,
    postpone ;

' 2* ' 1+ compose ( xt ) 3 swap execute . \ 7</lang>

Groovy

Solution: <lang groovy>def compose = { f, g -> { x -> f(g(x)) } }</lang>

Test program: <lang groovy>def sq = { it * it } def plus1 = { it + 1 } def minus1 = { it - 1 }

def plus1sqd = compose(sq,plus1) def sqminus1 = compose(minus1,sq) def identity = compose(plus1,minus1) def plus1sqdminus1 = compose(minus1,compose(sq,plus1)) def identity2 = compose(Math.&sin,Math.&asin)

println "(x+1)**2 = (0+1)**2 = " + plus1sqd(0) println "x**2-1 = 20**2-1 = " + sqminus1(20) println "(x+1)-1 = (12+1)-1 = " + identity(12) println "(x+1)**2-1 = (3+1)**2-1 = " + plus1sqdminus1(3) println "sin(asin(x)) = sin(asin(1)) = " + identity2(1)</lang>

Output:

(x+1)**2 = (0+1)**2 = 1
x**2-1 = 20**2-1 = 399
(x+1)-1 = (12+1)-1 = 12
(x+1)**2-1 = (3+1)**2-1 = 15
sin(asin(x)) = sin(asin(1)) = 1.0

Haskell

This is already defined as the . (dot) operator in Haskell. <lang haskell>compose f g x = f (g x)</lang> Example use: <lang haskell>Prelude> let compose f g x = f (g x) Prelude> let sin_asin = compose sin asin Prelude> sin_asin 0.5 0.5</lang>

J

Solution: <lang j>compose =: @</lang>

Example: <lang j>f compose g</lang>

Of course, given that @ is only one character long and is a built-in primitive, there is no need for the cover function compose. And @ is not the only composition primitive; composition is a very important concept in J. For more details, see the talk page.

Java

<lang java>public class Compose {

   // Java doesn't have function type so we define an interface
   // of function objects instead
   public interface Fun<A,B> {
       B call(A x);
   }
   public static <A,B,C> Fun<A,C> compose(final Fun<B,C> f, final Fun<A,B> g) {
       return new Fun<A,C>() {
           public C call(A x) {
               return f.call(g.call(x));
           }
       };
   }
   public static void main(String[] args) {
       Fun<Double,Double> sin = new Fun<Double,Double>() {
           public Double call(Double x) {
               return Math.sin(x);
           }
       };
       Fun<Double,Double> asin = new Fun<Double,Double>() {
           public Double call(Double x) {
               return Math.asin(x);
           }
       };
       Fun<Double,Double> sin_asin = compose(sin, asin);
       System.out.println(sin_asin.call(0.5)); // prints "0.5"
   }

}</lang>

JavaScript

<lang javascript>function compose(f, g) {

 return function(x) { return f(g(x)) }

}

var id = compose(Math.sin, Math.asin) print id(0.5) // 0.5</lang>

Joy

Composition is the default operation in Joy. The composition of two functions is the concatenation of those functions, in the order in which they are to be applied. <lang joy>g f</lang>

Mathematica

Built-in function that takes any amount of function-arguments: <lang Mathematica>Composition[f, g][x] Composition[f, g, h, i][x]</lang> gives back: <lang Mathematica>f[g[x]] f[g[h[i[x]]]]</lang> Custom function: <lang Mathematica>compose[f_, g_][x_] := f[g[x]] compose[Sin, Cos][r]</lang> gives back: <lang Mathematica>Sin[Cos[r]]</lang> Composition can be done in more than 1 way: <lang Mathematica>Composition[f,g,h][x] f@g@h@x x//h//g//f</lang> all give back: <lang Mathematica>f[g[h[x]]]</lang> The built-in function has a couple of automatic simplifications: <lang Mathematica>Composition[f, Identity, g] Composition[f, InverseFunction[f], h][x]</lang> becomes: <lang Mathematica>f[g[x]] h[x]</lang>

Objective-C

The FunctionComposer is able to compose any object that conforms to the protocol FunctionCapsule (a selector/method accepting any object as argument and returning another object, i.e. computing a "function" of an object). A FunctionCaps class thought to encapsulate a function returning a double and with a double as argument is shown; anyway, as said, any object conforming to FunctionCapsule protocol can be composed with another object conforming to the same protocol. Argument passed and returned can be of any object type.

<lang objc>#include <Foundation/Foundation.h>

// the protocol of objects that can behave "like function" @protocol FunctionCapsule -(id)computeWith: (id)x; @end


// a commodity for "encapsulating" double f(double) typedef double (*func_t)(double); @interface FunctionCaps : NSObject <FunctionCapsule> {

 func_t function;

} +(id)capsuleFor: (func_t)f; -(id)initWithFunc: (func_t)f; @end

@implementation FunctionCaps -(id)initWithFunc: (func_t)f {

 self = [self init];
 function = f;
 return self;

} +(id)capsuleFor: (func_t)f {

 return [[[self alloc] initWithFunc: f] autorelease];

} -(id)computeWith: (id)x {

 return [NSNumber numberWithDouble: function([x doubleValue])];

} @end


// the "functions" composer @interface FunctionComposer : NSObject <FunctionCapsule> {

 id funcA;
 id funcB;

} +(id) createCompositeFunctionWith: (id)A and: (id)B; -(id) initComposing: (id)A with: (id)B; -(id) init; -(id) dealloc; @end

@implementation FunctionComposer +(id) createCompositeFunctionWith: (id)A and: (id)B {

 return [[[self alloc] initComposing: A with: B] autorelease];

}

-(id) init {

 NSLog(@"FunctionComposer: init with initComposing!");
 funcA = nil; funcB = nil;
 return self;

}

-(id) initComposing: (id)A with: (id)B {

 self = [super init];
 if ( ([A conformsToProtocol: @protocol(FunctionCapsule)] == YES) &&
      ([B conformsToProtocol: @protocol(FunctionCapsule)] == YES) ) {
   [A retain]; [B retain];
   funcA = A; funcB = B;
   return self;
 }
 NSLog(@"FunctionComposer: cannot compose functions not responding to protocol FunctionCapsule!");
 return nil;

}

-(id)computeWith: (id)x {

 return [funcA computeWith: [funcB computeWith: x]];

} @end

-(void) dealloc {

 [funcA release];
 [funcB release];
 [super dealloc];

}


// functions outside... double my_f(double x) {

 return x+1.0;

}

double my_g(double x) {

 return x*x;

}


int main() {

 NSAutoreleasePool *pool = [[NSAutoreleasePool alloc] init];
 id funcf = [FunctionCaps capsuleFor: my_f];
 id funcg = [FunctionCaps capsuleFor: my_g];
 id composed = [FunctionComposer 

createCompositeFunctionWith: funcf and: funcg];

 printf("g(2.0) = %lf\n", [[funcg computeWith: [NSNumber numberWithDouble: 2.0]] doubleValue]);
 printf("f(2.0) = %lf\n", [[funcf computeWith: [NSNumber numberWithDouble: 2.0]] doubleValue]);
 printf("f(g(2.0)) = %lf\n", [[composed computeWith: [NSNumber numberWithDouble: 2.0]] doubleValue]);
 [pool release];
 return 0;

}</lang>

OCaml

<lang ocaml>let compose f g x = f (g x)</lang> Example use: <lang ocaml># let compose f g x = f (g x);; val compose : ('a -> 'b) -> ('c -> 'a) -> 'c -> 'b = <fun>

  1. let sin_asin = compose sin asin;;

val sin_asin : float -> float = <fun>

  1. sin_asin 0.5;;

- : float = 0.5</lang>

Octave

<lang octave>function r = compose(f, g)

 r = @(x) f(g(x));

endfunction

r = compose(@exp, @sin); r(pi/3)</lang>

Perl

<lang perl>sub compose

  {my ($f, $g) = @_;
   return sub {$f->($g->(@_))};}

use Math::Trig; print compose(sub {sin $_[0]}, \&asin)->(0.5), "\n";</lang>

Perl 6

Works with: Rakudo version #23 "Lisbon"

<lang perl6>sub infix:<∘> (&f, &g --> Block) {

   sub (*@args) { f g |@args };

}</lang>

Example of use:

<lang perl6>sub triple($n) { 3 * $n } my $f = &triple ∘ { $^x + 2 }; say $f(5); # Prints "21".</lang>

Python

<lang python>compose = lambda f, g: lambda x: f( g(x) )</lang> Example use: <lang python>>>> compose = lambda f, g: lambda x: f( g(x) ) >>> from math import sin, asin >>> sin_asin = compose(sin, asin) >>> sin_asin(0.5) 0.5 >>></lang>

R

<lang R>compose <- function(f,g) function(x) { f(g(x)) } r <- compose(sin, cos) print(r(.5))</lang>

Ruby

This compose method gets passed two Method objects <lang ruby>def compose(f,g)

 lambda {|x| f.call(g.call(x))}

end s = compose(Math.method('sin'), Math.method('cos')) s.call(0.5) # => 0.769196354841008

  1. verify

Math.sin(Math.cos(0.5)) # => 0.769196354841008</lang>

With this method, you pass two symbols <lang ruby>include Math def compose(f,g)

 lambda {|x| send(f, send(g, x))}

end s = compose(:sin, :cos) s.call(0.5) # => 0.769196354841008</lang>

Scala

<lang scala>def compose[A](f: A => A, g: A => A) = { x: A => f(g(x)) }

def add1(x: Int) = x+1 val add2 = compose(add1, add1)</lang>

We can achieve a more natural style by creating a container class for composable functions, which provides the compose method 'o':

<lang scala>class Composable[A](f: A => A) {

 def o (g: A => A) = compose(f, g)

}

implicit def toComposable[A](f: A => A) = new Composable(f)

val add3 = (add1 _) o add2</lang>

> (add2 o add3)(37)
res0: Int = 42

Scheme

<lang scheme>(define (compose f g) (lambda (x) (f (g x))))</lang> Example: <lang scheme> (display ((compose sin asin) 0.5)) (newline)</lang> Output: <lang>0.5</lang>

Slate

Function (method) composition is standard: <lang slate>[| :x | x + 1] ** [| :x | x squared] applyTo: {3}</lang>

Smalltalk

<lang smalltalk>| composer fg | composer := [ :f :g | [ :x | f value: (g value: x) ] ]. fg := composer value: [ :x | x + 1 ]

              value: [ :x | x * x ].

(fg value:3) displayNl.</lang>

Standard ML

This is already defined as the o operator in Standard ML. <lang sml>fun compose (f, g) x = f (g x)</lang> Example use: <lang sml>- fun compose (f, g) x = f (g x); val compose = fn : ('a -> 'b) * ('c -> 'a) -> 'c -> 'b - val sin_asin = compose (Math.sin, Math.asin); val sin_asin = fn : real -> real - sin_asin 0.5; val it = 0.5 : real</lang>

Tcl

Works with: Tcl version 8.5


This creates a compose procedure that returns an anonymous function term that should be expanded as part of application to its argument. <lang tcl>package require Tcl 8.5 namespace path {::tcl::mathfunc}

proc compose {f g} {

   list apply [list {f g x} {{*}$f [{*}$g $x]}] $f $g]

}

set sin_asin [compose sin asin] {*}$sin_asin 0.5 ;# ==> 0.5 {*}[compose abs int] -3.14 ;# ==> 3</lang>

Clojure

Composition is built in to Clojure. Simply call the comp function

<lang lisp>user=>(comp rest rest);returns a closure that takes the rest of rest.</lang>

Ursala

Functional composition is a built in operation expressible as f+g for functions f and g, hence hardly worth defining. However, it could be defined without using the operator like this. <lang Ursala>compose("f","g") "x" = "f" "g" "x"</lang> test program: <lang Ursala>#import nat

  1. cast %n

test = compose(successor,double) 3</lang> output:

7