Jensen's Device

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Jensen's Device
You are encouraged to solve this task according to the task description, using any language you may know.

This task is an exercise in call by name.

Jensen's Device is a computer programming technique devised by Danish computer scientist Jensen after studying the ALGOL 60 Report.

The following program was proposed to illustrate the technique. It computes the 100th harmonic number:

begin
   integer i;
   real procedure sum (i, lo, hi, term);
      value lo, hi;
      integer i, lo, hi;
      real term;
      comment term is passed by-name, and so is i;
   begin
      real temp;
      temp := 0;
      for i := lo step 1 until hi do
         temp := temp + term;
      sum := temp
   end;
   comment note the correspondence between the mathematical notation and the call to sum;
   print (sum (i, 1, 100, 1/i))
end

The above exploits call by name to produce the correct answer (5.187...). It depends on the assumption that an expression passed as an actual parameter to a procedure would be re-evaluated every time the corresponding formal parameter's value was required. If the last parameter to sum had been passed by value, and assuming the initial value of i were 1, the result would have been 100 × 1/1 = 100.

Moreover, the first parameter to sum, representing the "bound" variable of the summation, must also be passed by name, otherwise it would not be possible to compute the values to be added. (On the other hand, the global variable does not have to use the same identifier, in this case i, as the formal parameter.)

Donald Knuth later proposed the Man or Boy Test as a more rigorous exercise.

Ada

<ada> with Ada.Text_IO; use Ada.Text_IO;

procedure Jensen_Device is

  function Sum
           (  I : not null access Float;
              Lo, Hi : Float;
              F : access function return Float
           )  return Float is
     Temp : Float := 0.0;
  begin
     I.all := Lo;
     while I.all <= Hi loop
        Temp := Temp + F.all;
        I.all := I.all + 1.0;
     end loop;
     return Temp;
  end Sum;

  I : aliased Float;
  function Inv_I return Float is
  begin
     return 1.0 / I;
  end Inv_I;

begin

  Put_Line (Float'Image (Sum (I'Access, 1.0, 100.0, Inv_I'Access)));

end Jensen_Device; </ada>

 5.18738E+00

ALGOL 68

Translation of: ALGOL 60

BEGIN
   INT i;
   PROC sum  = (REF INT i, INT lo, hi, PROC REAL term)REAL:
      COMMENT term is passed by-name, and so is i COMMENT
   BEGIN
      REAL temp := 0;
      i := lo;
      WHILE i <= hi DO           # ALGOL 68 has a "for" loop but it creates a distinct #
         temp +:= term;          # variable which would not be shared with the passed "i" #
         i +:= 1                 # Here the actual passed "i" is incremented. #
      OD;
      temp
   END;
   COMMENT note the correspondence between the mathematical notation and the call to sum COMMENT
   print (sum (i, 1, 100, REAL: 1/i))
END

Output: +5.18737751763962e +0

C

<cpp>#include <stdio.h>

int i; double sum(int *i, int lo, int hi, double (*term)()) {

   double temp = 0;
   for (*i = lo; *i <= hi; (*i)++)
       temp += term();
   return temp;

}

double term_func() { return 1.0 / i; }

int main () {

   printf("%f\n", sum(&i, 1, 100, term_func));
   return 0;

}</cpp> Output: 5.18738

C++

<cpp>#include <iostream>

int i; double sum(int &i, int lo, int hi, double (*term)()) {

   double temp = 0;
   for (i = lo; i <= hi; i++)
       temp += term();
   return temp;

}

double term_func() { return 1.0 / i; }

int main () {

   std::cout << sum(i, 1, 100, term_func) << std::endl;
   return 0;

}</cpp> Output: 5.18738

E

In E, the distinct mutable locations behind assignable variables can be reified as Slot objects. The E language allows a variable name (noun) to be bound to a particular slot, and the slot of an already-bound noun to be extracted, using the & operator.

(The definition of the outer i has been moved down to emphasize that it is unrelated to the i inside of sum.)

pragma.enable("one-method-object") # "def _.get" is experimental shorthand
def sum(&i, lo, hi, &term) {   # bind i and term to passed slots
    var temp := 0
    i := lo
    while (i <= hi) {          # E has numeric-range iteration but it creates a distinct
        temp += term           # variable which would not be shared with the passed i
        i += 1
    }
    return temp
}
{ 
    var i := null
    sum(&i, 1, 100, def _.get() { return 1/i })
}

1/i is not a noun, so there is no slot associated with it; so we use def _.get() { return 1/i } to define a slot object which does the computation when it is read as a slot.

The value returned by the above program (expression) is 5.187377517639621.

This emulation of the original call-by-name is of course unidiomatic; a natural version of the same computation would be:

def sum(lo, hi, f) {
    var temp := 0
    for i in lo..hi { temp += f(i) }
    return temp
}
sum(1, 100, fn i { 1/i })

Haskell

import Control.Monad
import Control.Monad.ST
import Data.STRef

sum' ref_i lo hi term =
  return sum `ap`
         mapM (\i -> writeSTRef ref_i i >> term) [lo..hi]

foo = runST $ do
        i <- newSTRef undefined -- initial value doesn't matter
        sum' i 1 100 $ return recip `ap` readSTRef i

main = print foo

Output: 5.187377517639621

OCaml

<ocaml>let i = ref 42 (* initial value doesn't matter *)

let sum' i lo hi term =

 let result = ref 0. in
   i := lo;
   while !i <= hi do
     result := !result +. term ();
     incr i
   done;
   !result

let () =

 Printf.printf "%f\n" (sum' i 1 100 (fun () -> 1. /. float !i))</ocaml>

Output: 5.187378

Python

<python>class Ref(object):

 def __init__(self, value=None):
   self.value=value

i = Ref() def sum(i, lo, hi, term):

  # term is passed by-name, and so is i
  temp = 0
  i.value = lo
  while i.value <= hi:          # Python has a "for" loop but it creates a distinct
     temp += term()             # variable which would not be shared with the passed "i"
     i.value += 1               # Here the actual passed "i" is incremented.
  return temp

1. note the correspondence between the mathematical notation and the call to sum

print (sum (i, 1, 100, lambda: 1.0/i.value))</python> Output: 5.18737751764