Jensen's Device

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Task
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 Jørn 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.

Contents

[edit] 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;
 5.18738E+00

[edit] 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

[edit] AppleScript

set i to 0
 
on jsum(i, lo, hi, term)
set {temp, i's contents} to {0, lo}
repeat while i's contents ≤ hi
set {temp, i's contents} to {temp + (term's f(i)), (i's contents) + 1}
end repeat
return temp
end jsum
 
script term_func
on f(i)
return 1 / i
end f
end script
 
return jsum(a reference to i, 1, 100, term_func)

Output: 5.18737751764

[edit] BBC BASIC

      PRINT FNsum(j, 1, 100, FNreciprocal)
END
 
DEF FNsum(RETURN i, lo, hi, RETURN func)
LOCAL temp
FOR i = lo TO hi
temp += FN(^func)
NEXT
= temp
 
DEF FNreciprocal = 1/i

Output:

5.18737752

[edit] C

#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;
}

Output: 5.18738

Works with: gcc

Alternatively, C's macros provide a closer imitation of ALGOL's call-by-name semantics:

#include <stdio.h>
 
int i;
 
#define sum(i, lo_byname, hi_byname, term) \
({ \
int lo = lo_byname; \
int hi = hi_byname; \
\
double temp = 0; \
for (i = lo; i <= hi; ++i) \
temp += term; \
temp; \
})

 
int main () {
printf("%f\n", sum(i, 1, 100, 1.0 / i));
return 0;
}

Output: 5.187378

[edit] C#

Can be simulated via lambda expressions:

using System;
 
class JensensDevice
{
public static double Sum(ref int i, int lo, int hi, Func<double> term)
{
double temp = 0.0;
for (i = lo; i <= hi; i++)
{
temp += term();
}
return temp;
}
 
static void Main()
{
int i = 0;
Console.WriteLine(Sum(ref i, 1, 100, () => 1.0 / i));
}
}

[edit] Clipper

With hindsight Algol60 provided this feature in a way that is terrible for program maintenance, because the calling code looks innocuous.

// Jensen's device in Clipper (or Harbour)
// A fairly direct translation of the Algol 60
// John M Skelton 11-Feb-2012
 
function main()
local i
? transform(sum(@i, 1, 100, {|| 1 / i}), "##.###############")
// @ is the quite rarely used pass by ref, {|| ...} is a
// code block (an anonymous function, here without arguments)
// The @i makes it clear that something unusual is occurring;
// a called function which modifies a parameter is commonly
// poor design!
return 0
 
function sum(i, lo, hi, bFunc)
local temp := 0
for i = lo to hi
temp += eval(bFunc)
next i
return temp
 

[edit] Common Lisp

Common Lisp does not have call-by-name for functions; however, it can be directly simulated by a macro wrapping selected parameters in lambdas.

(declaim (inline %sum))
 
(defun %sum (lo hi func)
(loop for i from lo to hi sum (funcall func i)))
 
(defmacro sum (i lo hi term)
`(%sum ,lo ,hi (lambda (,i) ,term)))
CL-USER> (sum i 1 100 (/ 1 i))
14466636279520351160221518043104131447711/2788815009188499086581352357412492142272
CL-USER> (float (sum i 1 100 (/ 1 i)))
5.1873775

[edit] D

There are better ways to do this in D, but this is closer to the original Algol version:

double sum(ref int i, in int lo, in int hi, lazy double term)
pure @safe nothrow {
double result = 0.0;
for (i = lo; i <= hi; i++)
result += term();
return result;
}
 
void main() {
import std.stdio;
 
int i;
sum(i, 1, 100, 1.0/i).writeln;
}
Output:
5.18738

[edit] DWScript

Must use a "while" loop, as "for" loop variables are restricted to local variable for code clarity, and this indeed a case where any kind of extra clarity helps.

function sum(var i : Integer; lo, hi : Integer; lazy term : Float) : Float;
begin
i:=lo;
while i<=hi do begin
Result += term;
Inc(i);
end;
end;
 
var i : Integer;
 
PrintLn(sum(i, 1, 100, 1.0/i));

Output: 5.187...

[edit] C++

#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;
}

Output: 5.18738

[edit] 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 })

[edit] Erlang

No call by name, no macros, so I use a fun(ction). Actually, the the macro part is a lie. Somebody else, that knows how, could do a parse transform.

 
-module( jensens_device ).
 
-export( [task/0] ).
 
task() ->
sum( 1, 100, fun (I) -> 1 / I end ).
 
 
 
sum( I, High, _Term ) when I > High -> 0;
sum( I, High, Term ) ->
Temp = Term( I ),
Temp + sum( I + 1, High, Term ).
 
Output:
4> jensens_device:task().
5.1873775176396215


[edit] Forth

This version passes i on the stack:

: sum 0 s>f 1+ swap ?do i over execute f+ loop drop ;
:noname s>f 1 s>f fswap f/ ; 1 100 sum f.

Output: 5.18737751763962

The following version passes i and 1/i as execution tokens and is thus closer to the original, but less idiomatic:

fvariable ii \ i is a Forth word that we need
: sum ( xt1 lo hi xt2 -- r )
0e swap 1+ rot ?do ( addr xt r1 )
i s>f over execute f! dup execute f+
loop 2drop ;
' ii 1 100 :noname 1e ii f@ f/ ; sum f.

[edit] Groovy

Translation of: JavaScript

Solution:

def sum = { i, lo, hi, term ->
(lo..hi).sum { i.value = it; term() }
}
def obj = [:]
println (sum(obj, 1, 100, { 1 / obj.value }))

Output:

5.1873775176

[edit] 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

[edit] Icon and Unicon

Traditional call by name and reference are not features of Icon/Unicon. Procedures parameters are passed by value (immutable types) and reference (mutable types). However, a similar effect may be accomplished by means of co-expressions. The example below was selected for cleanliness of calling.

record mutable(value)   # record wrapper to provide mutable access to immutable types
 
procedure main()
A := mutable()
write( sum(A, 1, 100, create 1.0/A.value) )
end
 
procedure sum(A, lo, hi, term)
temp := 0
every A.value := lo to hi do
temp +:= @^term
return temp
end

Refreshing the co-expression above is more expensive to process but to avoid it requires unary alternation in the call.

    write( sum(A, 1, 100, create |1.0/A.value) )
...
temp +:= @term

Alternately, we can use a programmer defined control operator (PDCO) approach that passes every argument as a co-expression. Again the refresh co-expression/unary iteration trade-off can be made. The call is cleaner looking but the procedure code is less clear. Additionally all the parameters are passed as individual co-expressions.

    write( sum{A.value, 1, 100, 1.0/A.value} )
...
procedure sum(X)
...
every @X[1] := @X[2] to @X[3] do
temp +:= @^X[4]

[edit] J

Solution:

jensen=: monad define
'name lo hi expression'=. y
temp=. 0
for_n. lo+i.1+hi-lo do.
(name)=. n
temp=. temp + ".expression
end.
)

Example:

   jensen 'i';1;100;'1%i'
5.18738

Note, however, that in J it is reasonably likely that the expression (or an obvious variation on the expression) can deal with the looping itself. And in typical use this often simplifies to entering the expression and data directly on the command line.

And another obvious variation here would be turning the expression into a named entity (if it has some lasting usefulness).

[edit] Java

This is Java 8.

import java.util.function.*;
import java.util.stream.*;
 
public class Jensen {
static double sum(int lo, int hi, IntToDoubleFunction f) {
return IntStream.rangeClosed(lo, hi).mapToDouble(f).sum();
}
 
public static void main(String args[]) {
System.out.println(sum(1, 100, (i -> 1.0/i)));
}
}
 

The program prints '5.187377517639621'.

Java 7 is more verbose, but under the hood does essentially the same thing:

public class Jensen2 {
 
interface IntToDoubleFunction {
double apply(int n);
}
 
static double sum(int lo, int hi, IntToDoubleFunction f) {
double res = 0;
for (int i = lo; i <= hi; i++)
res += f.apply(i);
return res;
 
}
public static void main(String args[]) {
System.out.println(
sum(1, 100,
new IntToDoubleFunction() {
public double apply(int i) { return 1.0/i;}
}));
}
}
 

[edit] JavaScript

Translation of: C

Uses an object o instead of integer pointer i, as the C example does.

var obj;
 
function sum(o, lo, hi, term) {
var tmp = 0;
for (o.val = lo; o.val <= hi; o.val++)
tmp += term();
return tmp;
}
 
obj = {val: 0};
alert(sum(obj, 1, 100, function() {return 1 / obj.val}));

The alert shows us '5.187377517639621'.

[edit] Joy

100 [0] [[1.0 swap /] dip +] primrec.

Joy does not have named parameters. Neither i nor 1/i are visible in the program.

[edit] jq

The technique used in the Javascript example can also be used in jq, but in jq it is more idiomatic to use "." to refer to the current term. For example, using sum/3 defined below, we can write: sum(1; 100; 1/.) to perform the task.

def sum(lo; hi; term):
reduce range(lo; hi+1) as $i (0; . + ($i|term));
 
# The task:
sum(1;100;1/.)
Output:
$ jq -n -f jensen.jq
5.187377517639621

[edit] Lua

 
function sum(var, a, b, str)
local ret = 0
for i = a, b do
ret = ret + setfenv(loadstring("return "..str), {[var] = i})()
end
return ret
end
print(sum("i", 1, 100, "1/i"))
 

[edit] M4

define(`for',
`ifelse($#,0,``$0'',
`ifelse(eval($2<=$3),1,
`pushdef(`$1',$2)$4`'popdef(`$1')$0(`$1',incr($2),$3,`$4')')')')
define(`sum',
`pushdef(`temp',0)`'for(`$1',$2,$3,
`define(`temp',eval(temp+$4))')`'temp`'popdef(`temp')')
sum(`i',1,100,`1000/i')

Output:

5142

[edit] Mathematica

sum[term_, i_, lo_, hi_] := Block[{temp = 0},
Do[temp = temp + term, {i, lo, hi}];
temp];
SetAttributes[sum, HoldFirst];

Output:

In[2]:= sum[1/i, i, 1, 100]
Out[2]= 14466636279520351160221518043104131447711/2788815009188499086581352357412492142272

In[3]:=N[sum[1/i, i, 1, 100]]
Out[3]:=5.18738

[edit] Maxima

mysum(e, v, lo, hi) := block([s: 0], for i from lo thru hi do s: s + subst(v=i, e), s)$
 
mysum(1/n, n, 1, 10);
7381/2520
 
/* compare with builtin sum */
sum(1/n, n, 1, 10);
7381/2520
 
/* what if n is assigned a value ? */
n: 200$
 
/* still works */
mysum(1/n, n, 1, 10);
7381/2520

[edit] NetRexx

 
import COM.ibm.netrexx.process.
 
class JensensDevice
 
properties static
interpreter=NetRexxA
exp=Rexx ""
termMethod=Method
 
method main(x=String[]) static
say sum('i',1,100,'1/i')
 
method sum(i,lo,hi,term) static SIGNALS IOException,NoSuchMethodException,IllegalAccessException,InvocationTargetException
sum=0
loop iv=lo to hi
sum=sum+termeval(i,iv,term)
end
return sum
 
method termeval(i,iv,e) static returns Rexx SIGNALS IOException,NoSuchMethodException,IllegalAccessException,InvocationTargetException
if e\=exp then interpreter=null
exp=e
 
if interpreter=null then do
termpgm='method term('i'=Rexx) static returns rexx;return' e
fw=FileWriter("termpgm.nrx")
fw.write(termpgm,0,termpgm.length)
fw.close
interpreter=NetRexxA()
interpreter.parse([String 'termpgm.nrx'],[String 'nocrossref'])
termClass=interpreter.getClassObject(null,'termpgm')
classes=[interpreter.getClassObject('netrexx.lang', 'Rexx', 0)]
termMethod=termClass.getMethod('term', classes)
end
 
return Rexx termMethod.invoke(null,[iv])
 
 

[edit] Nimrod

var i: int
 
proc harmonicSum(i: var int, lo, hi, term): float =
i = lo
while i <= hi:
result += term()
inc i
 
echo harmonicSum(i, 1, 100, proc: float = 1.0 / float(i))

Output:

5.1873775176396206e+00

[edit] Objeck

 
bundle Default {
class Jensens {
i : static : Int;
 
function : Sum(lo : Int, hi : Int, term : () ~ Float) ~ Float {
temp := 0.0;
 
for(i := lo; i <= hi; i += 1;) {
temp += term();
};
 
return temp;
}
 
function : term() ~ Float {
return 1.0 / i;
}
 
function : Main(args : String[]) ~ Nil {
Sum(1, 100, term() ~ Float)->PrintLine();
}
}
}
 

Output: 5.18738

[edit] 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))

Output: 5.187378

[edit] Oz

Translation using mutable references and an anonymous function:

declare
fun {Sum I Lo Hi Term}
Temp = {NewCell 0.0}
in
I := Lo
for while:@I =< Hi do
Temp := @Temp + {Term}
I := @I + 1
end
@Temp
end
I = {NewCell unit}
in
{Show {Sum I 1 100 fun {$} 1.0 / {Int.toFloat @I} end}}

Idiomatic code:

declare
fun {Sum Lo Hi F}
{FoldL {Map {List.number Lo Hi 1} F} Number.'+' 0.0}
end
in
{Show {Sum 1 100 fun {$ I} 1.0/{Int.toFloat I} end}}

[edit] PARI/GP

GP does not have pass-by-reference semantics for user-generated functions, though some predefined functions do. PARI programming allows this, though such a solution would essentially be identical to the C solution above.

[edit] Perl

my $i;
sub sum {
my ($i, $lo, $hi, $term) = @_;
my $temp = 0;
for ($$i = $lo; $$i <= $hi; $$i++) {
$temp += $term->();
}
return $temp;
}
 
print sum(\$i, 1, 100, sub { 1 / $i }), "\n";

Output: 5.18737751763962

Or you can take advantage of the fact that elements of the @_ are aliases of the original:

my $i;
sub sum {
my (undef, $lo, $hi, $term) = @_;
my $temp = 0;
for ($_[0] = $lo; $_[0] <= $hi; $_[0]++) {
$temp += $term->();
}
return $temp;
}
 
print sum($i, 1, 100, sub { 1 / $i }), "\n";

Output: 5.18737751763962

[edit] Perl 6

Rather than playing tricks like Perl 5 does, the declarations of the formal parameters are quite straightforward in Perl 6:

sub sum($i is rw, $lo, $hi, &term) {
my $temp = 0;
loop ($i = $lo; $i <= $hi; $i++) {
$temp += term;
}
return $temp;
}
 
my $i;
say sum $i, 1, 100, { 1 / $i };

Note that the C-style "for" loop is pronounced "loop" in Perl 6, and is the only loop statement that actually requires parens.

[edit] PHP

$i;
function sum (&$i, $lo, $hi, $term) {
$temp = 0;
for ($i = $lo; $i <= $hi; $i++) {
$temp += $term();
}
return $temp;
}
 
echo sum($i, 1, 100, create_function('', 'global $i; return 1 / $i;')), "\n";
//Output: 5.18737751764 (5.1873775176396)
 
function sum ($lo,$hi)
{
$temp = 0;
for ($i = $lo; $i <= $hi; $i++)
{
$temp += (1 / $i);
}
return $temp;
}
echo sum(1,100);
 
//Output: 5.1873775176396
 

[edit] PicoLisp

(scl 6)
 
(de jensen (I Lo Hi Term)
(let Temp 0
(set I Lo)
(while (>= Hi (val I))
(inc 'Temp (Term))
(inc I) )
Temp ) )
 
(let I (box) # Create indirect reference
(format
(jensen I 1 100 '(() (*/ 1.0 (val I))))
*Scl ) )

Output:

-> "5.187383"

[edit] PureBasic

Translation of: C
Prototype.d func()
 
Global i
 
Procedure.d Sum(*i.Integer, lo, hi, *term.func)
Protected Temp.d
For i=lo To hi
temp + *term()
Next
ProcedureReturn Temp
EndProcedure
 
Procedure.d term_func()
ProcedureReturn 1/i
EndProcedure
 
Answer.d = Sum(@i, 1, 100, @term_func())

[edit] Python

class Ref(object):
def __init__(self, value=None):
self.value = value
 
def harmonic_sum(i, lo, hi, term):
# term is passed by-name, and so is i
temp = 0
i.value = lo
while i.value <= hi: # Python "for" loop creates a distinct which
temp += term() # would not be shared with the passed "i"
i.value += 1 # Here the actual passed "i" is incremented.
return temp
 
i = Ref()
 
# note the correspondence between the mathematical notation and the
# call to sum it's almost as good as sum(1/i for i in range(1,101))
print harmonic_sum(i, 1, 100, lambda: 1.0/i.value)

Output: 5.18737751764

[edit] R

R uses a call by need evaluation strategy where function inputs are evaluated on demand and then cached; functions can bypass the normal argument evaluation by using functions substitute and match.call to access the parse tree of the as-yet-unevaluated arguments, and using parent.frame to access the scope of the caller. There are some proposed conventions to do this in a way that is less confusing to the user of a function; however, ignoring conventions we can come disturbingly close to the ALGOL call-by-name semantics.

sum <- function(var, lo, hi, term)
eval(substitute({
.temp <- 0;
for (var in lo:hi) {
.temp <- .temp + term
}
.temp
}, as.list(match.call()[-1])),
enclos=parent.frame())
 
sum(i, 1, 100, 1/i) #prints 5.187378
 
##and because of enclos=parent.frame(), the term can involve variables in the caller's scope:
x <- -1
sum(i, 1, 100, i^x) #5.187378


[edit] Racket

Racket happens to have an Algol 60-language, so Jensen's Device can be written just as Jørn Jensen did at Regnecentralen.

 
#lang algol60
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;
printnln (sum (i, 1, 100, 1/i))
end
 

But of course you can also use the more boring popular alternative of first class functions:

 
#lang racket/base
(define (sum lo hi f)
(for/sum ([i (in-range lo (add1 hi))]) (f i)))
(sum 1 100 (λ(i) (/ 1.0 i)))
 

[edit] Rascal

public num Jenssen(int lo, int hi, num (int i) term){
temp = 0;
while (lo <= hi){
temp += term(lo);
lo += 1;}
return temp;
}

With as output:

rascal>Jenssen(1, 100, num(int i){return 1.0/i;})
num: 5.18737751763962026080511767565825315790897212670845165317653395662

[edit] REXX

/*REXX program to demonstrate Jensen's device  (call sub, arg by name). */
numeric digits 50 /*might as well get some accuracy*/
 
say sum( 'i', "1", '100', "1/i" ) /*invoke SUM (100th harmonic num)*/
 
exit /*stick a fork in it, we're done.*/
 
/*──────────────────────────────────SUM subroutine──────────────────────*/
sum: procedure; parse arg i,start,finish,term; sum=0
interpret 'do' i '=' start 'to' finish'; sum=sum+'term'; end'
return sum

output

5.1873775176396202608051176756582531579089721267080

[edit] Ruby

Here, setting the variable and evaluating the term are truly executed in the "outer" context:

def sum(var, lo, hi, term, context)
sum = 0.0
lo.upto(hi) do |n|
sum += eval "#{var} = #{n}; #{term}", context
end
sum
end
p sum "i", 1, 100, "1.0 / i", binding # => 5.18737751763962

But here is the Ruby way to do it:

def sum2(lo, hi)
lo.upto(hi).inject(0.0) {|sum, n| sum += yield n}
end
p sum2(1, 100) {|i| 1.0/i} # => 5.18737751763962

[edit] Scala

Actually, the i parameter needs to be passed by reference, as done in so many examples here, so that changes made to it reflect on the parameter that was passed. Scala supports passing parameters by name, but not by reference, which means it can't change the value of any parameter passed. The code below gets around that by creating a mutable integer class, which is effectively the same as passing by reference.

class MyInt { var i: Int = _ }
val i = new MyInt
def sum(i: MyInt, lo: Int, hi: Int, term: => Double) = {
var temp = 0.0
i.i = lo
while(i.i <= hi) {
temp = temp + term
i.i += 1
}
temp
}
sum(i, 1, 100, 1.0 / i.i)

Result:

res2: Double = 5.187377517639621

[edit] Scheme

Scheme procedures do not support call-by-name. Scheme macros, however, do:

 
(define-syntax sum
(syntax-rules ()
((sum var low high . body)
(let loop ((var low)
(result 0))
(if (> var high)
result
(loop (+ var 1)
(+ result . body)))))))
 
(exact->inexact (sum i 1 100 (/ 1 i)))
5.18737751763962

[edit] Seed7

Seed7 supports call-by-name with function parameters:

 
$ include "seed7_05.s7i";
include "float.s7i";
 
var integer: i is 0;
 
const func float: sum (inout integer: i, in integer: lo, in integer: hi,
ref func float: term) is func
result
var float: sum is 0.0
begin
for i range lo to hi do
sum +:= term;
end for;
end func;
 
const proc: main is func
begin
writeln(sum(i, 1, 100, 1.0/flt(i)) digits 6);
end func;
 

Output:

5.187378

[edit] Standard ML

val i = ref 42 (* initial value doesn't matter *)
 
fun sum' (i, lo, hi, term) = let
val result = ref 0.0
in
i := lo;
while !i <= hi do (
result := !result + term ();
i := !i + 1
);
 !result
end
 
val () =
print (Real.toString (sum' (i, 1, 100, fn () => 1.0 / real (!i))) ^ "\n")

Output: 5.18737751764

[edit] Swift

var i = 42 // initial value doesn't matter
 
func sum(inout i: Int, lo: Int, hi: Int, term: @autoclosure () -> Double) -> Double {
var result = 0.0
for i = lo; i <= hi; i++ {
result += term()
}
return result
}
 
println(sum(&i, 1, 100, 1 / Double(i)))

Output: 5.187378

[edit] Tcl

Here, we set the value of the passed variable in the caller's frame. We then evaluate the passed term there too.

proc sum {var lo hi term} {
upvar 1 $var x
set sum 0.0
for {set x $lo} {$x < $hi} {incr x} {
set sum [expr {$sum + [uplevel 1 [list expr $term]]}]
}
return $sum
}
puts [sum i 1 100 {1.0/$i}] ;# 5.177377517639621

However, the solution is expressed more simply like this

proc sum2 {lo hi lambda} {
set sum 0.0
for {set n $lo} {$n < $hi} {incr n} {
set sum [expr {$sum + [apply $lambda $n]}]
}
return $sum
}
puts [sum2 1 100 {i {expr {1.0/$i}}}] ;# 5.177377517639621

[edit] zkl

zkl doesn't support call by name/address but does have reference objects. Using an explicit call to term:

fcn sum(ri, lo,hi, term){
temp:=0.0; ri.set(lo);
do{ temp += term(ri); } while(ri.inc() < hi); // inc return previous value
return(temp);
}
sum(Ref(0), 1,100, fcn(ri){ 1.0 / ri.value }).println();

Using function application/deferred(lazy) objects, we can make the function call implicit (addition forces evaluation of the LHS):

fcn sum2(ri, lo,hi, term){
temp:=0.0; ri.set(lo);
do{ temp=term+temp; } while(ri.inc() < hi); // inc return previous value
return(temp);
}
ri:=Ref(0);
sum2(ri, 1,100, 'wrap(){ 1.0 / ri.value }).println();

In this case, we can call sum or sum2 and it does the same thing (the ri parameter will be ignored).

Of course, as others have pointed out, this can be expressed very simply:

fcn sum3(lo,hi, term){ [lo..hi].reduce('wrap(sum,i){sum+term(i) },0.0) }
sum3(1,100, fcn(i){ 1.0 / i }).println();
Output:
5.187378
5.187378
5.187378
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