# 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 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 in the caller's context 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 (or at least by reference), otherwise changes to it (made within sum) would not be visible in the caller's context when computing each of 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.

`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
```

## ALGOL 60

Honor given where honor is due. In Algol 60, 'call by name' is the default argument evaluation.

```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
```

## 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

## 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 tempend jsum script term_func	on f(i)		return 1 / i	end fend script return jsum(a reference to i, 1, 100, term_func)`

Output: 5.18737751764

## 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
```

## Bracmat

`( ( sum  =   I lo hi Term temp    .   !arg:((=?I),?lo,?hi,(=?Term))      & 0:?temp      & !lo:?!I      &   whl        ' ( !!I:~>!hi          & !temp+!Term:?temp          & 1+!!I:?!I          )      & !temp  )& sum\$((=i),1,100,(=!i^-1)));`

Output:

`14466636279520351160221518043104131447711/2788815009188499086581352357412492142272`

## 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

## 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

## 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));    }}`

## 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 := 0for i = lo to hi   temp += eval(bFunc)next ireturn temp `

## 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))14466636279520351160221518043104131447711/2788815009188499086581352357412492142272CL-USER> (float (sum i 1 100 (/ 1 i)))5.1873775`

## 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 @nogc*/ {    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`

## 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...

## 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 shorthanddef 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 })`

## Elixir

Translation of: Erlang
`defmodule JensenDevice do  def task, do: sum( 1, 100, fn i -> 1 / i end )   defp sum( i, high, _term ) when i > high, do: 0  defp sum( i, high, term ) do    temp = term.( i )    temp + sum( i + 1, high, term )  endend IO.puts JensenDevice.task`
Output:
```5.1873775176396215
```

## 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
```

## Factor

Similar to the Java and Kotlin examples:

`: sum ( lo hi term -- x ) [ [a,b] ] dip map-sum ; inline 1 100 [ recip ] sum .`

This version is a bit closer to the original, as it increments `i` in the caller's namespace.

`SYMBOL: i : sum ( i lo hi term -- x )    [ [a,b] ] dip pick [ inc ] curry compose map-sum nip ;    inline i 1 100 [ recip ] sum .`
Output:
```5+522561233577855727314756256041670736351/2788815009188499086581352357412492142272
```

## 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 [email protected] f/ ; sum f.`

## Fortran

Fortran does not offer call-by-name in the manner of the Algol language. It passes parameters by reference (i.e. by passing the storage address) and alternatively uses copy-in, copy-out to give the same effect, approximately, as by reference. If a parameter is an arithmetic expression, it will be evaluated and its value stored in a temporary storage area, whose address will be passed to the routine. This evaluation is done once only for each call, thus vitiating the repeated re-evaluation required by Jensen's device every time within the routine that the parameter is accessed. So, this will not work
`      FUNCTION SUM(I,LO,HI,TERM)        SUM = 0        DO I = LO,HI          SUM = SUM + TERM        END DO      END FUNCTION SUM      WRITE (6,*) SUM(I,1,100,1.0/I)      END`

Here, type declarations have been omitted to save space because they won't help - until there appears a "BY NAME" or some such phrasing. Although variable `I` in the calling routine will have its value adjusted as the DO-loop in SUM proceeds (the parameter being passed by reference), this won't affect the evaluation of 1.0/I, which will be performed once using whatever value is in the caller's variable (it is uninitialised, indeed, undeclared also and so by default an integer) then the function is invoked with the address of the location containing that result. The function will make many references to that result, obtaining the same value each time. The fact that the caller's `I` will be changed each time doesn't matter.

Fortran does offer a facility to pass a function as a parameter using the EXTERNAL declaration, as follows - SUM is a F90 library function, so a name change to SUMJ:
`      FUNCTION SUMJ(I,LO,HI,TERM)	!Attempt to follow Jensen's Device...       INTEGER I	!Being by reference is workable.       INTEGER LO,HI	!Just as any other parameters.       EXTERNAL TERM	!Thus, not a variable, but a function.        SUMJ = 0        DO I = LO,HI	!The specified span.          SUMJ = SUMJ + TERM(I)	!Number and type of parameters now apparent.        END DO		!TERM will be evaluated afresh, each time.      END FUNCTION SUMJ	!So, almost there.       FUNCTION THIS(I)	!A function of an integer.       INTEGER I        THIS = 1.0/I	!Convert to floating-point.      END		!Since 1/i will mostly give zero.       PROGRAM JENSEN	!Aspiration.      EXTERNAL THIS	!Thus, not a variable, but a function.      INTEGER I		!But this is a variable, not a function.       WRITE (6,*) SUMJ(I,1,100,THIS)	!No statement as to the parameters of THIS.      END`

The result of this is 5.187378, however it does not follow the formalism of Jensen's Device. The invocation statement SUMJ(I,1,100,THIS) does not contain the form of the function but only its name, and the function itself is defined separately. This means that the convenience of different functions via the likes of SUM(I,1,100,1.0/I**2) is unavailable, a separately-defined function with its own name must be defined for each such function. Further, the SUM routine must invoke TERM(I) itself, explicitly supplying the appropriate parameter. And the fact that variable `I` is a parameter to SUM is an irrelevance, and might as well be omitted from SUMJ.

Incidentally, a subroutine such as TEST(A,B) invoked as TEST(X,X) enables the discovery of copy-in, copy-out parameter passing. Within the routine, modify the value of A and look to see if B suddenly has a new value also.

## Go

`package main import "fmt" var i int func sum(i *int, lo, hi int, term func() float64) float64 {    temp := 0.0    for *i = lo; *i <= hi; (*i)++ {        temp += term()    }    return temp} func main() {    fmt.Printf("%f\n", sum(&i, 1, 100, func() float64 { return 1.0 / float64(i) }))}`
Output:
```5.187378
```

## 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`

`import Control.Monadimport Control.Monad.STimport 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

## Huginn

`harmonic_sum( i, lo, hi, term ) {        temp = 0.0;        i *= 0.0;        i += lo;        while ( i <= hi ) {                temp += term();                i += 1.0;        }        return ( temp );} main() {        i = 0.0;        print( "{}\n".format( harmonic_sum( i, 1.0, 100.0, @[i](){ 1.0 / i; } ) ) );}`
Output:
`5.18737751764`

## 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 tempend`

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]`

## 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).

## 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;}                }));    }} `

## 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}));`

## Joy

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

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

## 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
```

## Julia

Works with: Julia version 0.6
Translation of: C
`macro sum(i, loname, hiname, term)    return quote        lo = \$loname        hi = \$hiname        tmp = 0.0        for i in lo:hi            tmp += \$term        end        return tmp    endend i = 0@sum(i, 1, 100, 1.0 / i)`

## Kotlin

`fun sum(lo: Int, hi: Int, f: (Int) -> Double) = (lo..hi).sumByDouble(f) fun main(args: Array<String>) = println(sum(1, 100, { 1.0 / it }))`

## 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 retendprint(sum("i", 1, 100, "1/i")) `

## M2000 Interpreter

The definition of the lazy function has two statements. First statement is a Module with one argument, the actual name of Jensen`s_Device, which make the function to get the same scope as module Jensen`s_Device, and the second statement is =1/i which return the expression.

` Module Jensen`s_Device {      Def double i      Report Lazy\$(1/i)  ' display the definition of the lazy function      Function Sum (&i, lo, hi, &f()) {            def double temp            For i= lo to hi {                  temp+=f()            }            =temp      }      Print Sum(&i, 1, 100, Lazy\$(1/i))==5.1873775176392  ' true      Print i=101 ' true}Jensen`s_Device   `

Using Decimal for better accuracy. change &i to &any to show that: when any change, change i, so f() use this i.

` Module Jensen`s_Device {      Def decimal i      Function Sum (&any, lo, hi, &f()) {            def decimal temp            For any= lo to hi {                  temp+=f()            }            =temp      }      Print Sum(&i, 1, 100, Lazy\$(1/i))[email protected]  ' true      Print i=101 ' true}Jensen`s_Device  `

Many other examples use single float. So this is one for single.

` Module Jensen`s_Device {      Def single i      Function Sum (&any, lo, hi, &f()) {            def single temp            For any= lo to hi {                  temp+=f()            }            =temp      }      Print Sum(&i, 1, 100, Lazy\$(1/i))=5.187378~  ' true      Print i=101 ' true}Jensen`s_Device `

## 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
```

## Mathematica / Wolfram Language

`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
```

## 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`

## 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])  `

## Nim

`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`

## 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

## 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

## Oforth

`: mysum(lo, hi, term)  | i | 0 lo hi for: i [ i term perform + ] ;`
Output:
```mysum(1, 100, #inv) println
5.18737751763962

mysum(1, 100, #[ sq inv ]) println
1.63498390018489
```

## 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}  endin  {Show {Sum 1 100 fun {\$ I} 1.0/{Int.toFloat I} end}}`

## 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.

## Pascal

`{\$MODE objFPC}type  tTerm = function(i: integer):real; function term(i:integer):real;Begin  term := 1/i;end; function sum(var i: LongInt;              lo,hi: integer;              term:tTerm):real; Begin  result := 0;  i := lo;  while i<=hi do begin    result := result+term(i);    inc(i);    end;end;      var  i : LongInt;Begin  writeln(sum(i,1,100,@term));end. `

Out

` 5.1873775176396206E+000`

## 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

## 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.

## Phix

Not really as asked for (implicit assumption replaced with explicit parameter) but this gives the required result.
I could also have done what C and PHP are doing, though in Phix I'd have to explicitly assign the static var within the loop.
I wholeheartedly agree with the comment on the Clipper example.

`function sumr(integer lo, hi, rid)    atom res = 0    for i=lo to hi do        res += call_func(rid,{i})    end for    return resend function function reciprocal(atom i) return 1/i end function ?sumr(1, 100, routine_id("reciprocal"))`
Output:
```5.187377518
```

## 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 `

## 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"`

## 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 TempEndProcedure Procedure.d term_func()  ProcedureReturn 1/iEndProcedure Answer.d = Sum(@i, 1, 100, @term_func())`

## 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

## 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 <- -1sum(i, 1, 100, i^x) #5.187378`

## Racket

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

` #lang algol60begin   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))) `

## 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`

## REXX

Note:   the 2nd and 3rd arguments for the   sum   function needn't be enclosed in quotes   (as they're numeric);
they were enclosed just to be consistent with the other arguments.

`/*REXX program demonstrates   Jensen's device   (via call subroutine, and args by name).*/parse arg d .;   if d=='' | d==","  then d=100   /*Not specified?  Then use the default.*/numeric digits d                                 /*use  D  decimal digits (9 is default)*/say 'using '    d    " decimal digits:"          /*display what's being used for digits.*/saysay sum( 'i',   "1",   '100',   "1/i" )          /*invoke  SUM  (100th harmonic number).*/exit                                             /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/sum: procedure;   parse arg j,start,finish,exp;              \$=0      interpret   'do'    j    "="   start   'to'   finish";   \$=\$+"    exp    ';   end'            /*   ────    ═    ───   ═════   ────   ══════──────────    ═══    ───────── */            /*    lit   var   lit    var     lit     var   literal     var     literal  */      return \$`
output   when using the default input:
```using  100  decimal digits:

5.187377517639620260805117675658253157908972126708451653176533956587219557532550496605687768923120415
```
output   when using the input:   1000

(Shown at three-quarter size   and   with 200 characters per line.)

```using  1000  decimal digits:

5.187377517639620260805117675658253157908972126708451653176533956587219557532550496605687768923120413552951372900080959485764334902003859251284547479399606488677719356437701034351417501628003612133813
93634033610397170258150385609229760925775852490242015786454123413833660918987060275907253504512582948807527866739590394714709377905509971663909084580816222756304901297019081913723833776150679344482592
19985786828216280140988475651174867766685160764730429716983310052063466701008405663630740646670436720827975050329078640945579952223172461998152578702106818073281191723171032278163615245743308956980821
10786794204451169328900410057940565163334352244388766863157323818250401277131246550164879348955299573048040410736739783727083287179928615106959660501145265658411572959372901925824344377263363761945330
17905075097606740175205276891748232922334187250177881689092871712673549165589217457070884105311065936887252732260150280756519586504475363590572034459636088593436136141078274322996362525543164325745468
2
```

## Ring

` # Project : Jensen's Device decimals(14)i = 100see sum(i,1,100,"1/n") + nl func sum(i,lo,hi,term)        temp = 0        for n = lo to hi step 1             eval("num = " + term)             temp = temp + num        next        return temp `

Output:

```5.18737751763962
```

## 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  sumendp 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}endp sum2(1, 100) {|i| 1.0/i}  # => 5.18737751763962`

Even more concise: (requires ruby >= 2.4)

` def sum lo, hi, &term  (lo..hi).sum(&term)endp sum(1,100){|i| 1.0/i}   # => 5.187377517639621# or using Rational:p sum(1,100){|i| Rational(1,i)}  # => 14466636279520351160221518043104131447711 / 2788815009188499086581352357412492142272 `

## 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 MyIntdef 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
```

## 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
```

## 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
```

## Sidef

`var i;func sum (i, lo, hi, term) {    var temp = 0;    for (*i = lo; *i <= hi; (*i)++) {        temp += term.run;    };    return temp;};say sum(\i, 1, 100, { 1 / i });`
Output:
`5.18737751763962026080511767565825315790899`

## Simula

Translation of: algol60
Works with: SIMULA-67
Compare with Algol 60, in Simula 67 'call by name' is specified with name. It is a true 'call by name' evaluation not a 'procedure parameter' emulation.
`comment Jensen's Device;begin   integer i;   real procedure sum (i, lo, hi, term);      name i, term;      value lo, hi;      integer i, lo, hi;      real term;      comment term is passed by-name, and so is i;   begin      integer j;      real temp;      temp := 0;      for j := lo step 1 until hi do      begin         i := j;         temp := temp + term      end;      sum := temp   end;   comment note the correspondence between the mathematical notation and the call to sum;   outreal (sum (i, 1, 100, 1/i), 7, 14)end`
Output:
``` 5.187378&+000
```

## Standard ML

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

Output: 5.18737751764

## Swift

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

(Prior to Swift 1.2, replace `@autoclosure term: () -> Double` with `term: @autoclosure () -> Double`.)

Output:
`5.187378`

## VBA

` Private Function sum(i As String, ByVal lo As Integer, ByVal hi As Integer, term As String) As Double    Dim temp As Double    For k = lo To hi        temp = temp + Evaluate(Replace(term, i, k))    Next k    sum = tempEnd FunctionSub Jensen_Device()    Debug.Print sum("i", 1, 100, "1/i")    Debug.Print sum("i", 1, 100, "i*i")    Debug.Print sum("j", 1, 100, "sin(j)")End Sub `
Output:
``` 5,18737751763962
338350
-0,12717101366042
```

## 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`

## 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
```

## ZX Spectrum Basic

`10 DEF FN r(x)=1/x20 LET f\$="FN r(i)"30 LET lo=1: LET hi=10040 GO SUB 100050 PRINT temp60 STOP 1000 REM Evaluation1010 LET temp=01020 FOR i=lo TO hi1030 LET temp=temp+VAL f\$1040 NEXT i1050 RETURN  `
Output:
```5.1873775
```