# Loops/with multiple ranges

Loops/with multiple ranges
You are encouraged to solve this task according to the task description, using any language you may know.

Some languages allow multiple loop ranges, such as the PL/I example (snippet) below.

`                                       /* all variables are DECLARED as integers. */          prod=  1;                    /*start with a product of unity.           */           sum=  0;                    /*  "     "  "   sum    " zero.            */             x= +5;             y= -5;             z= -2;           one=  1;         three=  3;         seven=  7;                                       /*(below)  **  is exponentiation:  4**3=64 */           do j=   -three  to     3**3        by three   ,                   -seven  to   +seven        by   x     ,                      555  to      550 - y               ,                       22  to      -28        by -three  ,                     1927  to     1939                   ,                        x  to        y        by   z     ,                    11**x  to    11**x + one;                                                        /* ABS(n) = absolute value*/           sum= sum + abs(j);                           /*add absolute value of J.*/           if abs(prod)<2**27 & j¬=0  then prod=prod*j; /*PROD is small enough & J*/           end;                                         /*not 0, then multiply it.*/                     /*SUM and PROD are used for verification of J incrementation.*/         display (' sum= ' ||  sum);                    /*display strings to term.*/         display ('prod= ' || prod);                    /*   "       "     "   "  */`

Simulate/translate the above PL/I program snippet as best as possible in your language,   with particular emphasis on the   do   loop construct.

The   do   index must be incremented/decremented in the same order shown.

If feasible, add commas to the two output numbers (being displayed).

Show all output here.

`      A simple PL/I   DO  loop  (incrementing or decrementing)  has the construct of:             DO variable = start_expression    {TO ending_expression]       {BY increment_expression} ;                 ---or---            DO variable = start_expression    {BY increment_expression}    {TO ending_expression]    ;         where it is understood that all expressions will have a value.  The  variable  is normally a       scaler variable,  but need not be  (but for this task, all variables and expressions are declared      to be scaler integers).   If the   BY   expression is omitted,  a   BY   value of unity is used.      All expressions are evaluated before the   DO   loop is executed,  and those values are used      throughout the   DO   loop execution   (even though, for instance,  the value of   Z   may be      changed within the   DO   loop.    This isn't the case here for this task.         A multiple-range   DO   loop can be constructed by using a comma (,) to separate additional ranges      (the use of multiple   TO   and/or   BY   keywords).     This is the construct used in this task.       There are other forms of   DO   loops in PL/I involving the  WHILE  clause,  but those won't be       needed here.    DO  loops without a   TO   clause might need a   WHILE   clause  or some other       means of exiting the loop  (such as  LEAVE,  RETURN,  SIGNAL,  GOTO,  or  STOP),  or some other       (possible error) condition that causes transfer of control outside the  DO  loop.       Also, in PL/I, the check if the   DO   loop index value is outside the range is made at the       "head"  (start)  of the   DO  loop,  so it's possible that the   DO   loop isn't executed,  but       that isn't the case for any of the ranges used in this task.        In the example above, the clause:                    x    to y       by z           will cause the variable   J   to have to following values  (in this order):  5  3  1  -1  -3  -5       In the example above, the clause:                 -seven  to +seven  by x        will cause the variable   J   to have to following values  (in this order):  -7  -2   3  `

## ALGOL 60

Works with: MARST
`begin  integer prod, sum, x, y, z, one, three, seven;  integer j;  prod := 1;  sum := 0;  x := 5; y := -5; z := -2;  one := 1;  three := 3;  seven := 7;   for j := -three  step  three  until 3^3    ,           -seven  step      x  until seven  ,              555  step      1  until 550 - y,               22  step -three  until -28    ,             1927  step      1  until 1939   ,                x  step      z  until y      ,             11^x  step      1  until 11^x + one  do begin    sum := sum + iabs(j);    if iabs(prod) < 2^27 & j != 0 then prod := prod*j  end;   outstring(1, " sum= "); outinteger(1, sum);  outstring(1, "\n");  outstring(1, "prod= "); outinteger(1, prod); outstring(1, "\n")end `
Output:
``` sum= 348173
prod= -793618560
```

## ALGOL W

As with most of the other languages, Algol W doesn't support multiple loop ranges, so a sequence pf loops is used instead.

`begin    % translation of task PL/1 code, with minimal changes, semicolons required by      %    % PL/1 but redundant in Algol W retained ( technically they introduce empty        %    % statements after the "if" in the loop body and before the final "end" )          %    % Note that in Algol W, the loop counter is a local variable to the loop and       %    % the value of j is not available outside the loops                                %    procedure loopBody ( integer value j );  %(below)  **  is exponentiation:  4**3=64 %    begin sum := sum + abs(j);                                %add absolute value of J.%          if abs(prod)<2**27 and j not = 0 then prod := prod*j; %PROD is small enough & J%                                                              % ABS(n) = absolute value%          end;                                                %not 0, then multiply it.%                           %SUM and PROD are used for verification of J incrementation.%      integer prod, sum, x, y, z, one, three, seven;          prod :=  1;                        %start with a product of unity.           %           sum :=  0;                        %  "     "  "   sum    " zero.            %             x := +5;             y := -5;             z := -2;           one :=  1;         three :=  3;         seven :=  7;         for j :=   -three  step  three until round( 3**3 )        do loopBody( j );         for j :=   -seven  step  x     until    +seven            do loopBody( j );         for j :=      555              until    550 - y           do loopBody( j );         for j :=       22  step -three until   -28                do loopBody( j );         for j :=     1927              until  1939                do loopBody( j );         for j :=        x  step  z     until     y                do loopBody( j );         for j := round( 11**x )        until round( 11**x ) + one do loopBody( j );         write(s_w := 0, " sum= ",  sum);                    %display strings to term.%         write(s_w := 0, "prod= ", prod);                    %   "       "     "   "  %end.`
Output:
``` sum=         348173
prod=     -793618560
```

## C

`#include <stdio.h>#include <stdlib.h>#include <locale.h> long prod = 1L, sum = 0L; void process(int j) {    sum += abs(j);    if (labs(prod) < (1 << 27) && j) prod *= j;} long ipow(int n, uint e) {    long pr = n;    int i;    if (e == 0) return 1L;    for (i = 2; i <= e; ++i) pr *= n;    return pr;} int main() {    int j;    const int x = 5, y = -5, z = -2;    const int one = 1, three = 3, seven = 7;    long p = ipow(11, x);    for (j = -three; j <= ipow(3, 3); j += three) process(j);    for (j = -seven; j <= seven; j += x) process(j);    for (j = 555; j <= 550 - y; ++j) process(j);    for (j = 22; j >= -28; j -= three) process(j);    for (j = 1927; j <= 1939; ++j) process(j);    for (j = x; j >= y; j -= -z) process(j);    for (j = p; j <= p + one; ++j) process(j);    setlocale(LC_NUMERIC, "");    printf("sum  = % 'ld\n", sum);    printf("prod = % 'ld\n", prod);    return 0;}`
Output:
```sum  =  348,173
prod = -793,618,560
```

## Factor

Factor doesn't have any special support for this sort of thing, but we can store iterable `range` objects in a collection and loop over them.

`USING: formatting kernel locals math math.functions math.rangessequences sequences.generalizations tools.memory.private ; [let                            ! Allow lexical variables.     1 :> prod!                 ! Start with a product of unity.     0 :> sum!                  !   "     "  "   sum    " zero.     5 :> x    -5 :> y    -2 :> z     1 :> one     3 :> three     7 :> seven     three neg 3 3 ^ three <range>              ! Create array    seven neg seven x     <range>              ! of 7 ranges.    555 550 y -             [a,b]    22 -28 three neg      <range>    1927 1939               [a,b]    x y z                 <range>    11 x ^ 11 x ^ 1 +       [a,b] 7 narray     [        [            :> j j abs sum + sum!            prod abs 2 27 ^ < j zero? not and            [ prod j * prod! ] when        ] each                      ! Loop over range.    ] each                          ! Loop over array of ranges.     ! SUM and PROD are used for verification of J incrementation.    sum prod [ commas ] [email protected] " sum=  %s\nprod= %s\n" printf]`
Output:
``` sum=  348,173
prod= -793,618,560
```

## Go

Nothing fancy from Go here (is there ever?), just a series of individual for loops.

`package main import "fmt" func pow(n int, e uint) int {    if e == 0 {        return 1    }    prod := n    for i := uint(2); i <= e; i++ {        prod *= n    }    return prod} func abs(n int) int {    if n >= 0 {        return n    }    return -n} func commatize(n int) string {    s := fmt.Sprintf("%d", n)    if n < 0 {        s = s[1:]    }    le := len(s)    for i := le - 3; i >= 1; i -= 3 {        s = s[0:i] + "," + s[i:]    }    if n >= 0 {        return " " + s    }    return "-" + s} func main() {    prod := 1    sum := 0    const (        x     = 5        y     = -5        z     = -2        one   = 1        three = 3        seven = 7    )    p := pow(11, x)    var j int     process := func() {        sum += abs(j)        if abs(prod) < (1<<27) && j != 0 {            prod *= j        }    }     for j = -three; j <= pow(3, 3); j += three {        process()    }    for j = -seven; j <= seven; j += x {        process()    }    for j = 555; j <= 550-y; j++ {        process()    }    for j = 22; j >= -28; j -= three {        process()    }    for j = 1927; j <= 1939; j++ {        process()    }    for j = x; j >= y; j -= -z {        process()    }    for j = p; j <= p+one; j++ {        process()    }    fmt.Println("sum  = ", commatize(sum))    fmt.Println("prod = ", commatize(prod))}`
Output:
```sum  =   348,173
prod =  -793,618,560
```

## Groovy

Solution:

`def (prod, sum, x, y, z, one, three, seven) = [1, 0, +5, -5, -2, 1, 3, 7] for (    j in (        ((-three) .. (3**3)       ).step(three)      + ((-seven) .. (+seven)     ).step(x)      + (555      .. (550-y)      )      + (22       .. (-28)        ).step(three)    // This is correct!      // Groovy interprets positive step size as stride through the LIST ELEMENTS as ordered      // and negative step size as stride through the REVERSED LIST ELEMENTS as ordered      //   so step(-3) gives:   -28, -25, -22, ... ,  20      //   while step(3) gives:  22,  19,  16, ... , -26      + (1927     .. 1939         )      + (x        .. y            ).step(z)      + (11**x    .. (11**x + one))    )) {     sum = sum + j.abs()    if ( prod.abs() < 2**27 && j != 0) prod *= j} println " sum= \${sum}"println "prod= \${prod}"`

Output:

``` sum= 348177
prod= -793618560```

## J

J uses the names x, y, m, n, u, v to pass arguments into explicit definitions. Treating these as reserved names is reasonable practice. Originally these had been x. , y. etceteras however the dots must have been deemed "noisy".

We've passed the range list argument literally for evaluation in local scope. Verb f evaluates and concatenates the ranges, then perhaps the ensuing for. loop looks somewhat like familiar code.

` NB. http://rosettacode.org/wiki/Loops/Wrong_ranges#JNB. define range as a linear polynomialstart =: 0&{stop =: 1&{increment =: 2&{ :: 1:  NB. on error use 1range =: (start , increment) p. [: i. [: >: [: <. (stop - start) % increment f =: 3 :0 input =. y 'prod sum x y z one three seven' =. 1 0 5 _5 _2 1 3 7 J =. ([: ; range&.>) ". input for_j. J do.  sum =. sum + | j  if. ((|prod)<2^27) *. (0 ~: j) do.   prod =. prod * j  end. end. sum , prod) `
```   ] A =: f '((-three), (3^3), three); ((-seven),seven,x); (555 , 550-y); (22 _28, -three); 1927 1939; (x,y,z); (0 1 + 11^x)'
348173 _7.93619e8

20j0 ": A
348173          _793618560
```

## Julia

Julia allows concatenation of iterators with the ; iterator within a vector. An attempt was made to preserve the shape of the PL/1 code.

`using Formatting function PL1example()                                     # all variables are DECLARED as integers.    prod  =  1;                     # start with a product of unity.    sum   =  0;                     #   "     "  "   sum    " zero.    x     = +5;    y     = -5;    z     = -2;    one   =  1;    three =  3;    seven =  7;                                    # (below)  **  is exponentiation:  4**3=64    for j in [           -three   :  three :  3^3           ;                         -seven   :   x    :  +seven        ;                            555            :  550 - y       ;                             22   : -three :  -28           ;                           1927            :  1939          ;                              x   :  z     :  y             ;                           11^x            :   11^x + one   ]                                                        # ABS(n) = absolute value        sum = sum + abs(j);                             # add absolute value of J.        if abs(prod) < 2^27 && j !=0 prod = prod*j      # PROD is small enough & J        end;                                            # not 0, then multiply it.    end             # SUM and PROD are used for verification of J incrementation.    println(" sum = \$(format(sum, commas=true))");      # display strings to term.    println("prod = \$(format(prod, commas=true))");     #   "       "     "   "end PL1example() `
Output:
```
sum = 348,173
prod = -793,618,560

```

## Kotlin

Nothing special here, just a series of individual for loops.

`// Version 1.2.70 import kotlin.math.abs infix fun Int.pow(e: Int): Int {    if (e == 0) return 1    var prod = this    for (i in 2..e) {        prod *= this    }    return prod} fun main(args: Array<String>) {    var prod = 1    var sum = 0    val x = 5    val y = -5    val z = -2    val one = 1    val three = 3    val seven = 7    val p = 11 pow x    fun process(j: Int) {        sum += abs(j)        if (abs(prod) < (1 shl 27) && j != 0) prod *= j    }     for (j in -three..(3 pow 3) step three) process(j)    for (j in -seven..seven step x) process(j)    for (j in 555..550-y) process(j)    for (j in 22 downTo -28 step three) process(j)    for (j in 1927..1939) process(j)    for (j in x downTo y step -z) process(j)    for (j in p..p + one) process(j)    System.out.printf("sum  = % ,d\n", sum)    System.out.printf("prod = % ,d\n", prod)}`
Output:
```sum  =  348,173
prod = -793,618,560
```

## Perl

`use constant   one =>  1;use constant three =>  3;use constant seven =>  7;use constant     x =>  5;use constant    yy => -5; # 'y' conflicts with use as equivalent to 'tr' operator (a carry-over from 'sed')use constant     z => -2; my \$prod = 1; sub from_to_by {    my(\$begin,\$end,\$skip) = @_;    my \$n = 0;    grep{ !(\$n++ % abs \$skip) } \$begin <= \$end ? \$begin..\$end : reverse \$end..\$begin;} sub commatize {    (my \$s = reverse shift) =~ s/(.{3})/\$1,/g;    \$s =~ s/,(-?)\$/\$1/;    \$s = reverse \$s;} for my \$j (    from_to_by(-three,3**3,three),    from_to_by(-seven,seven,x),    555 .. 550 - yy,    from_to_by(22,-28,-three),    1927 .. 1939,    from_to_by(x,yy,z),    11**x .. 11**x+one,   ) {     \$sum  += abs(\$j);     \$prod *= \$j if \$j and abs(\$prod) < 2**27;} printf "%-8s %12s\n", 'Sum:',     commatize \$sum;printf "%-8s %12s\n", 'Product:', commatize \$prod;`
Output:
```Sum:          348,173
Product: -793,618,560```

## Perl 6

This task is really conflating two separate things, (at least in Perl 6). Sequences and loops are two different concepts and may be considered / implemented separately from each other.

Yes, you can generate a sequence with a loop, and a loop can use a sequence for an iteration value, but the two are somewhat orthogonal and don't necessarily overlap.

Sequences are first class objects in Perl 6. You can (and typically do) generate a sequence using the (appropriately enough) sequence operator and can assign it to a variable and/or pass it as a parameter; the entire sequence, not just it's individual values. It may be used in a looping construct, but it is not necessary to do so.

Various looping constructs often do use sequences as their iterator but not exclusively, possibly not even in the majority.

Displaying the j sequence as well since it isn't very large.

`sub comma { (\$^i < 0 ?? '-' !! '') ~ \$i.abs.flip.comb(3).join(',').flip } my \x     =  5;my \y     = -5;my \z     = -2;my \one   =  1;my \three =  3;my \seven =  7; my \$j = flat  ( -three, *+three … 3³         ),  ( -seven, *+x     …^ * > seven ),  ( 555   .. 550 - y             ),  ( 22,     *-three …^ * < -28   ),  ( 1927  .. 1939                ),  ( x,      *+z     …^ * < y     ),  ( 11**x .. 11**x + one         ); put 'j sequence: ', \$j;put '       Sum: ', comma [+] \$j».abs;put '   Product: ', comma ([\*] \$j.grep: so +*).first: *.abs > 2²⁷; # Or, an alternate method for generating the 'j' sequence, employing user-defined# operators to preserve the 'X to Y by Z' layout of the example code.# Note that these operators will only work for monotonic sequences. sub infix:<to> { \$^a ... \$^b }sub infix:<by> { \$^a[0, \$^b.abs ... *] } \$j = cache flat    -three  to          3**3  by  three ,    -seven  to         seven  by      x ,       555  to     (550 - y)            ,        22  to           -28  by -three ,      1927  to          1939  by    one ,         x  to             y  by      z ,     11**x  to (11**x + one)            ; put "\nLiteral minded variant:";put '       Sum: ', comma [+] \$j».abs;put '   Product: ', comma ([\*] \$j.grep: so +*).first: *.abs > 2²⁷;`
Output:
```j sequence: -3 0 3 6 9 12 15 18 21 24 27 -7 -2 3 555 22 19 16 13 10 7 4 1 -2 -5 -8 -11 -14 -17 -20 -23 -26 1927 1928 1929 1930 1931 1932 1933 1934 1935 1936 1937 1938 1939 5 3 1 -1 -3 -5 161051 161052
Sum: 348,173
Product: -793,618,560

Literal minded variant:
Sum: 348,173
Product: -793,618,560
```

## Phix

`integer prod =  1,       total =  0,  -- (renamed as sum is a Phix builtin)           x = +5,           y = -5,           z = -2,         one =  1,       three =  3,       seven =  7 sequence loopset = {{     -three,        power(3,3), three },                    {     -seven,            +seven,   x   },                    {        555,           550 - y,   1   },                    {         22,               -28, -three},                    {       1927,              1939,   1   },                    {          x,                 y,   z   },                    {power(11,x), power(11,x) + one,   1   }} for i=1 to length(loopset) do    integer {f,t,s} = loopset[i]    for j=f to t by s do        total += abs(j)        if abs(prod)<power(2,27) and j!=0 then            prod *= j        end if    end forend forprintf(1," sum = %,d\n",total)printf(1,"prod = %,d\n",prod)`
Output:
``` sum = 348,173
prod = -793,618,560
```

## Prolog

Prolog does not have the richness of some other languages where it comes to loops, variables and the like, but does have some rather interesting features such as difference lists and backtracking for generating solutions.

`for(Lo,Hi,Step,Lo)  :- Step>0, Lo=<Hi.for(Lo,Hi,Step,Val) :- Step>0, plus(Lo,Step,V), V=<Hi, !, for(V,Hi,Step,Val).for(Hi,Lo,Step,Hi)  :- Step<0, Lo=<Hi.for(Hi,Lo,Step,Val) :- Step<0, plus(Hi,Step,V), Lo=<V, !, for(V,Lo,Step,Val). sym(x,5).                 % symbolic lookups for valuessym(y,-5).sym(z,-2).sym(one,1).sym(three,3).sym(seven,7). range(-three,3^3,three).  % as close as we can syntactically getrange(-seven,seven,x).range(555,550-y,1).range(22,-28, -three).range(1927,1939,1).range(x,y,z).range(11^x,11^x+one,1). translate(V, V)   :- number(V), !.    % difference list based parsertranslate(S, V)   :- sym(S,V), !.translate(-S, V)  :- translate(S,V0), !, V is -V0.translate(A+B, V) :- translate(A,A0), translate(B, B0), !, V is A0+B0.translate(A-B, V) :- translate(A,A0), translate(B, B0), !, V is A0-B0.translate(A^B, V) :- translate(A,A0), translate(B, B0), !, V is A0^B0. range_value(Val) :-             % enumerate values for all ranges in order	range(From,To,Step), 	translate(From,F), translate(To,T), translate(Step,S), 	for(F,T,S,Val). calc_values([], S, P, S, P).    % calculate all values in generated ordercalc_values([J|Js], S, P, Sum, Product) :-  S0 is S + abs(J), ((abs(P)< 2^27, J \= 0) -> P0 is P * J; P0=P),  !, calc_values(Js, S0, P0, Sum, Product). calc_values(Sum, Product) :-    % Find the sum and product	findall(V, range_value(V), Values),	calc_values(Values, 0, 1, Sum, Product).`
```?- calc_values(Sum, Product).
Sum = 348173,
Product = -793618560.```

## PureBasic

`#X = 5 : #Y = -5 : #Z = -2#ONE   = 1 : #THREE = 3 : #SEVEN = 7Define j.iGlobal prod.i = 1, sum.i = 0 Macro ipow(n, e)  Int(Pow(n, e))EndMacro Macro ifn(x)  FormatNumber(x,0,".",",")EndMacro Macro loop_for(start, stop, step_for=1)  For j = start To stop Step step_for    proc(j)  NextEndMacro Procedure proc(j.i)  sum + Abs(j)  If (Abs(prod) < ipow(2 , 27)) And (j<>0)    prod * j  EndIfEndProcedure loop_for(-#THREE, ipow(3, 3), #THREE)loop_for(-#SEVEN, #SEVEN, #X)loop_for(555, 550 - #Y)loop_for(22, -28, -#THREE)loop_for(1927, 1939)loop_for(#X, #Y, #Z)loop_for(ipow(11, #X), ipow(11, #X) + 1) If OpenConsole("Loops/with multiple ranges")  PrintN("sum  = " + ifn(sum))  PrintN("prod = " + ifn(prod))  Input()EndIf`
Output:
```sum  = 348,173
prod = -793,618,560```

## Python

Pythons range function does not include the second argument hence the definition of _range()

`from itertools import chain prod, sum_, x, y, z, one,three,seven = 1, 0, 5, -5, -2, 1, 3, 7 def _range(x, y, z=1):    return range(x, y + (1 if z > 0 else -1), z) print(f'list(_range(x, y, z)) = {list(_range(x, y, z))}')print(f'list(_range(-seven, seven, x)) = {list(_range(-seven, seven, x))}') for j in chain(_range(-three, 3**3, three), _range(-seven, seven, x),                _range(555, 550 - y), _range(22, -28, -three),               _range(1927, 1939), _range(x, y, z),               _range(11**x, 11**x + 1)):    sum_ += abs(j)    if abs(prod) < 2**27 and (j != 0):        prod *= jprint(f' sum= {sum_}\nprod= {prod}')`
Output:
```list(_range(x, y, z)) = [5, 3, 1, -1, -3, -5]
list(_range(-seven, seven, x)) = [-7, -2, 3]
sum= 348173
prod= -793618560```

## REXX

Programming note:   the (sympathetic) trailing semicolons (;) after each REXX statement are optional,   they are only there to mimic what the PL/I language requires after each statement.

The technique used by this REXX version is to "break up" the various   do   iterating clauses (ranges) into separate   do   loops,   and have them invoke a subroutine to perform the actual computations.

`/*REXX program emulates a multiple─range  DO  loop  (all variables can be any numbers). */ prod=  1;  sum=  0;    x= +5;    y= -5;    z= -2;  one=  1;three=  3;seven=  7;       do j=   -three  to      3**3      by three  ;      call meat;      end;      do j=   -seven  to    seven       by   x    ;      call meat;      end;      do j=      555  to      550 - y             ;      call meat;      end;      do j=       22  to      -28       by -three ;      call meat;      end;      do j=     1927  to     1939                 ;      call meat;      end;      do j=        x  to        y       by   z    ;      call meat;      end;      do j=    11**x  to    11**x + one           ;      call meat;      end; say ' sum= ' || commas( sum);                          /*display   SUM   with commas.   */say 'prod= ' || commas(prod);                          /*   "     PROD     "     "      */exit;                                                  /*stick a fork in it, we're done.*//*──────────────────────────────────────────────────────────────────────────────────────*/commas: procedure; parse arg _;     n= _'.9';     #= 123456789;     b= verify(n, #, "M")                                    e= verify(n, #'0', , verify(n, #"0.", 'M') )  - 4          do j=e  to b  by -3;      _= insert(',', _, j);   end;                  return _/*──────────────────────────────────────────────────────────────────────────────────────*/meat:  sum= sum + abs(j);       if abs(prod)<2**27 & j\==0  then prod= prod * j;       return;`
output   when using the same variable values:
``` sum= 348,173
prod= -793,618,560
```

## Ruby

Uses chaining of enumerables, which was introduced with Ruby 2.6

`x, y, z, one, three, seven = 5, -5, -2, 1, 3, 7 enums = (-three).step(3**3, three) +        (-seven).step(seven, x) +        555     .step(550-y, -1) +        22      .step(-28, -three) +        (1927..1939) +                # just toying, 1927.step(1939) is fine too        x       .step(y, z) +        (11**x) .step(11**x + one)# enums is an enumerator, consisting of a bunch of chained enumerators,# none of which has actually produced a value. puts "Sum of absolute numbers:  #{enums.sum(&:abs)}"prod = enums.inject(1){|prod, j| ((prod.abs < 2**27) && j!=0) ? prod*j : prod}puts "Product (but not really): #{prod}" `
Output:
```Sum of absolute numbers:  348173
Product (but not really): -793618560
```

## VBA

`Dim prod As Long, sum As LongPublic Sub LoopsWithMultipleRanges()    Dim x As Integer, y As Integer, z As Integer, one As Integer, three As Integer, seven As Integer, j As Long    prod = 1    sum = 0    x = 5    y = -5    z = -2    one = 1    three = 3    seven = 7    For j = -three To pow(3, 3) Step three: Call process(j): Next j    For j = -seven To seven Step x: Call process(j): Next j    For j = 555 To 550 - y: Call process(j): Next j    For j = 22 To -28 Step -three: Call process(j): Next j    For j = 1927 To 1939: Call process(j): Next j    For j = x To y Step z: Call process(j): Next j    For j = pow(11, x) To pow(11, x) + one: Call process(j): Next j    Debug.Print " sum= " & Format(sum, "#,##0")    Debug.Print "prod= " & Format(prod, "#,##0")End SubPrivate Function pow(x As Long, y As Integer) As Long    pow = WorksheetFunction.Power(x, y)End FunctionPrivate Sub process(x As Long)    sum = sum + Abs(x)    If Abs(prod) < pow(2, 27) And x <> 0 Then prod = prod * xEnd Sub`
Output:
``` sum= 348.173
prod= -793.618.560```

## zkl

`prod,sum := 1,0;  /* start with a product of unity, sum of 0 */x,y,z := 5, -5, -2;one,three,seven := 1,3,7;foreach j in (Walker.chain([-three..(3).pow(3),three], // do these sequentially               [-seven..seven,x], [555..550 - y], [22..-28,-three], #[start..last,step]               [1927..1939], [x..y,z], [(11).pow(x)..(11).pow(x) + one])){   sum+=j.abs();	/* add absolute value of J */   if(prod.abs()<(2).pow(27) and j!=0) prod*=j; /* PROD is small enough & J */}/* SUM and PROD are used for verification of J incrementation */println("sum  = %,d\nprod = %,d".fmt(sum,prod));`
Output:
```sum  = 348,173
prod = -793,618,560
```