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.
<lang pli> /* 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); /* " " " " */</lang>
- Task
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. <lang> 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 </lang>
- Related tasks
- Loop over multiple arrays simultaneously
- Loops/Break
- Loops/Continue
- Loops/Do-while
- Loops/Downward for
- Loops/For
- Loops/For with a specified step
- Loops/Foreach
- Loops/Increment loop index within loop body
- Loops/Infinite
- Loops/N plus one half
- Loops/Nested
- Loops/While
- Loops/with multiple ranges
- Loops/Wrong ranges
ALGOL 60
<lang algol60>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 </lang>
- Output:
sum= 348173 prod= -793618560
ALGOL 68
As with most of the other languages, Algol 68 doesn't support multiple loop ranges, so a sequence pf loops is used instead. <lang algol68>BEGIN
# translation of task PL/1 code, with minimal changes, semicolons required by #
# PL/1 but not allowed in Algol 68 removed, unecessary rounding removed #
# Note that in Algol 68, the loop counter is a local variable to the loop and #
# the value of j is not available outside the loops #
PROC loop body = ( INT j )VOID: #(below) ** is exponentiation: 4**3=64 #
BEGIN sum +:= ABS j; #add absolute value of J.#
IF ABS prod<2**27 AND j /= 0 THEN prod *:= j FI #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.#
INT prod := 1; #start with a product of unity. #
INT sum := 0; # " " " sum " zero. #
INT x := +5;
INT y := -5;
INT z := -2;
INT one := 1;
INT three := 3;
INT seven := 7;
FOR j FROM -three BY three TO ( 3**3 ) DO loop body( j ) OD;
FOR j FROM -seven BY x TO +seven DO loop body( j ) OD;
FOR j FROM 555 TO 550 - y DO loop body( j ) OD;
FOR j FROM 22 BY -three TO -28 DO loop body( j ) OD;
FOR j FROM 1927 TO 1939 DO loop body( j ) OD;
FOR j FROM x BY z TO y DO loop body( j ) OD;
FOR j FROM ( 11**x ) TO ( 11**x ) + one DO loop body( j ) OD;
print((" sum= ", whole( sum,0), newline)); #display strings to term.#
print(("prod= ", whole(prod,0), newline)) # " " " " #
END </lang>
- 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. <lang algolw>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.</lang>
- Output:
sum= 348173 prod= -793618560
C
<lang 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;
}</lang>
- Output:
sum = 348,173 prod = -793,618,560
Common Lisp
Using raw code and DO iterator <lang lisp> (let ((prod 1) ; Initialize aggregator
(sum 0)
(x 5) ; Initialize variables
(y -5)
(z -2)
(one 1)
(three 3)
(seven 7))
(flet ((loop-body (j) ; Set the loop function
(incf sum (abs j)) (if (and (< (abs prod) (expt 2 27)) (/= j 0)) (setf prod (* prod j)))))
(do ((i (- three) (incf i three))) ; Just a serie of individual loops
((> i (expt 3 3)))
(loop-body i)) (do ((i (- seven) (incf i x)))
((> i seven))
(loop-body i)) (do ((i 555 (incf i -1)))
((< i (- 550 y)))
(loop-body i)) (do ((i 22 (incf i (- three))))
((< i -28))
(loop-body i)) (do ((i 1927 (incf i)))
((> i 1939))
(loop-body i)) (do ((i x (incf i z)))
((< i y))
(loop-body i)) (do ((i (expt 11 x) (incf i)))
((> i (+ (expt 11 x) one)))
(loop-body i)))
(format t "~&sum = ~14<~:d~>" sum) (format t "~&prod = ~14<~:d~>" prod))
</lang> or with loop ranges and increments as list to dolist <lang lisp> (let ((prod 1)
(sum 0)
(x 5)
(y -5)
(z -2)
(one 1)
(three 3)
(seven 7))
(flet ((loop-body (j) ; Set the loop function
(incf sum (abs j)) (if (and (< (abs prod) (expt 2 27)) (/= j 0)) (setf prod (* prod j)))))
(dolist (lst `((,(- three) ,(expt 3 3) ,three)
(,(- seven) ,seven ,x) (555 ,(- 550 y) -1) (22 -28 ,(- three)) (1927 1939 1) (,x ,y ,z) (,(expt 11 x) ,(+ (expt 11 x) one) 1)))
(do ((i (car lst) (incf i (caddr lst))))
((if (plusp (caddr lst)) (> i (cadr lst)) (< i (cadr lst)))) (loop-body i))))
(format t "~&sum = ~14<~:d~>" sum) (format t "~&prod = ~14<~:d~>" prod))
</lang>
- 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.
<lang factor>USING: formatting kernel locals math math.functions math.ranges
sequences 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 ] bi@ " sum= %s\nprod= %s\n" printf
]</lang>
- Output:
sum= 348,173 prod= -793,618,560
Go
Nothing fancy from Go here (is there ever?), just a series of individual for loops. <lang go>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))
}</lang>
- Output:
sum = 348,173 prod = -793,618,560
Groovy
Solution: <lang groovy>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}"</lang>
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.
<lang j> NB. http://rosettacode.org/wiki/Loops/Wrong_ranges#J NB. define range as a linear polynomial start =: 0&{ stop =: 1&{ increment =: 2&{ :: 1: NB. on error use 1 range =: (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
) </lang>
] 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. <lang julia>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()
</lang>
- Output:
sum = 348,173 prod = -793,618,560
Kotlin
Nothing special here, just a series of individual for loops. <lang scala>// 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)
}</lang>
- Output:
sum = 348,173 prod = -793,618,560
Perl
<lang 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 $prod;</lang>
- Output:
Sum: 348,173 Product: -793,618,560
Perl 6
This task is really conflating two separate things, (at least in Perl 6). Sequences and loops are two different concepts and may be considered / implemented separately from each other.
Yes, you can generate a sequence with a loop, and a loop can use a sequence for an iteration value, but the two are somewhat orthogonal and don't necessarily overlap.
Sequences are first class objects in Perl 6. You can (and typically do) generate a sequence using the (appropriately enough) sequence operator and can assign it to a variable and/or pass it as a parameter; the entire sequence, not just it's individual values. It may be used in a looping construct, but it is not necessary to do so.
Various looping constructs often do use sequences as their iterator but not exclusively, possibly not even in the majority.
Displaying the j sequence as well since it isn't very large.
<lang perl6>sub comma { ($^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;</lang>
- 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
<lang 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 for
end for printf(1," sum = %,d\n",total) printf(1,"prod = %,d\n",prod)</lang>
- 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. <lang prolog>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 values sym(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 get range(-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 parser translate(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 order calc_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).</lang>
?- calc_values(Sum, Product). Sum = 348173, Product = -793618560.
PureBasic
<lang purebasic>#X = 5 : #Y = -5 : #Z = -2
- ONE = 1 : #THREE = 3 : #SEVEN = 7
Define j.i Global 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) Next
EndMacro
Procedure proc(j.i)
sum + Abs(j) If (Abs(prod) < ipow(2 , 27)) And (j<>0) prod * j EndIf
EndProcedure
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</lang>
- Output:
sum = 348,173 prod = -793,618,560
Python
Pythons range function does not include the second argument hence the definition of _range() <lang python>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 *= j
print(f' sum= {sum_}\nprod= {prod}')</lang>
- 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. <lang rexx>/*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;</lang>
- 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 <lang Ruby>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}" </lang>
- Output:
Sum of absolute numbers: 348173 Product (but not really): -793618560
Vala
<lang vala>const int CHARBIT = 8; long prod = 1; long sum = 0;
long labs(long n) {
long mask = n >> ((long)sizeof(long) * CHARBIT - 1); return ((n + mask) ^ mask);
}
long lpow(long base_num, long exp) {
long result = 1;
while (true)
{
if ((exp & 1) != 0) result *= base_num;
exp >>= 1;
if (exp == 0) break;
base_num *= base_num;
}
return result;
}
void process(long j) {
sum += labs(j); if (labs(prod) < (1 << 27) && j != 0) prod *= j;
}
void main() {
const int x = 5;
const int y = -5;
const int z = -2;
const int one = 1;
const int three = 3;
const int seven = 11;
long p = lpow(11, x);
for (int j = -three; j <= lpow(3, 3); j += three ) process(j);
for (int j = -seven; j <= seven; j += x) process(j);
for (int j = 555; j <= 550 - y; ++j) process(j);
for (int j = 22; j >= -28; j -= three) process(j);
for (int j = 1928; j <= 1939; ++j) process(j);
for (int j = x; j >= y; j -= -z) process(j);
for (long j = p; j <= p + one; ++j) process(j);
stdout.printf("sum = %10ld\n", sum);
stdout.printf("prod = %10ld\n", prod);
}</lang>
- Output:
sum = 346265 prod = -793618560
VBA
<lang VB>Dim prod As Long, sum As Long Public 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 Sub Private Function pow(x As Long, y As Integer) As Long
pow = WorksheetFunction.Power(x, y)
End Function Private Sub process(x As Long)
sum = sum + Abs(x) If Abs(prod) < pow(2, 27) And x <> 0 Then prod = prod * x
End Sub</lang>
- Output:
sum= 348.173 prod= -793.618.560
Visual Basic .NET
VB.NET loops can't have multiple ranges, so this implementation will use the For Each loop and demonstrate various functions that produce concatenated ranges.
Composite formatting is used to add digit separators.
Using the following to provide the functionality of the For loop as a function, <lang vbnet>Partial Module Program
' Stop and Step are language keywords and must be escaped with brackets.
Iterator Function Range(start As Integer, [stop] As Integer, Optional [step] As Integer = 1) As IEnumerable(Of Integer)
For i = start To [stop] Step [step]
Yield i
Next
End Function
End Module</lang>
and Enumerable.Concat (along with extension method syntax) to splice the ranges, the program ends up looking like this:
<lang vbnet>Imports System.Globalization
Partial Module Program
Sub Main()
' All variables are inferred to be of type Integer.
Dim prod = 1,
sum = 0,
x = +5,
y = -5,
z = -2,
one = 1,
three = 3,
seven = 7
' The exponent operator compiles to a call to Math.Pow, which returns Double, and so must be converted back to Integer.
For Each j In Range(-three, CInt(3 ^ 3), 3 ).
Concat(Range(-seven, +seven, x )).
Concat(Range(555, 550 - y )).
Concat(Range(22, -28, -three)).
Concat(Range(1927, 1939 )).
Concat(Range(x, y, z )).
Concat(Range(CInt(11 ^ x), CInt(11 ^ x) + one ))
sum = sum + Math.Abs(j)
If Math.Abs(prod) < 2 ^ 27 AndAlso j <> 0 Then prod = prod * j
Next
' The invariant format info by default has two decimal places.
Dim format As New NumberFormatInfo() With {
.NumberDecimalDigits = 0
}
Console.WriteLine(String.Format(format, " sum= {0:N}", sum))
Console.WriteLine(String.Format(format, "prod= {0:N}", prod))
End Sub
End Module</lang>
To improve the program's appearance, a ConcatRange method can be defined to combine the two method calls, <lang vbnet> <Runtime.CompilerServices.Extension>
Function ConcatRange(source As IEnumerable(Of Integer), start As Integer, [stop] As Integer, Optional [step] As Integer = 1) As IEnumerable(Of Integer)
Return source.Concat(Range(start, [stop], [step]))
End Function</lang>
which results in a loop that looks like this: <lang vbnet> For Each j In Range(-three, CInt(3 ^ 3), 3 ).
ConcatRange(-seven, +seven, x ).
ConcatRange(555, 550 - y ).
ConcatRange(22, -28, -three).
ConcatRange(1927, 1939 ).
ConcatRange(x, y, z ).
ConcatRange(CInt(11 ^ x), CInt(11 ^ x) + one )
Next</lang>
An alternative to avoid the repeated method calls would be to make a Range function that accepts multiple ranges, in this case as a parameter array of tuples. <lang vbnet> Function Range(ParamArray ranges() As (start As Integer, [stop] As Integer, [step] As Integer)) As IEnumerable(Of Integer)
' Note: SelectMany is equivalent to bind, flatMap, etc.
Return ranges.SelectMany(Function(r) Range(r.start, r.stop, r.step))
End Function</lang>
resulting in: <lang vbnet> For Each j In Range((-three, CInt(3 ^ 3), 3 ),
(-seven, +seven, x ),
(555, 550 - y, 1 ),
(22, -28, -three ),
(1927, 1939, 1 ),
(x, y, z ),
(CInt(11 ^ x), CInt(11 ^ x) + one, 1 ))
Next</lang>
Note, however, that the inability to have a heterogenous array means that specifying the step is now mandatory. Using a parameter array of arrays is slightly less clear but results in the tersest loop. <lang vbnet> Function Range(ParamArray ranges As Integer()()) As IEnumerable(Of Integer)
Return ranges.SelectMany(Function(r) Range(r(0), r(1), If(r.Length < 3, 1, r(2)))) End Function</lang>
<lang vbnet> For Each j In Range({-three, CInt(3 ^ 3), 3 },
{-seven, +seven, x },
{555, 550 - y },
{22, -28, -three },
{1927, 1939 },
{x, y, z },
{CInt(11 ^ x), CInt(11 ^ x) + one })
Next</lang>
- Output (for all variations):
sum= 348,173 prod= -793,618,560
zkl
<lang 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));</lang>
- Output:
sum = 348,173 prod = -793,618,560