Loops/Continue
You are encouraged to solve this task according to the task description, using any language you may know.
Show the following output using one loop.
1, 2, 3, 4, 5 6, 7, 8, 9, 10
Try to achieve the result by forcing the next iteration within the loop upon a specific condition, if your language allows it.
Ada
Ada has no continue reserved word, nor does it need one. The continue reserved word is only syntactic sugar for operations that can be achieved without it as in the following example.
<lang ada>with Ada.Text_Io; use Ada.Text_Io;
procedure Loop_Continue is begin
for I in 1..10 loop
Put(Integer'Image(I));
if I mod 5 = 0 then
New_Line;
else
Put(",");
end if;
end loop;
end Loop_Continue;</lang>
Aikido
<lang aikido>foreach i 1..10 {
print (i)
if ((i % 5) == 0) {
println()
continue
}
print (", ")
}</lang>
ALGOL 68
ALGOL 68 has no continue reserved word, nor does it need one. The continue reserved word is only syntactic sugar for operations that can be achieved without it as in the following example: <lang algol68>FOR i FROM 1 TO 10 DO
print ((i,
IF i MOD 5 = 0 THEN
new line
ELSE
","
FI
))
OD</lang> Output:
+1, +2, +3, +4, +5
+6, +7, +8, +9, +10
AutoHotkey
<lang autohotkey>Loop, 10 {
Delimiter := (A_Index = 5) || (A_Index = 10) ? "`n":", " Index .= A_Index . Delimiter
} MsgBox %Index%</lang>
AWK
<lang awk>BEGIN {
for(i=1; i <= 10; i++) {
printf("%d", i)
if ( i % 5 == 0 ) {
print
continue
}
printf(", ")
}
}</lang>
C
<lang c>for(int i = 1;i <= 10; i++){
printf("%d", i);
if(i % 5 == 0){
printf("\n");
continue;
}
printf(", ");
}</lang>
C++
<lang cpp>for(int i = 1;i <= 10; i++){
cout << i;
if(i % 5 == 0){
cout << endl;
continue;
}
cout << ", ";
}</lang>
C#
<lang csharp>using System;
class Program {
static void Main(string[] args) {
for (int i = 1; i <= 10; i++) {
Console.Write(i);
if (i % 5 == 0) {
Console.WriteLine();
continue;
}
Console.Write(", ");
}
}
}</lang>
Clojure
Clojure doesn't have a continue keyword. It has a recur keyword, although I prefer to work with ranges in this case. <lang clojure>(doseq
[n (range 1 11)] (do (print n) (if (= (rem n 5) 0) (print "\n") (print ", "))))</lang>
ColdFusion
Remove the leading space from the line break tag. <lang cfm><cfscript>
for( i = 1; i <= 10; i++ )
{
writeOutput( i );
if( 0 == i % 5 )
{
writeOutput( "< br />" );
continue;
}
writeOutput( "," );
}
</cfscript></lang>
Common Lisp
Common Lisp doesn't have a continue keyword, but the do iteration construct does use an implicit tagbody, so it's easy to go to any label. Four solutions follow. The first pushes the conditional (whether to print a comma and a space or a newline) into the format string. The second uses the implicit tagbody and go. The third is a do loop with conditionals outside of the output functions.
<lang lisp>(do ((i 1 (1+ i))) ((> i 10))
(format t "~a~:[, ~;~%~]" i (zerop (mod i 5))))
(do ((i 1 (1+ i))) ((> i 10))
(write i) (when (zerop (mod i 5)) (terpri) (go end)) (write-string ", ") end)
(do ((i 1 (1+ i))) ((> i 10))
(write i) (if (zerop (mod i 5)) (terpri) (write-string ", ")))</lang>
These use the loop iteration form, which does not contain an implicit tagbody (though one could be explicitly included). The first uses an explicit condition to omit the rest of the loop; the second uses block/return-from to obtain the effect of skipping the rest of the code in the block which makes up the entire loop body.
<lang lisp>(loop for i from 1 to 10
do (write i)
if (zerop (mod i 5)) do (terpri)
else do (write-string ", "))
(loop for i from 1 to 10 do
(block continue
(write i)
(when (zerop (mod i 5))
(terpri)
(return-from continue))
(write-string ", ")))</lang>
D
<lang d>for(int i = 1;i <= 10; i++){
writef(i);
if(i % 5 == 0){
writefln();
continue;
}
writef(", ");
}</lang>
E
<lang e>for i in 1..10 {
print(i)
if (i %% 5 == 0) {
println()
continue
}
print(", ")
}</lang>
Factor
There is no built-in continue in Factor.
<lang factor>1 10 [a,b] [
[ number>string write ] [ 5 mod 0 = "\n" ", " ? write ] bi
] each</lang>
Fantom
While and for loops support continue to jump back to begin the next iteration of the loop.
<lang fantom> class LoopsContinue {
public static Void main ()
{
for (Int i := 1; i <= 10; ++i)
{
Env.cur.out.print (i)
if (i % 5 == 0)
{
Env.cur.out.printLine ("")
continue
}
Env.cur.out.print (", ")
}
Env.cur.out.printLine ("")
}
} </lang>
Forth
Although this code solves the task, there is no portable equivalent to "continue" for either DO-LOOPs or BEGIN loops. <lang forth>: main
11 1 do i dup 1 r. 5 mod 0= if cr else [char] , emit space then loop ;</lang>
Fortran
<lang fortran>do i = 1, 10
write(*, '(I0)', advance='no') i
if ( mod(i, 5) == 0 ) then
write(*,*)
cycle
end if
write(*, '(A)', advance='no') ', '
end do</lang>
GML
<lang GML>for(i = 1; i <= 10; i += 1)
{
show_message(string(i))
i += 1
if(i <= 10)
continue
}</lang>
Go
<lang go>for i := 1; i <= 10; i++ {
fmt.Printf("%d", i)
if i % 5 == 0 {
fmt.Printf("\n")
continue
}
fmt.Printf(", ")
}</lang>
Haskell
As a functional language, it is not idiomatic to have true loops - recursion is used instead. Below is one of many possible implementations of the task. The below code uses a guard (| symbol) to compose functions differently for the two alternative output paths, instead of using continue like in an imperative language.
<lang haskell>import Control.Monad (forM) main = forM [1..10] out
where
out x | (x `mod` 5 == 0) = (putStrLn . show) x
| otherwise = (putStr . (++", ") . show) x</lang>
HicEst
<lang hicest>DO i = 1, 10
IF( MOD(i, 5) == 1 ) THEN
WRITE(Format="i3") i
ELSE
WRITE(APPend, Format=" ',', i3 ") i
ENDIF
ENDDO </lang>
Icon and Unicon
The following code demonstrates the use of 'next' (the reserved word for 'continue'): <lang Icon>procedure main() every writes(x := 1 to 10) do {
if x % 5 = 0 then {
write()
next
}
writes(", ")
}
end</lang> However, the output sequence can be written without 'next' and far more succinctly as: <lang Icon>every writes(x := 1 to 10, if x % 5 = 0 then "\n" else ", ")</lang>
J
J is array-oriented, so there is very little need for loops. For example, one could satisfy this task this way:
<lang j>_2}."1'lq<, >'8!:2>:i.2 5</lang>
J does support loops for those times they can't be avoided (just like many languages support gotos for those time they can't be avoided). <lang j>3 : 0 ] 10
z=.
for_i. 1 + i.y do.
z =. z , ": i
if. 0 = 5 | i do.
z 1!:2 ]2
z =.
continue.
end.
z =. z , ', '
end.
i.0 0
)</lang>
Though it's rare to see J code like this.
Java
<lang java>for(int i = 1;i <= 10; i++){
System.out.print(i);
if(i % 5 == 0){
System.out.println();
continue;
}
System.out.print(", ");
}</lang>
JavaScript
Using the print() function from Rhino or SpiderMonkey.
<lang javascript>var output = "";
for (var i = 1; i <= 10; i++) {
output += i;
if (i % 5 == 0) {
print(output);
output = "";
continue;
}
output += ", ";
}</lang>
Lisaac
<lang Lisaac>1.to 10 do { i : INTEGER;
i.print;
(i % 5 = 0).if { '\n'.print; } else { ','.print; };
};</lang>
Lua
<lang Lua>for i = 1, 10 do
io.write( i )
if i % 5 == 0 then
io.write( "\n" )
else
io.write( ", " )
end
end</lang>
Mathematica
<lang Mathematica>tmp = ""; For[i = 1, i <= 10, i++,
tmp = tmp <> ToString[i]; If[Mod[i, 5] == 0, tmp = tmp <> "\n"; , tmp = tmp <> ", "; ]; ];
Print[tmp]</lang>
MAXScript
<lang maxscript>for i in 1 to 10 do (
format "%" i
if mod i 5 == 0 then
(
format "\n"
continue
) continue
format ", "
)</lang>
Metafont
Metafont has no a continue (or similar) keyword. As the Ada solution, we can complete the task just with conditional.
<lang metafont>string s; s := ""; for i = 1 step 1 until 10: if i mod 5 = 0:
s := s & decimal i & char10;
else:
s := s & decimal i & ", "
fi; endfor message s; end</lang>
Since message append always a newline at the end, we need to build a string and output it at the end, instead of writing the output step by step.
Note: mod is not a built in; like TeX, "bare Metafont" is rather primitive, and normally a set of basic macros is preloaded to make it more usable; in particular mod is defined as
<lang metafont>primarydef x mod y = (x-y*floor(x/y)) enddef;</lang>
Modula-3
Modula-3 defines the keyword RETURN as an exception, but when it is used with no arguments it works just like continue in C.
Note, however, that RETURN only works inside a procedure or a function procedure; use EXIT otherwise.
Module code and imports are omitted. <lang modula3>FOR i := 1 TO 10 DO
IO.PutInt(i);
IF i MOD 5 = 0 THEN
IO.Put("\n");
RETURN;
END;
IO.Put(", ");
END;</lang>
MOO
<lang moo>s = ""; for i in [1..10]
s += tostr(i); if (i % 5 == 0) player:tell(s); s = ""; continue; endif s += ", ";
endfor</lang>
OCaml
There is no continue statement for for loops in OCaml, but it is possible to achieve the same effect with an exception. <lang ocaml># for i = 1 to 10 do
try
print_int i;
if (i mod 5) = 0 then raise Exit;
print_string ", "
with Exit ->
print_newline()
done
;;
1, 2, 3, 4, 5 6, 7, 8, 9, 10 - : unit = ()</lang> Though even if the continue statement does not exist, it is possible to add it with camlp4.
Octave
<lang octave>v = ""; for i = 1:10
v = sprintf("%s%d", v, i);
if ( mod(i, 5) == 0 )
disp(v)
v = "";
continue
endif
v = sprintf("%s, ", v);
endfor</lang>
Oz
By using the "continue" feature of the for-loop, we bind C to a nullary procedure which, when invoked, immediately goes on to the next iteration of the loop. <lang oz>for I in 1..10 continue:C do
{System.print I}
if I mod 5 == 0 then
{System.printInfo "\n"}
{C}
end
{System.printInfo ", "}
end</lang>
PARI/GP
<lang parigp>for(n=1,10,
print1(n);
if(n%5 == 0, print();continue);
print1(", ")
)</lang>
Perl
<lang perl>foreach (1..10) {
print $_;
if ($_ % 5 == 0) {
print "\n";
next;
}
print ', ';
}</lang>
Perl 6
<lang perl6>for 1 .. 10 {
.print;
if $_ %% 5 {
print "\n";
next;
}
print ', ';
}</lang>
or without using a loop:
<lang perl6>$_.join(", ").say for [1..5], [6..10];</lang>
PHP
<lang php>for ($i = 1; $i <= 10; $i++) {
echo $i;
if ($i % 5 == 0) {
echo "\n";
continue;
}
echo ', ';
}</lang>
PicoLisp
PicoLisp doesn't have an explicit 'continue' functionality. It can always be emulated with a conditional expression. <lang PicoLisp>(for I 10
(print I)
(if (=0 (% I 5))
(prinl)
(prin ", ") ) )</lang>
Pike
<lang pike>int main(){
for(int i = 1; i <= 10; i++){
write(sprintf("%d",i));
if(i % 5 == 0){
write("\n");
continue;
}
write(", ");
}
}</lang>
PL/I
<lang PL/I>loop: do i = 1 to 10;
put edit (i) (f(3));
if mod(i,5) = 0 then do; put skip; iterate loop; end;
put edit (', ') (a);
end;</lang>
Pop11
<lang pop11>lvars i; for i from 1 to 10 do
printf(i, '%p');
if i rem 5 = 0 then
printf('\n');
nextloop;
endif;
printf(', ')
endfor;</lang>
PowerShell
<lang powershell>for ($i = 1; $i -le 10; $i++) {
Write-Host -NoNewline $i
if ($i % 5 -eq 0) {
Write-Host
continue
}
Write-Host -NoNewline ", "
}</lang>
PureBasic
<lang purebasic>OpenConsole()
For i.i = 1 To 10
Print(Str(i))
If i % 5 = 0
PrintN("")
Continue
EndIf
Print(",")
Next
Repeat: Until Inkey() <> ""</lang>
Python
<lang python>for i in xrange(1,11):
if i % 5 == 0:
print i
continue
print i, ",",</lang>
R
<lang R>for(i in 1:10) {
cat(i)
if(i %% 5 == 0)
{
cat("\n")
next
}
cat(", ")
}</lang>
REBOL
<lang REBOL>REBOL [ Title: "Loop/Continue" Author: oofoe Date: 2010-01-05 URL: http://rosettacode.org/wiki/Loop/Continue ]
- REBOL does not provide a 'continue' word for loop constructs,
- however, you may not even miss it
print "One liner (compare to ALGOL 68 solution):" repeat i 10 [prin rejoin [i either 0 = mod i 5 [crlf][", "]]]
print [crlf "Port of ADA solution:"] for i 1 10 1 [ prin i either 0 = mod i 5 [ prin newline ][ prin ", " ] ]</lang>
Output:
One liner (compare to ALGOL 68 solution): 1, 2, 3, 4, 5 6, 7, 8, 9, 10 Port of ADA solution: 1, 2, 3, 4, 5 6, 7, 8, 9, 10
REXX
(Remember that there exists implementations of the REXX language that needs that the source begins with /*, i.e. with a comment) <lang rexx>do i = 1 to 10
call charout ,i", " if i//5 = 0 then do say iterate end
end</lang>
Ruby
<lang ruby>for i in 1..10 do
print i
if i % 5 == 0 then
puts
next
end
print ', '
end</lang> The "for" look could be written like this: <lang ruby>(1..10).each do |i| ... 1.upto(10) do |i| ... 10.times do |n| i=n+1; ...</lang> Without meeting the criteria (showing loop continuation), this task could be written as: <lang ruby>1.upto(10) {|i| print "%d%s" % [i, i%5==0 ? "\n" : ", "]}</lang>
Sather
There's no continue! in Sather. The code solve the task without forcing a new iteration.
<lang sather>class MAIN is
main is
i:INT;
loop i := 1.upto!(10);
#OUT + i;
if i%5 = 0 then
#OUT + "\n";
else
#OUT + ", ";
end;
end;
end;
end;</lang>
Scheme
<lang scheme>(define (loop i)
(if (> i 10)
'done
(begin
(display i)
(if (zero? (modulo i 5))
(begin (newline) (loop (1+ i)))
(begin (display ", ") (loop (1+ i)))))))</lang>
Suneido
<lang Suneido>ob = Object() for (i = 1; i <= 10; ++i)
{
ob.Add(i)
if i is 5
{
Print(ob.Join(','))
ob = Object()
}
}
Print(ob.Join(','))</lang>
Output: <lang Suneido>1,2,3,4,5 6,7,8,9,10 ok</lang>
Tcl
<lang tcl>for {set i 1} {$i <= 10} {incr i} {
puts -nonewline $i
if {$i % 5 == 0} {
puts ""
continue
}
puts -nonewline ", "
}</lang>
TI-89 BASIC
<lang ti-89>count() Prgm
""→s
For i,1,10
s&string(i)→s
If mod(i,5)=0 Then
Disp s
""→s
Cycle
EndIf
s&", "→s
EndFor
EndPrgm</lang>
Ti-89 lacks support for multi-argument display command or controlling the print position so that one can print several data on the same line. The display command (Disp) only accepts one argument and prints it on a single line (causing a line a feed at the end, so that the next Disp command will print in the next line). The solution is appending data to a string (s), using the concatenator operator (&), by converting numbers to strings, and then printing the string at the end of the line.
TUSCRIPT
<lang tuscript> $$ MODE TUSCRIPT numbers="" LOOP n=1,10 numbers=APPEND (numbers,", ",n) rest=n%5 IF (rest!=0) CYCLE
PRINT numbers numbers=""
ENDLOOP </lang> Output:
1, 2, 3, 4, 5 6, 7, 8, 9, 10
UnixPipes
<lang bash>yes \ | cat -n | head -n 10 | xargs -n 5 echo | tr ' ' ,</lang>
UNIX Shell
<lang bash>Z=1 while (( Z<=10 )); do
echo -e "$Z\c"
if (( Z % 5 != 0 )); then
echo -e ", \c"
else
echo -e ""
fi
(( Z++ ))
done</lang>
<lang bash>for ((i=1;i<=10;i++)); do
echo -n $i if [ $((i%5)) -eq 0 ]; then echo continue fi echo -n ", "
done</lang>
Vedit macro language
<lang vedit>for (#1 = 1; #1 <= 10; #1++) {
Num_Type(#1, LEFT+NOCR)
if (#1 % 5 == 0) {
Type_Newline
Continue
}
Message(", ")
}</lang>
Visual Basic .NET
<lang vbnet>For i = 1 To 10
Console.Write(i)
If i Mod 5 = 0 Then
Console.WriteLine()
Else
Console.Write(", ")
End If
Next</lang>
- Programming Tasks
- Iteration
- Ada
- Aikido
- ALGOL 68
- AutoHotkey
- AWK
- C
- C++
- C sharp
- Clojure
- ColdFusion
- Common Lisp
- D
- E
- Factor
- Fantom
- Forth
- Fortran
- GML
- Go
- Haskell
- HicEst
- Icon
- Unicon
- J
- Java
- JavaScript
- Lisaac
- Lua
- Mathematica
- MAXScript
- Metafont
- Modula-3
- MOO
- OCaml
- Octave
- Oz
- PARI/GP
- Perl
- Perl 6
- PHP
- PicoLisp
- Pike
- PL/I
- Pop11
- PowerShell
- PureBasic
- Python
- R
- REBOL
- REXX
- Ruby
- Sather
- Scheme
- Suneido
- Tcl
- TI-89 BASIC
- TUSCRIPT
- UnixPipes
- UNIX Shell
- Vedit macro language
- Visual Basic .NET
- M4/Omit