Loops/Continue

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.

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;

Aikido

foreach i 1..10 {
    print (i)
    if ((i % 5) == 0) {
        println()
        continue
    } 
    print (", ")
}

ALGOL 68

Works with: ALGOL 68 version Revision 1 - no extensions to language used

Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny

Works with: ELLA ALGOL 68 version Any (with appropriate job cards) - tested with release 1.8-8d

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:

FOR i FROM 1 TO 10 DO
  print ((i, 
    IF i MOD 5 = 0 THEN
      new line
    ELSE
      ","
    FI
  ))
OD

Output:

         +1,         +2,         +3,         +4,         +5
         +6,         +7,         +8,         +9,        +10

AutoHotkey

Loop, 10 {
  Delimiter := (A_Index = 5) || (A_Index = 10) ? "`n":", "
  Index .= A_Index . Delimiter
}
MsgBox %Index%

AWK

BEGIN {
  for(i=1; i <= 10; i++) {
    printf("%d", i)
    if ( i % 5 == 0 ) {
      print
      continue
    }
    printf(", ")
  }
}

BBC BASIC

BBC BASIC doesn't have a 'continue' statement so the remainder of the loop must be made conditional.

      FOR i% = 1 TO 10
        PRINT ; i% ;
        IF i% MOD 5 = 0 PRINT ELSE PRINT ", ";
      NEXT

bc

Requires a bc with the print and continue statements. POSIX bc has not these statements.

Works with: OpenBSD bc

for (i = 1; i <= 10; i++) {
	print i
	if (i % 5) {
		print ", "
		continue
	}
	print "\n"
}
quit

C

Translation of: C++

for(int i = 1;i <= 10; i++){
   printf("%d", i);
   if(i % 5 == 0){
      printf("\n");
      continue;
   }
   printf(", ");
}

C++

Translation of: Java

for(int i = 1;i <= 10; i++){
   cout << i;
   if(i % 5 == 0){
      cout << endl;
      continue;
   }
   cout << ", ";
}

C#

Translation of: Java

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(", ");
        }
    }
}

Clojure

Clojure doesn't have a continue keyword. It has a recur keyword, although I prefer to work with ranges in this case.

(doseq
  [n (range 1 11)]
  (do
    (print n)
    (if (= (rem n 5) 0) (print "\n") (print ", "))))

ColdFusion

Remove the leading space from the line break tag.

<cfscript>
  for( i = 1; i <= 10; i++ )
  {
    writeOutput( i );
    if( 0 == i % 5 )
    {
      writeOutput( "< br />" );
      continue;
    }
    writeOutput( "," );
  }
</cfscript>

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.

(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 ", ")))

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.

(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 ", ")))

D

import std.stdio;
 
void main() {
    foreach (i; 1 .. 11) {
        write(i);
        if (i % 5 == 0) {
            writeln();
            continue;
        }
        write(", ");
    }
}

Output:

1, 2, 3, 4, 5
6, 7, 8, 9, 10

dc

The four commands # n J M are special to OpenBSD dc. The # command starts a comment. The n command prints a number without a newline.

Translation of: bc

Works with: OpenBSD dc

1 si		# i = 1
[2Q]sA		# A = code to break loop
[[, ]P 1J]sB	# B = code to print comma, continue loop
[
 li n		# print i
 li 5 % 0 !=B	# call B if i % 5
 [
]P              # print newline
 M		# mark from calling B
 li 1 + si	# i += 1
 li 10!<C	# continue loop if 10 >= i
]sC li 10!<C	# enter loop if 10 >= i

This program uses J and M to force the next iteration of a loop. The nJ command breaks n levels of brackets (like nQ does so), but then skips to the next M command. One can place M at the end of the iteration.

Delphi

program DoLoop(output);
var
  i: integer;
begin
  for i := 1 to 10 do
  begin
    write(i);
    if i mod 5 = 0 then
    begin
      writeln;
      continue;
    end;
    write(', ');
  end;
end.

Output:

1, 2, 3, 4, 5
6, 7, 8, 9, 10

DWScript

var i : Integer;
 
for i := 1 to 10 do begin
   Print(i);
   if i mod 5 = 0 then begin
      PrintLn('');
      continue;
   end;
   Print(', ');
end;

Ela

Direct Approach

open console imperative
 
loop n | n > 10 = ()
       | else = rec write (show n) f `seq` loop (n+1)
                where f | n % 5 == 0 = "\r\n"
                        | else = ", "
 
loop 1

Function rec creates a cyclic version of a given function, it is defined in imperative module as 'rec f x = f x `seq` rec f' where 'seq' is a sequencing operator.

Using list

open console imperative
 
loop [] = ()
loop (x::xs) = rec (write << show) x c `seq` loop xs
               where c | x % 5 == 0 = "\r\n"
                       | else = ", "
 
loop [1..10]

This version is more generic and can work for any given range of values.

Erlang

%% Implemented by Arjun Sunel
-module(continue).
-export([main/0, for_loop/1]).
 
main() ->
	for_loop(1).    
 
 for_loop(N)  when N /= 5 , N <10 ->
	io:format("~p, ",[N] ),
	for_loop(N+1);	
 
for_loop(N) when N >=10->
 	if N=:=10 ->
		io:format("~p\n",[N] )
	end;
 
 for_loop(N) ->
 	if N=:=5 ->
		io:format("~p\n",[N] ),
		for_loop(N+1)
	end.
 

Output:

1, 2, 3, 4, 5
6, 7, 8, 9, 10
ok

Euphoria

Works with: Euphoria version 4.0.3, 4.0.0 or later

include std\console.e --only for any_key to make running command window easier on windows
 
for i = 1 to 10 do
    if remainder(i,5) = 0 then
        printf(1, "%d\n", i)
        else
            printf(1,"%d, ", i)
            continue
    end if
end for
any_key()

Version without newline after 10 below.

include std\console.e --only for any_key to make running command window easier on windows
 
for i = 1 to 10 do
    if remainder(i,5) = 0 then
        switch i do 
            case 10 then
                printf(1,"%d ",i)
                break --new to euphoria 4.0.0+
            case else
                printf(1,"%d\n", i)
        end switch
 
        else
            printf(1,"%d, ", i)
            continue --new to euphoria 4.0.0+
    end if
end for
any_key()
 

Factor

There is no built-in continue in Factor.

1 10 [a,b] [ 
    [ number>string write ]
    [ 5 mod 0 = "\n" ", " ? write ] bi
] each

Fantom

While and for loops support continue to jump back to begin the next iteration of the loop.

 
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 ("")
  }
}
 

Forth

Although this code solves the task, there is no portable equivalent to "continue" for either DO-LOOPs or BEGIN loops.

: main
  11 1 do
    i dup 1 r.
    5 mod 0= if cr else [char] , emit space then
  loop ;

Fortran

Works with: Fortran version 90 and later

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

Works with: Fortran version 77 and later

C     WARNING: This program is not valid ANSI FORTRAN 77 code. It uses
C     one nonstandard character on the line labelled 5001. Many F77
C     compilers should be okay with it, but it is *not* standard.
C
C     It is also worth noting that FORTRAN 77 uses the command CONTINUE,
C     but not in the semantic, looping sense of the word. In FORTRAN,
C     CONTINUE means "do absolutely nothing." It is a placeholder. If
C     anything, it means "continue to the next line."
C
C     Python does the same thing with `pass`; C and its family of
C     languages, with `{/* do nothing */}`. Write CONTINUE when you need
C     to write something but have nothing to write.
C
C     This page on Rosetta Code is about a very different "continue"
C     statement that tells a loop to go back to the beginning. In
C     FORTRAN, we use (you guessed it!) a GOTO to accomplish this.
      PROGRAM CONTINUELOOP
        INTEGER I
 
        DO 10 I = 1, 10
C         Is it five or ten?
          IF (MOD(I, 5) .EQ. 0) THEN
C           If it is, write a newline and no comma.
            WRITE (*,5000) I
 
C           Continue the loop; that is, skip to the end of the loop.
            GOTO 10
          ENDIF
 
C         Write I with a comma and no newline.
          WRITE (*,5001) I
 
C       Again, in this case, CONTINUE is completely unrelated to the
C       semantic, looping sense of the word.
   10   CONTINUE
 
        STOP
 
C       This will print an integer and a newline (no comma).
 5000   FORMAT (I3)
 
C       Standard FORTRAN 77 is completely incapable of completing a
C       WRITE statement without printing a newline. If you want to print
C       five integers in standard code, you have to do something like
C       this:
C
C           FORMAT (I3, ',', I3, ',', I3, ',', I3, ',', I3)
C
C       Writing `1, 2, 3, 4, 5` and then `6, 7, 8, 9, 10` to that format
C       would produce the following two lines:
C
C             1,  2,  3,  4,  5
C             6,  7,  8,  9, 10
C
C       However, this code exists to demonstrate continuing a FORTRAN 77
C       loop and not to demonstrate how to get around its rigidity about
C       newlines.
C
C       The dollar sign at the end of the format is a nonstandard
C       character. It tells the compiler not to print a newline. If you
C       are actually using FORTRAN 77, you should figure out what your
C       particular compiler accepts. If you are actually using Fortran
C       90 or later, you should replace this line with the commented
C       line that follows it.
 5001   FORMAT (I3, ',', $)
C5001   FORMAT (I3, ',', ADVANCE='NO')
      END

GAP

for i in [1 .. 11] do
    if RemInt(i, 5) = 0 then
        Print(i, "\n");
        continue;
    fi;
    Print(i, ", ");
od;
 
# 1, 2, 3, 4, 5
# 6, 7, 8, 9, 10

GML

for(i = 1; i <= 10; i += 1)
    {
    show_message(string(i))
    i += 1
    if(i <= 10)
        continue
    }

Go

package main
 
import "fmt"
 
func main() {
    for i := 1; i <= 10; i++ {
        fmt.Printf("%d", i)
        if i%5 == 0 {
            fmt.Printf("\n")
            continue
        }
        fmt.Printf(", ")
    }
}

Output:

1, 2, 3, 4, 5
6, 7, 8, 9, 10

Groovy

for (i in 1..10) {
    print i
    if (i % 5 == 0) {
        println ()
        continue
    }
    print ', '
}

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.

import Control.Monad (forM)
main = forM [1..10] out
    where
      out x | x `mod` 5 == 0 = print x
            | otherwise = (putStr . (++", ") . show) x

HicEst

DO i = 1, 10
  IF( MOD(i, 5) == 1 ) THEN
      WRITE(Format="i3") i
    ELSE
      WRITE(APPend, Format=" ',', i3 ") i
    ENDIF
ENDDO 

Icon and Unicon

The following code demonstrates the use of 'next' (the reserved word for 'continue'):

procedure main()
every writes(x := 1 to 10) do {
   if x % 5 = 0 then {
      write()
      next         
      }
   writes(", ")
   }
end

However, the output sequence can be written without 'next' and far more succinctly as:

every writes(x := 1 to 10, if x % 5 = 0 then "\n" else ", ")

J

J is array-oriented, so there is very little need for loops. For example, one could satisfy this task this way:

_2}."1'lq<, >'8!:2>:i.2 5

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).

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
   )

Though it's rare to see J code like this.

Java

for(int i = 1;i <= 10; i++){
   System.out.print(i);
   if(i % 5 == 0){
      System.out.println();
      continue;
   }
   System.out.print(", ");
}

JavaScript

Using the print() function from Rhino or SpiderMonkey.

var output = "";
for (var i = 1; i <= 10; i++) {
  output += i; 
  if (i % 5 == 0) {
    print(output);
    output = "";
    continue;
  } 
  output += ", ";
}

Liberty BASIC

 
for i =1 to 10
    if i mod 5 <>0 then print i; ", "; else print i
next i
end
 

Lisaac

1.to 10 do { i : INTEGER;
  i.print;
  (i % 5 = 0).if { '\n'.print; } else { ','.print; };
};

Lua

for i = 1, 10 do
    io.write( i )
    if i % 5 == 0 then
        io.write( "\n" )
    else
    	io.write( ", " ) 
    end
end

Maple

for i from 1 to 10 do
        printf( "%d", i );
        if irem( i, 5 ) = 0 then
                printf( "\n" );
                next
        end if;
        printf( ", " )
end do:

This can also be done as follows, but without the use of "next".

for i to 10 do
        printf( "%d%s", i, `if`( irem( i, 5 ) = 0, "\n", ", " ) )
end do:

Mathematica

tmp = "";
For[i = 1, i <= 10, i++,
  tmp = tmp <> ToString[i];
  If[Mod[i, 5] == 0,
   tmp = tmp <> "\n";
   ,
   tmp = tmp <> ", ";
   ];
  ];
Print[tmp]

MATLAB / Octave

Loops are considered slow in Matlab and Octave, it is preferable to vectorize the code.

disp([1:5; 6:10])

or

disp(reshape([1:10],5,2)')

A non-vectorized version of the code is shown below in Octave

for i = 1:10
  printf(' %2d',  i);
  if ( mod(i, 5) == 0 ) 
    printf('\n');
    continue
  end
end

Maxima

/* There is no "continue" in Maxima, the easiest is using a "if" instead */
block(
   [s: ""],
   for n thru 10 do (
      s: sconcat(s, n),
      if mod(n, 5) = 0 then (
         ldisp(s),
         s: ""
      ) else (
         s: sconcat(s, ", ")
      )
   )
)$

MAXScript

for i in 1 to 10 do
(
    format "%" i
    if mod i 5 == 0 then
    (
        format "\n"
        continue
    )   continue
    format ", "
)

Insert non-formatted text here

Metafont

Metafont has no a continue (or similar) keyword. As the Ada solution, we can complete the task just with conditional.

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

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

primarydef x mod y = (x-y*floor(x/y)) enddef;

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.

FOR i := 1 TO 10 DO
  IO.PutInt(i);
  IF i MOD 5 = 0 THEN
    IO.Put("\n");
    RETURN;
  END;
  IO.Put(", ");
END;

MOO

s = "";
for i in [1..10]
  s += tostr(i);
  if (i % 5 == 0)
    player:tell(s);
    s = "";
    continue;
  endif
  s += ", ";
endfor

Nemerle

Translation of: C#

using System;
using System.Console;
using Nemerle.Imperative;
 
module Continue
{
    Main() : void
    {
        foreach (i in [1 .. 10])
        {
            Write(i);
            when (i % 5 == 0) {WriteLine(); continue;}
            Write(", ");
        }
    }
}

NetRexx

/* NetRexx */
options replace format comments java crossref savelog symbols nobinary
 
  say
  say 'Loops/Continue'
 
  nul = '\-'
  loop i_ = 1 to 10
    say i_.right(2) || nul
    if i_ // 5 = 0 then do
      say
      iterate i_
      end
    say ', ' || nul
 
    end i_
 

OCaml

There is no continue statement for for loops in OCaml, but it is possible to achieve the same effect with an exception.

# 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 = ()

Though even if the continue statement does not exist, it is possible to add it with camlp4.

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

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.

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

PARI/GP

for(n=1,10,
  print1(n);
  if(n%5 == 0, print();continue);
  print1(", ")
)

Pascal

See Delphi

Perl

foreach (1..10) {
    print $_;
    if ($_ % 5 == 0) {
        print "\n";
        next;
    }
    print ', ';
}

It is also possible to use a goto statement to jump over the iterative code section for a particular loop:

foreach (1..10) {
    print $_;
    if ($_ % 5 == 0) {
        print "\n";
        goto MYLABEL;
    }
    print ', ';
MYLABEL:
}

Perl 6

Translation of: Perl

Works with: Rakudo Star version 2010.08

for 1 .. 10 {
    .print;
    if $_ %% 5 {
        print "\n";
        next;
    }
    print ', ';
}

or without using a loop:

$_.join(", ").say for [1..5], [6..10];

PHP

for ($i = 1; $i <= 10; $i++) {
    echo $i;
    if ($i % 5 == 0) {
        echo "\n";
        continue;
    }
    echo ', ';
}

PicoLisp

PicoLisp doesn't have an explicit 'continue' functionality. It can always be emulated with a conditional expression.

(for I 10
   (print I)
   (if (=0 (% I 5))
      (prinl)
      (prin ", ") ) )

Pike

int main(){
   for(int i = 1; i <= 10; i++){
      write(sprintf("%d",i));
      if(i % 5 == 0){
         write("\n");
         continue;
      }
      write(", ");
   }
}

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;

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;

PowerShell

Translation of: C

for ($i = 1; $i -le 10; $i++) {
    Write-Host -NoNewline $i
    if ($i % 5 -eq 0) {
        Write-Host
        continue
    }
    Write-Host -NoNewline ", "
}

PureBasic

OpenConsole()
 
For i.i = 1 To 10
  Print(Str(i))
  If i % 5 = 0
    PrintN("")
    Continue
  EndIf
  Print(",")
Next
 
Repeat: Until Inkey() <> ""

Python

for i in xrange(1,11):
    if i % 5 == 0:
        print i
        continue
    print i, ",",

R

Translation of: C++

for(i in 1:10)
{
   cat(i)
   if(i %% 5 == 0) 
   {
      cat("\n")
      next
   }
   cat(", ")
}

Racket

It is possible to skip loop iterations in Racket, but an explicit continue construct is rarely used:

 
#lang racket
 
;; Idiomatic way
(for ([i (in-range 1 11)])
  (if (= (remainder i 5) 0)
      (printf "~a~n" i)
      (printf "~a, " i)))
 
;; Forces a skip, but not idiomatic because
;; the logic is less obvious
(for ([i (in-range 1 11)]
      #:unless (and (= (remainder i 5) 0)
                    (printf "~a~n" i)))
  (printf "~a, " i))
 

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 ", "
	]
]

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

/*REXX program to illustrate  DO  loop with an  ITERATE  (continue). */
 
    do j=1 to 10
    call charout ,  j", "
 
    if j//5==0 then do
                    say
                    iterate
                    end
    end   /*j*/

Ruby

for i in 1..10 do
   print i
   if i % 5 == 0 then
      puts
      next
   end
   print ', '
end

The "for" look could be written like this:

(1..10).each do |i| ...
1.upto(10) do |i| ...
10.times do |n| i=n+1; ...

Without meeting the criteria (showing loop continuation), this task could be written as:

1.upto(10) {|i| print "%d%s" % [i, i%5==0 ? "\n" : ", "]}

Run BASIC

for i = 1 to 10
    if i mod 5 <> 0 then print i;", "; else print i
next i

Salmon

iterate (x; [1...10])
  {
    print(x);
    if (x % 5 == 0)
      {
        print("\n");
        continue;
      };
    print(", ");
  };

Sather

There's no continue! in Sather. The code solve the task without forcing a new iteration.

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;

Scala

Scala doesn't have a continue keyword. if could be used here.

for(i <- 1 to 10) {
  print(i)
  if (i%5==0) println() else print(", ")
}

Scheme

(define (loop i)
  (if (> i 10) 'done
      (begin
       (display i)
       (cond ((zero? (modulo i 5))
              (newline) (loop (+ 1 i)))
             (else (display ", ")
                   (loop (+ 1 i)))))))

Suneido

ob = Object()
for (i = 1; i <= 10; ++i)
    {
    ob.Add(i)
    if i is 5
        {
        Print(ob.Join(','))
        ob = Object()
        }
    }
Print(ob.Join(','))

Output:

1,2,3,4,5
6,7,8,9,10
ok

Tcl

for {set i 1} {$i <= 10} {incr i} {
   puts -nonewline $i
   if {$i % 5 == 0} {
      puts ""
      continue
   }
   puts -nonewline ", "
}

TI-89 BASIC

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

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

 
$$ MODE TUSCRIPT
numbers=""
LOOP n=1,10
numbers=APPEND (numbers,", ",n)
rest=n%5
IF (rest!=0) CYCLE
 PRINT numbers
 numbers=""
ENDLOOP
 

Output:

1, 2, 3, 4, 5
6, 7, 8, 9, 10

UnixPipes

yes \ | cat -n | head -n 10 | xargs -n 5 echo | tr ' ' ,

UNIX Shell

Z=1
while (( Z<=10 )); do
    echo -e "$Z\c"
    if (( Z % 5 != 0 )); then
        echo -e ", \c"
    else
        echo -e ""
    fi
    (( Z++ ))
done

Works with: Bash

for ((i=1;i<=10;i++)); do
  echo -n $i
  if [ $((i%5)) -eq 0 ]; then
    echo
    continue
  fi
  echo -n ", "
done

Vedit macro language

for (#1 = 1; #1 <= 10; #1++) {
    Num_Type(#1, LEFT+NOCR)
    if (#1 % 5 == 0) {
        Type_Newline
        Continue
    }
    Message(", ")
}

Visual Basic .NET

For i = 1 To 10
    Console.Write(i)
    If i Mod 5 = 0 Then
        Console.WriteLine()
    Else
        Console.Write(", ")
    End If
Next

XPL0

Like Ada and ALGOL there's no 'continue' command. The task is solved very simply anyway. The commands 'int' and 'rem' are shown spelled out here. Only the first three characters of a command are required.

code CrLf=9, IntOut=11, Text=12;
integer N;
for N:= 1 to 10 do
        [IntOut(0, N); if remainder(N/5) \#0\ then Text(0, ", ") else CrLf(0)]

Output:

1, 2, 3, 4, 5
6, 7, 8, 9, 10