Loops/Continue

From Rosetta Code

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.

Contents 1 Ada 2 Aikido 3 ALGOL 68 4 AutoHotkey 5 AWK 6 C 7 C++ 8 C# 9 Clojure 10 ColdFusion 11 Common Lisp 12 D 13 E 14 Factor 15 Forth 16 Fortran 17 Go 18 Haskell 19 HicEst 20 Icon and Unicon 20.1 Icon 20.2 Unicon 21 J 22 Java 23 JavaScript 24 Lisaac 25 Mathematica 26 MAXScript 27 Metafont 28 Modula-3 29 MOO 30 OCaml 31 Octave 32 Oz 33 Perl 34 Perl 6 35 PHP 36 PicoLisp 37 Pike 38 PL/I 39 Pop11 40 PowerShell 41 PureBasic 42 Python 43 R 44 REBOL 45 REXX 46 Ruby 47 Sather 48 Scheme 49 Suneido 50 Tcl 51 TI-89 BASIC 52 UnixPipes 53 Vedit macro language 54 Visual Basic .NET

[edit] 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;

[edit] Aikido

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

[edit] 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

[edit] AutoHotkey

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

[edit] AWK

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

[edit] C

Translation of: C++

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

[edit] C++

Translation of: Java

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

[edit] 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(", ");
        }
    }
}

[edit] 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 ", "))))
 

[edit] 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>

[edit] 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 ", ")))

[edit] D

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

[edit] E

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

[edit] Factor

There is no built-in continue in Factor.

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

[edit] 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 ;

[edit] 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

[edit] Go

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

[edit] 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) = (putStrLn . show) x
            | otherwise = (putStr . (++", ") . show) x

[edit] HicEst

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

[edit] Icon and Unicon

The Icon and Unicon reserved word for 'continue' is 'next'.

[edit] Icon

The following code demonstrates the use of 'next':

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 ", ")
 

[edit] Unicon

The Icon solution works in Unicon.

[edit] 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.

[edit] Java

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

[edit] 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 += ", ";
}

[edit] Lisaac

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

[edit] Mathematica

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

[edit] MAXScript

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

[edit] 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;

[edit] 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;

[edit] MOO

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

[edit] 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.

[edit] 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

[edit] 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

[edit] Perl

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

[edit] 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];

[edit] PHP

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

[edit] 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 ", ") ) )

[edit] Pike

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

[edit] 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;
 

[edit] 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;

[edit] 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 ", "
}

[edit] PureBasic

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

[edit] Python

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

[edit] R

Translated from C++.

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

[edit] 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

[edit] REXX

(Remember that there exists implementations of the REXX language that needs that the source begins with /*, i.e. with a comment)

do i = 1 to 10
  call charout ,i", "
  if i//5 = 0 then do
    say
    iterate
  end
end

[edit] 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" : ", "]}

[edit] 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;

[edit] 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)))))))

[edit] 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

Bold text

[edit] Tcl

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

[edit] 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.

[edit] UnixPipes

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

[edit] Vedit macro language

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

[edit] 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