Reinvent the For loop

Question

For loops are used extensively in many languages, but what would you do if no languages supported them?

Create a way to execute a basic for loop without using any repetition structures (for, foreach, while, do, etc).

The basic for loop that you need to replicate is set up like this

for(i=0;i<1000;i++)

You must be able to replicate this without using repetition structures. It must also execute code in your language that would be in the body of the loop. Any form of eval is allowed, though it may not execute the for loop on its own.

You can test your code by having it print i with a space 100 times, add this test case to the end of your answer to verify your code with others.

There are no restrictions on what exactly it has to be, all it needs to do is replicate the for loop.

Winner will be decided based on upvotes at the time that it is chosen.

I find the problem too undefined. You say “without using any repetition structures” and only give examples (but not an extensive list) of such prohibited structures. The answers provided so far use recursion or goto, both of which I would classify as “repetition structures”, but it can be debated. Without a proper definition of what is allowed and what isn’t, this question is not interesting.

houbysoft · Accepted Answer · 2011-02-19 22:12:29Z

C, 45 chars without goto, if, for, ...

Not the smallest solution, but I find this way of doing it quite interesting in C :)

Doesn't use goto, if, for, or any other kind of control structures.

The function by itself (45 chars):

m(){int (*a[])()={m,exit};i++;(*a[i>999])();}

A compilable, working program:

#include <stdio.h>
#include <stdlib.h>

int i=0;
m(){int (*a[])()={m,exit};i++;printf("%i ",i);(*a[i>999])();}

int main()
{
  m();
}

@Denis Bueno: No, a control structure is stuff like while, do, for, if...

2 revs, 2 users 78% Eelvex · Answer 2 · 2011-03-25 03:59:58Z

up vote 10 down vote

C, 21 characters

f(){if(c++<999)f();}

For example:

#include <stdio.h>
int c = 0;
f()
{
    printf("%d ", c);
    if (c++<999) f();
}
int main(void)
{
    f();
}

Outputs:

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ... 998 999

edited Mar 25 '11 at 3:59

community wiki

2 revs, 2 users 78%
Eelvex

Without a main() in the first block, the code won't link. Therefore I think it should be included in the entry. – Nathan Osman Feb 19 '11 at 6:11

4

@George: I think it's fair omitting main() in this case because the question says "Create a way to execute" not "make a complete program" as in other problems. – Eelvex Feb 19 '11 at 15:46

1

remove void then it will be 26 characters only. – Prince John Wesley Mar 25 '11 at 3:45

R. Martinho Fernandes · Answer 3 · 2011-02-19 19:42:26Z

up vote 7 down vote

Haskell, 33 chars

This is more like inventing the for loop, because there is no such nonsense in Haskell :P.

mapM_(putStr.(' ':).show)[0..999]

answered Feb 19 '11 at 19:42

community wiki

R. Martinho Fernandes

1

forM_ is just mapM_ with the arguments flipped, so this might be cheating. ;) – Dan Mar 11 '11 at 5:29

J B · Answer 4 · 2011-02-19 00:14:02Z

up vote 6 down vote

x86 assembly, 12 11 bytes' worth of instructions

66 31 c0                xor    %ax,%ax
<body>                ; involving %ax
66 40                   inc    %ax
66 3d  e8 03            cmp    $1000,%ax
75 ??                   jne    <body>

answered Feb 19 '11 at 0:14

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J B

	A pity they're asked in ascending order... I'm pretty sure it would be shorter descending. – J B Feb 19 '11 at 0:28
	@J B: Yes -- in descending order, you'd normally use the `loop` instruction: `mov cx, 1000\nx: ... loop x` – Jerry Coffin Feb 20 '11 at 6:47

3 revs Nemo157 · Answer 5 · 2011-02-19 22:21:22Z

Ruby 1.9 - 51 characters

def f a,b,c,&d
a[]
k=->{d[];c[];redo if b[]}
k[]end

This is a lot larger than other entries, but I think it captures the essence of the question better. In effect this allows you to write code almost exactly like the example:

def f a,b,c,&d
a[]
k=->{d[];c[];redo if b[]}
k[]end


i = 0 # Have to pre-declare i to have the closures work properly
f(proc{ i = 0 }, proc{ i < 1000 }, proc{ i += 1}) do
  p i
end

# Been a long time since I've actually used for's so took a long
# time to think of a non-trivial use.

j, sum = 0, 0 # Pre-declare variables for the closures

goal = 1000

f(proc{ j = 1; sum = 1 }, proc{ sum < goal }, proc{ j += 1 }) do
  sum = sum * j
end

puts "The next largest factorial after #{goal} is #{j-1}! at #{sum}"

3 revs stam · Answer 6 · 2011-03-24 10:53:02Z

up vote 5 down vote

Yet another Scala variant

Since Scala allows to use non-ASCII characters, we can implement real for :) (o is cyrillic).

def fоr(s:Unit,c: ⇒Boolean,e: ⇒Unit)(b: ⇒Unit):Unit=if(c){b;e;fоr(0,c,e)(b)}

Test:

var i:Int = _

fоr (i = 0, i < 1000, i += 1) {
  print(i + " ")
}

edited Mar 24 '11 at 10:53

community wiki

3 revs
stam

2 revs Joey Adams · Answer 7 · 2011-02-20 01:20:15Z

GCC - 106 95 chars

#define FOR(i,l,h)auto F();L(F,l,h);int F(i)
L(int f(int),int l,int h){l<=h?f(l),L(f,l+1,h):0;}

Unlike the other C solutions where you have to declare a callback, this one does it for you automagically:

FOR(i, 1, 10) {
    printf("%d\n", i);
}

It works by using GCC's nested function extension. Namely, it forward-declares the nested function F, passes it to the looping function L, and then starts the definition of F, leaving the braces for the user to add.

One beautiful thing about nested functions in GCC is that they support downward funargs, meaning the illusion is nearly complete:

long n = 1;
FOR(i, 1, 10) {
    n *= i;
}
printf("%ld\n", n); // 3628800

There is one major caveat: if you use FOR twice in the same scope, you'll get a conflict (namely, it will compile, but all the FOR loops will share one loop body). To allow multiple FOR loops in the same scope, we'll need 69 65 more characters:

~~175~~ 160 chars:

#define FOR(i,l,h)F(i,l,h,__COUNTER__)
#define F(i,l,h,f)auto N(f)();L(N(f),l,h);int N(f)(i)
#define N(n)F##n
L(int f(int),int l,int h){l<=h?f(l),L(f,l+1,h):0;}

2 revs, 2 users 63% Matt · Answer 8 · 2011-02-26 18:17:01Z

up vote 4 down vote

C: 33 chars for the base for loop (assuming goto is allowed)

int i=0;s:if(i<1000){/`*code inside for loop here`*/i++;goto s;}


int i=0;s:if(i<1000){printf("%d ",i);i++;goto s;}

edited Feb 26 '11 at 18:17

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2 revs, 2 users 63%
Matt

user732 · Answer 9 · 2011-03-06 21:13:05Z

up vote 4 down vote

Here's your loop using Brainfuck, which is probably cheating:

,[.,]

answered Mar 6 '11 at 21:13

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user732

Timwi · Answer 10 · 2011-03-08 15:14:38Z

C# — no recursion, no goto

(using the Y combinator from lambda calculus)

class Program
{
    delegate Func<TInput, TResult> Lambda<TInput, TResult>(Lambda<TInput, TResult> f);

    static Func<TInput, TResult> Y<TInput, TResult>(Func<Func<TInput, TResult>, Func<TInput, TResult>> f)
    {
        Lambda<TInput, TResult> y = r => n => f(r(r))(n);
        return y(y);
    }

    static void Main()
    {
        Func<int, int> fibonacci = Y<int, int>(f => n => n > 1 ? f(n - 1) + f(n - 2) : n);
        Func<int, int> factorial = Y<int, int>(f => n => n == 0 ? 1 : n * f(n - 1));

        // Executes “fibonacci” 10 times, yielding 55
        Console.WriteLine(fibonacci(10));
        // Executes “factorial” 5 times, yielding 120
        Console.WriteLine(factorial(5));
    }
}

4 revs, 2 users 95% zzzzBov · Answer 11 · 2011-03-24 20:28:04Z

JavaScript, 34 characters

This assumes the actual printing function is not part of the for loop.

parameterized recursion (43):

function f(i,m,c){if (i<m){c();f(++i,m,c)}}

function f(i, max, callback)
{
  if (i<max)
  {
    callback();
    f(++i,max,callback);
  }
}

reverse recursion (36):
assumes max >= 0

function f(m,c){if(m){c()
f(--m,c)}}

function f(max,callback)
{
  if(max)
  {
    callback();
    f(--max,callback);
  }
}

ternary operator (34)

function f(m,c)
{m?c()&f(--m,c):0}

function f(m,c){m&&c()&f(--m,c)} - You could make that 32 bytes :)

2 revs Sam · Answer 12 · 2011-02-19 00:56:35Z

up vote 3 down vote

Python

Specific case:

exec("print i,"+";")*1000

And in general:

exec("f"+";")*1000

where 'f' is your code. Proabably doesn't work for all cases.

edited Feb 19 '11 at 0:56

community wiki

2 revs
Sam

NameError: name 'i' is not defined – J B Feb 19 '11 at 0:45

I've kind of fixed it. Plus, I assumed that i was predefined. – user611 Feb 19 '11 at 0:57

2

If it's predefined, it's not going to change with the iteration. That's not how a for loop works. – J B Feb 19 '11 at 1:02

Fair point. I imagine it would need ;i=+1 after the comma. I did say it wouldn't work with all cases. – user611 Feb 19 '11 at 1:11

Ted Dziuba · Answer 13 · 2011-02-19 01:34:16Z

up vote 3 down vote

I shall bear great shame for this, but here it is in Linux shell:

cat -An /dev/urandom|head -n1000|cut -f1

40 characters.

answered Feb 19 '11 at 1:34

community wiki

Ted Dziuba

Prints linebreaks => disqualification! :) – user unknown Feb 19 '11 at 7:08

5 revs Justin E. Morgan · Answer 14 · 2012-04-24 21:00:16Z

C#, 70 57 characters

C# isn't the weapon of choice for code golf, but I thought I'd try it out:

void f(int i,Func<int,bool>j,int k){if(j(i)) f(i+k,j,k);}

This doesn't only perform the task of counting up to 1000; rather, it attempts to actually replicate for loop behavior for any task. That seemed somewhat closer to the intent of the challenge, but maybe that's just me.

Expanded:

void f(int i, Func<int, bool> j, int k)
{
    if (j(i))
    {
        f(i + k, j, k);
    }
}

Usage is very close to for-loop syntax:

f(0, i => i < 1000, 1);

PleaseStand · Answer 15 · 2011-02-19 00:58:06Z

JavaScript - 39 characters in test case

Chrome, Opera, and IE: eval(s="i<1e3&&eval(s,print(i++))",i=0). This fails with "call stack size exceeded" on Safari and "too much recursion" on Firefox.

Firefox: (function f()i<1e3&&f(print(i++)))(i=0). This uses Mozilla's non-standard "expression closure" feature to eliminate a pair of curly braces.

Note: Change print(i++) to alert(i++) if you need to. You can change 1e3 to 100 to reduce the number of iterations for Safari.

2 revs idealmachine · Answer 16 · 2011-02-19 01:36:16Z

up vote 2 down vote

QBasic (24 characters)

1?i:i=i+1:IF i<1E3THEN 1

expands to:

1 PRINT i: i = i + 1: IF i < 1000! THEN 1

Note: This assumes that i still has its original value of 0.

edited Feb 19 '11 at 1:36

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2 revs
idealmachine

3 revs Eelvex · Answer 17 · 2011-02-19 02:42:25Z

J, ?<10 chars

f^:1000

In J there are (almost) no loops; instead, one usually uses arrays and implicit loops through them. For example, to sum the integers 1..100, you apply (/) the "verb" plus (+) to the array 1..100 (i.101)

+/i.101
5050

To display the numbers 0..99, we just construct the array:

i.100
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 ...

^: is the "power" verb; f^:1000 is something like exec()*1000 in python.

2 revs Steve · Answer 18 · 2011-02-19 03:10:13Z

C#, 58 characters

Update: D'oh...appears this is a virtually identical answer to Eelvex who beat me by a few minutes.

Not very clever - just simple recursion in .NET.

void f(int c,int m){Console.Write(c+" ");if(c++<m)f(c,m);}

Or taking out the body of the "for loop" (35 characters):

void f(int c,int m)if(c++<m)f(c,m);

Allows to seed the initial value and the max value.

4 revs user300 · Answer 19 · 2011-02-19 11:45:23Z

Scala

Inspired from Nemo157's answer:

def myfor(test: =>Boolean, inc: =>Unit)(stmts: =>Unit) : Unit =
        if (test) {
                stmts
                inc
                myfor(test, inc)(stmts)
        }

And this can be used like this:

var i = 0
myfor(i<1000, i+=1) {
        print(i+" ")
}

Golfed (62) :

def f(t: =>Boolean,i: =>Unit)(s: =>Unit){if(t){s;i;f(t,i)(s)}}

Or:

def myfor[A](init: A, test: (A)=>Boolean, inc: (A)=>(A))(stmts: =>Unit) : Unit = 
        if (test(init)) {
                stmts
                myfor(inc(init), test, inc)(stmts)
        }

myfor[Int](0, _<10, _+1) {
        println("loop")
}

7 revs Ty Auvil · Answer 20 · 2011-02-19 16:48:34Z

PowerShell 18

filter p{$i=$_;$i}

1 - 10

PS C:\> (1..100 | p)-join' '
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 6
4 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100

Yeah, it's completely useless and you have to rewrite the function to do anything useful. But isn't that kind of the point? :)

Prints newlines as delimiters? ... print i with a space 1 ... - disqualification. :)

deiga · Answer 21 · 2011-02-21 01:36:20Z

up vote 2 down vote

Clojure
(map #(print (str % " ")) (range 1 (inc 100)))

Replace print with the function to execute.
Use (inc 100) or 101 for the last value.

answered Feb 21 '11 at 1:36

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deiga

2 revs, 2 users 75% J B · Answer 22 · 2011-02-26 18:16:47Z

up vote 2 down vote

Perl, 21 characters

map{print"$_ "}0..999

edited Feb 26 '11 at 18:16

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2 revs, 2 users 75%
J B

	Mapping is a type of looping. Don't cheat! :-P – Chris Jester-Young♦ Feb 19 '11 at 2:58
	This comment is nonsense, since we have to write something which is equivalent to a for-loop. If we met the criteria, it will be always cheating, or what will make the difference? Explicit gotos? Recursion? – user unknown Feb 19 '11 at 6:56
	@Chris well I explicitely asked the OP in the chat before answering that one. – J B Feb 19 '11 at 9:26

3 revs rubber boots · Answer 23 · 2011-03-06 21:49:37Z

Perl most simple - 18 chars (ok, cheating)

print"@{[0..999]}"

output: 0 1 2 3 ... 998 999

but:

Perl regex/goatse op (real pseudoloop) - 39 (16) chars

()="@{[0..999]}"=~/\d+(?{print"$& "})/g

the 'base loop' beeing 16 chars

()=             =~/\d+(?{          })/g

output: 0 1 2 3 ... 998 999

Regards

rbo

user612 · Answer 24 · 2011-02-19 01:25:25Z

;; for (int index=1, result=1; !cmp(result, max); index=incr(index)) result=body(index,result);
((lambda (index result cmp max body incr)
   ((lambda (f n i)
      (if (cmp i max) n
          (f f (body i n) (incr i))))
    (lambda (f n i)
      (if (cmp i max) n
          (f f (body i n) (incr i)))) result index)) 1 1 > 1000 (lambda (i n) (printf "i:~a~nn:~a~n" i n) (* n i)) (lambda (x) (+ x 1)))

2 revs Eelvex · Answer 25 · 2011-02-19 02:51:57Z

up vote 1 down vote

LISP, 91 characters

(defun m(x)(if(= 1000(length x))x(m(append x(list(1+(car(last x))))))))
(mapcar #'f(m'(0)))

It could be shorter with labels.

Example usage:

(defun f(x) (format t "~a " x))
(mapcar #'f (m '(0)))
0 1 2 3 4 5 6 7 8 9 10 11 12 ...  997 998 999

edited Feb 19 '11 at 2:51

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2 revs
Eelvex

user unknown · Answer 26 · 2011-02-19 06:58:10Z

up vote 1 down vote

bash:

echo {1..1000}

14 chars.

answered Feb 19 '11 at 6:58

community wiki

user unknown

gnibbler · Answer 27 · 2011-02-19 10:48:41Z

up vote 1 down vote

Python3 - 27 chars

any(map(print,range(1000)))

answered Feb 19 '11 at 10:48

community wiki

gnibbler

Cool. For the nice output: print(" ".join(map(str,range(1000)))) – wok Feb 19 '11 at 20:53

4 revs DavidM · Answer 28 · 2011-02-19 15:06:15Z

up vote 1 down vote

Groovy: (0..1000).each{ print "${it} " }

edited Feb 19 '11 at 15:06

community wiki

4 revs
DavidM

	each is foreach => disqualified :) println is printing with newline, not space => disqualified. ;) – user unknown Feb 19 '11 at 7:20
	Which is silly. Iteration is not a loop even though they contain 4 letters which are similar. Iteration = processing data in memory = what computers are built to do. – user615 Feb 19 '11 at 14:11

Joey · Answer 29 · 2011-02-19 17:15:28Z

up vote 1 down vote

Windows Batch File, 65

set I=%S%
:l
if %I% GEQ %E% goto :eof
call %C%
set /a I+=1
goto l

Test file:

@echo off
set S=0
set E=100
set "C=<nul set /p X=%%I%% "
call for.cmd

Can save two bytes if the loop always starts at 0.

answered Feb 19 '11 at 17:15

community wiki

Joey

3 revs, 2 users 57% hydralien · Answer 30 · 2011-02-26 18:16:26Z

up vote 1 down vote

Perl (23 chars.):

print join " ",(0..999);*

or same Perl but in simpler yet more lengthy and recursive words:

sub i {($s,$e)=@_; return if $s==$e; print "$s "; i($s+1,$e)}

i(0,1000);

edited Feb 26 '11 at 18:16

community wiki

3 revs, 2 users 57%
hydralien

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Reinvent the For loop

49 Answers

C, 45 chars without goto, if, for, ...

C, 21 characters

Haskell, 33 chars

x86 assembly, 12 11 bytes' worth of instructions

Ruby 1.9 - 51 characters

Yet another Scala variant

GCC - 106 95 chars

C# — no recursion, no goto

JavaScript, 34 characters

Python

C#, 70 57 characters

JavaScript - 39 characters in test case

QBasic (24 characters)

J, ?<10 chars

C#, 58 characters

Scala

PowerShell 18

Perl, 21 characters

Perl most simple - 18 chars (ok, cheating)

Perl regex/goatse op (real pseudoloop) - 39 (16) chars

the 'base loop' beeing 16 chars

LISP, 91 characters

Python3 - 27 chars

Windows Batch File, 65

Perl (23 chars.):

Your Answer

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Reinvent the For loop

49 Answers

C, 45 chars without goto, if, for, ...

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C, 21 characters

Haskell, 33 chars

x86 assembly, 12 11 bytes' worth of instructions

Ruby 1.9 - 51 characters

Yet another Scala variant

GCC - 106 95 chars

C# — no recursion, no goto

JavaScript, 34 characters

Python

C#, 70 57 characters

JavaScript - 39 characters in test case

QBasic (24 characters)

J, ?<10 chars

C#, 58 characters

Scala

PowerShell 18

Perl, 21 characters

Perl most simple - 18 chars (ok, cheating)

Perl regex/goatse op (real pseudoloop) - 39 (16) chars

the 'base loop' beeing 16 chars

LISP, 91 characters

Python3 - 27 chars

Windows Batch File, 65

Perl (23 chars.):

Your Answer

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