/prog/ Challenge #9002

The task:

Print strings from "a" to "zzzzz" without using any loop or conditional statements. Don't just write all 1000 permutations out by hand. The output should look like this:
a
b
c
...
aa
ab
ac
...
zzzzx
zzzzy
zzzzz

inb4 lipthfags and dead dogs using some obscure functionality of their obscure languages.

>>1
>The output should look like this:

ok

print """
a
b
c
...
aa
ab
ac
...
zzzzx
zzzzy
zzzzz
"""

>>21
By connecting SIGFPE to an exit function and dividing some dummy volatile value with ('z' - c0) + ('z' - c1) + ('z' - c2) + ('z' - c3) + ('z' - c4), you could make it terminate at the right time.

>>22
>>23
>>26
>>30
Ok, thanks for all your tips but I did it somewhat different now. Hers's the new code, with cheap ASM, function pointing and actual spaces instead of null characters:
#include <stdio.h>

char c0 = 0, c1 = -1, c2 = -1, c3 = -1, c4 = -1, done;
unsigned long t;
void (*f)();
void p(), x();

void p()
{
 printf("%c%c%c%c%c\n",
        ((c4 + 'a') & ((c4 >= 0) * 0xFF)) | (' ' & ((c4 < 0) * 0xFF)),
        ((c3 + 'a') & ((c3 >= 0) * 0xFF)) | (' ' & ((c3 < 0) * 0xFF)),
        ((c2 + 'a') & ((c2 >= 0) * 0xFF)) | (' ' & ((c2 < 0) * 0xFF)),
        ((c1 + 'a') & ((c1 >= 0) * 0xFF)) | (' ' & ((c1 < 0) * 0xFF)),
        (c0 + 'a'));
 c0++;
 c1 += c0 > 25;
 c2 += c1 > 25;
 c3 += c2 > 25;
 c4 += c3 > 25;
 done = (c0 > 25) & (c1 > 25) & (c2 > 25) & (c3 > 25) & (c4 > 25);
 c0 %= 26;
 c1 %= 26;
 c2 %= 26;
 c3 %= 26;
 c4 %= 26;
 f = (void*) ((p & ((1 - done) * 0xFFFFFFFF)) | (x & (done * 0xFFFFFFFF)));
 asm("movl %0, %%esp\n" : : "r"(t));
 f();
}

void x()
{
 /* YES, it's reached now. Feels good to return zero. */
 printf("DONE!!\n");
 exit(0);
}

int main(void)
{
 asm("movl %%esp, %0\n" : "r"(t) : );
 p();
}

...yes, this compiles for me. I'm using TCC/Win32 for compiling, so ASM is in GAS/AT&T format. The function pointing didn't work like this for me in GCC, though.
I also removed the return statement, since this might be interpreted as a jump/loop/whatever. Only calls now, no JMPs.

Check my dubs.

Solve the problem by converting to bases. But you run into the problem of the character set is base 27 not base 26.

So here's some hacky perl that doesn't use a lot of memory. For the rest of you, why did you over complicate it?

 my $i = 0;
 my $l = "";
 while ($l ne "zzzzz") {
     eval {
         $l = convert($i);
         print $l,$/;
     };
     $i++;
 }
 sub convert {
     my ($in) = @_;
     my $mod = $in%27;
     if ($mod == 0) { die "Fuckit"; }
     my $diff = $in - $mod;
     my $new = $diff / 27;
     my $c = ((' ','a'..'z')[$mod]);
     return ($new==0)?$c:convert($new).$c;
 }

Recursion not counting as a loop is just stupid.