if you know me or not, I'm completely unaware of other programming languages; I just know a little Perl but I found myself in the rare situation where I need to translate a little code from C to Perl.

Infact I rapidly (*) understood that is a task to be done by hand, and I've been told many times that these two languages share a lot in their syntax.

My attempt is below and does not produces the output I expected (**).

I looked a bit to some description of the C syntax to try to understand if there were some difference between the C operator and the correspective Perl's one. For example for ++ autoincrement or arithmentics ones. I found nothing relevant: many operators seems to act the same. Doubts remain about the C array syntax ( int A[len] ?? that I read as the elelment len of the array A is an int but..).

Here below my attempt: can someone be so kind to point me where I lost in the translation? After a plain translation, when I possibly end with some working Perl code I'll arrange it into a more perlish version.

use strict; use warnings;
# #include <math.h>
# #include <stdio.h>

# #define N 100
my $n=100;

# int len = floor(10 * N/3) + 1;
my $len = 1 + int (10 * $n / 3);

# int A[len];
my @a;
$#a=$len-1; #? -1 ????

# for(int i = 0; i < len; ++i) {A[i] = 2;}
for (my $i = 0; $i < $len; $i++){ $a[$i]= 2; }
# int nines    = 0;
my $nines = 0;

# int predigit = 0;
my $predigit = 0;

# for(int j = 1; j < N + 1; ++j) {
for (my $j = 1; $j < $n + 1; ++$j){
    # int q = 0;
    my $q = 0;
    # for(int i = len; i > 0; --i) {
    for(my $i = $len; $i > 0; $i--){
        # int x  = 10 * A[i-1] + q*i;
        my $x = 10 * $a[$i-1] + $q * $i;
        # A[i-1] = x % (2*i - 1);
        $a[$i-1] = $x % (2 * $i - 1);
        # q = x / (2*i - 1);     }
        $q = $x / (2 * $i - 1);
    }
    # A[0] = q%10;
    $a[0]=$q%10;
    # q  = q/10;
    $q=$q/10;
    # if (9 == q) {   ++nines;}
    if (9 == $q){ ++$nines;   }
    # else if (10 == q) {
    elsif(10 == $q){
          # printf("%d", predigit + 1);
          printf("%d", $predigit + 1);
          # for (int k = 0; k < nines; ++k) {  printf("%d", 0); }
          for (my $k = 0; $k < $nines; $k++){ printf("%d", 0);}
          #  predigit, nines = 0;
          $predigit = $nines = 0;
    # }
    }
    # else {
    else{
         # printf("%d", predigit);
         printf("%d", $predigit);
         #  predigit = q;
         $predigit = $q;
         # if (0 != nines) {
         if(0 != $nines){
              # for (int k = 0; k < nines; ++k) {
              for (my $k = 0; $k < $nines; $k++) {
                  # printf("%d", 9);
                  printf("%d", 9);
              # }
              }
              # nines = 0;
              $nines = 0;
        }
    }
}
# printf("%d", predigit);
printf("%d", $predigit);
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(*) Not so rapid: I found a thread here at PM with a link to a C to perl translator, but I missed the <ironic> tags and my hands called rapidly gcc -P -E file.c > file.pl using the compiler I have shipped within strawberry perl. Was not useful and the next line gcc -Larry -Wall file.c > file.pl revealed me it was an humoristic faq or well a iaq..

(**) Well i wanted to verify that the C code printed what I expected and I tried blindly using gcc to compile it (??) and using Inline::C but I had no success to not even install it on my strawberry perl.

If you need my C source you can, obviously perl -lne 'print if s/\s?#\s//' my_post.pl

Thanks

L*