On Sun, Apr 20, 2008 at 7:38 AM, Rob Dixon <rob.dixon@[EMAIL PROTECTED]
> wrote:
>
> Chas. Owens wrote:
> > On Sat, Apr 19, 2008 at 1:57 PM, Rob Dixon <rob.dixon@[EMAIL PROTECTED]
> wrote:
> >> Chas. Owens wrote:
> >> > On Sat, Apr 19, 2008 at 1:03 PM, Rob Dixon <rob.dixon@[EMAIL PROTECTED]
>
wrote:
> >> > snip
> >> >> foreach my $i (1 .. 9999) {
> >> >> my $zfill = substr("0000$i", -4);
> >> >> my $longer = "dbt${zfill}dsfg";
> >> >> print $longer, "\n";
> >> >> }
> >> >>
> >> >> Rob
> >> > snip
> >> >
> >> > TIMTOWTDI fight!
> >> >
> >> > for my $i (1 .. 9999) {
> >> > my $longer = "0" x (4 - length $i) . $i;
> >> > print "$longer\n";
> >> > }
> >>
> >> You're on!
> >>
> >>
> >> foreach my $i (1 .. 9999) {
> >> (my $longer = $i + 10000) =~ s/1(....)$/dbt$1dsfg/;
> >> print $longer, "\n";
> >> }
> >>
> >
> > #!/usr/bin/perl
> >
> > use strict;
> > use warnings;
> >
> > for my $i (1, 10, 100, 1000) {
> > chop(my $longer = reverse(10000 + $i));
> > print scalar reverse($longer), "\n";
> > }
>
> You've changed the problem! If we're allowed to solve that one instead,
> then I submit
I have not:
10000 + 1 = 10001, reverse(10001) = 10001, chop(10001) = a return of 1
and $longer holding 1000, scalar reverse(1000) = 0001
10000 + 10 = 10010, reverse(10010) = 01001, chop(01001) = a return of
1 and $longer holding 0100, scalar reverse(0100) = 0010
10000 + 100 = 10100, reverse(10100) = 00101, chop(00101) = a return of
1 and $longer holding 0010, scalar reverse(0010) = 0100
10000 + 1000 = 11000, reverse(11000) = 00011, chop(00011) = a return
of 1 and $longer holding 0001, scalar reverse(0001) = 1000
>
> my $longer = reverse '0000' | reverse "$i";
>
> Rob
>
for my $i (1, 10, 100, 1000) {
print "000$i" =~ /(....)$/, "\n";
}
--
Chas. Owens
wonkden.net
The most im****tant skill a programmer can have is the ability to read.
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