[Perl-beginners] Incrementing letters in for loop

In the above, when $_ gets to 'y' it is less than 'z' so it
auto increments to 'z', then when the next comparison is made, it is
equal to 'z' and so increments again.

I suspect auto increment is different for letters and numbers


Which goes to show that the thing after 'z' is 'aa' using the
auto increment




Owen

At which point it becomes "aa", which is less than or equal to "z". It
is not until it becomes "za" is it greater than "z".

Hi again.

Thanks for all the good info. I think I have a grasp on incrementing and
comparison, but now, what puzzles me, is the fact that when I use "le"
less than or equal, why does it actually increment the "z"? It is by
then passed the less than, and already at equal to "z". It should then
just exit the loop.
And just to add to my confusion, "ge" and "gt" does not give any output
at all. "u" is not greatier than "z", is it?

Note that in for loops, the 'increment' (i.e., $_++) occurs before the
condition ($_ le "z"). So when $_ is equal to "z", the condition
passes (i.e., $_ is less than or /equal/ to "z"). Then $_ is
"incremented" from "z" to "aa". This too is /less than/ or equal to
"z" using a string comparison so it continues on until it reaches
"za", which is not.

You may be thinking of lt instead of le. If you used lt then when $_
was "z" it would fail the << $_ lt "z" >> condition and exit out of
the loop right there. Because you're using le instead, you enable the
string ordering to "wrap" back around to "aa", which starts a long
process of "increments" before reaching something not less than or
equal to the string "z".

The reason gt and ge won't give any output at all is because $_ starts
at "u", which is not greater than /nor/ equal to "z" so the condition
fails right away.

Forgive me if I'm misread you.

Hi Magne,
"le" is the string equivalent of "<=" (less than or equal). You probably want
"lt" instead here. A program using it yields the following:

{{
u v w x y z

u v w x y

u v w x y z aa
}}

Regards,

Shlomi Fish

Hi.

So, what happened to the e in le, less than or "equal" to. In my
opinion, "z" == "yz", would be incorrect, even tough that is what
trigers the exit loop response. Try using "eq", and hopefully you
understand my frustration.
There has to be something I have misunderstood.

Brgds
Magne.

Hi Magne,
"eq" will exit immediately:

{{{
shlomi:~$ cat t2.pl
#!/usr/bin/perl

use warnings;
use strict;

for ($_="u"; $_ eq "z"; $_++){
print " $_ ";
}
print "\n";
shlomi:~$ perl t2.pl

shlomi:~$
}}}

The C-style for-loop continues as long as the condition is true and stops once
it is false. "le" evaluates to true for all the strings in "u".."z" and beyond
("aa", etc.) and so the loop doesn't terminate there.

Regards,

Shlomi Fish

Hi again.

Does this mean this is a bug?
My point is to get "z" at the end. Using "eq" or lt" wont work.
Brgds
Magne

On Tuesday 13 Apr 2010 11:17:12 Magne Sandøy wrote:
[SNIP]
You need to evaluate the condition at the end of the loop instead of at the
beginning. Try doing something like that (tested):

{{{
#!/usr/bin/perl

use warnings;
use strict;

sub process_letter
{
my $letter = shift;

print "$letter\n";
}

my $letter = "u";

LETTERS_LOOP:
while (1)
{
process_letter($letter);

if ($letter eq "z")
{
last LETTERS_LOOP;
}
}
continue
{
$letter++;
}

}}}

It's a bit ugly and I've erred a bit on using pseudo-Hungarian Notation ( see
http://en.wikipedia.org/wiki/Hungarian_notation ) - for the sake of the
demonstration, but it works.

Hope it helps.

Regards,

Shlomi Fish

This is the trickiness of magical autoincrement. See: perldoc perlop.

Once, you increment $_ = 'z' in your loop:

perl -le "$_='z';$_++; print" ---> prints 'aa';

And, the incrementing continues until, finally, you reach 'yz':

perl -le "$_='yz';$_++; print $_" ---> prints 'za'

Why did it stop at 'yz'... you ask... Because the string 'za'
would generate a value 'longer than z' itself .

Similiarly, if your range had been 'uu'..'zz', the increment
would have stopped at 'zyz' because incrementing would
then produce 'zza'.

Clear as mud? Did you say 'Hell, no'...? Go then and
meditate on autoincrement magic, grasshopper. When
enlightenment comes, please report back and explain it
to us too...

Actually, it is because string-comparison operators order strings
differently than auto-increment.

String ordering by auto-increment:
"a"
"b"
...
"y"
"z"
"aa"
"ab"
...
"yy"
"yz"
"za"
"zb"
...

String ordering by string-comparison operators:
"a"
"aa"
"aaa"
...
"y"
"ya"
"yaa"
...
"yz"
"yzz"
"yzzz"
...
"z"
"za"
"zaa"
...

Notice that "z" appears in very different place in each.

Yes, you're right. But I mentioned "enlightenment" because
the auto-increment algorithm itself is somewhat mysterious.
And, here's the doozy for me as I tried remembering:

If the final value specified is not in the sequence that the
magical increment would produce, the sequence continues
until the next value is longer than the final value specified.
^^^^^^

So, in the OP's 'u'..'z' example, the expansion stops at 'yz'
because another increment would be 'za' which is 'longer'
than the final value specified'; whereas, 'yz' isn't:

'z' 122
'aa' 97 97
...
'yz' 121 122 ---> 'shorter' than 122
'za' 122 97 ----> 'longer' than 122

In other words, the sequencing continues until there's
carry past the final 'z'. I think that's the 'long and short
of it... maybe I've auto-enlightened myself..

Actually, no. "z" is in the sequence, so it stops there.

Try:

for ( 'u' .. ' ' ){
print "$_ ";
}
print "\n";

No, the OP was talking about the *second* 'for' loop of course --
not
the first loop which obviously works.


OP> In the following code, the second for loop actually counts way
OP> passed what I expected, and actually stops at "yz" and not "z"
OP> as expected.


The 'u'..'z' mis-identifies the loop but the context is clear as to
which
loop was being discussed.