Shlomi Fish wrote:
Hi Magne,
On Tuesday 13 Apr 2010 10:37:15 Magne Sandøy wrote:
Shlomi Fish wrote:
Hi Magne,
On Tuesday 13 Apr 2010 06:35:51 Magne Sandøy wrote:
Hi.
I'm new to perl, and I stumbled across a strange behavior in my for
loop. In the following code, the second for loop actually counts way
passed what I expected, and actually stops at "yz" and not "z" as
expected. As shown by the third for loop, incrementing the letters,
seems to give me the desired output in each loop.
What is going on here?
#!/usr/bin/perl
use warnings;
use strict;
my $letter = "u";
for ("u".."z"){
print " $_ ";
}
print "\n\n";
for ($_="u"; $_ le "z"; $_++){
print " $_ ";
}
"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
print "\n\n";
for(1..7){
print " $letter ";
$letter++;
}
Hi.
So, what happened to the e in le, less than or "equal" to. In my
opinion, "z" == "yz", would be incorrect, even tough that is what
trigers the exit loop response. Try using "eq", and hopefully you
understand my frustration.
"eq" will exit immediately:
{{{
shlomi:~$ 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
There has to be something I have misunderstood.
Brgds
Magne.
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
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