Why perl cannot detect zero value?

[Hon Seng Phuah]

When I run on the below statements, I do not understand why it zero
and here. The $a contains value zero. Why can't perl skip the print
statement?

#!/usr/intel/bin/perl
$count = "17.0";
if ($count =~ /\./)
{
($before_decimal, $a) = split(/\./, $count);
if (!$before_decimal || $before_decimal =~ /\D/)
{
exit;
}

if (!$a)
{
print "$a";
}
if (!$aft_decimal || (!$aft_decimal =~ /0$/ && $aft_decimal =~
/\D/))
{
print "Here";
exit;
}
}

 

[Gunnar Hjalmarsson]

When I run on the below statements, I do not understand why it zero
and here.

Do you mean why it prints "0Here"? That's because you coded it to do so.

Maybe you should try to explain in English what it is you are *trying*
to do. I'm not able to figure it out from the code.

 

[Joe Smith]

if (!$b || $b =~ /\D/)

That's the same as
if (
(not defined $b) || ($b eq "") || ($b eq "0") || ($b == 0) || ($b =~ /\D/)
)

if (!$a)
{
print "$a";
}

That's the same as
print "$a" if (not defined $a or $a eq "" or $a eq "0" or $a == 0);
so it will print "0" with the example given.

if (!$aft_decimal || (!$aft_decimal =~ /0$/ && $aft_decimal =~
/\D/)) { print "Here"; }

Since $aft_decimal is undef, !$aft_decimal is true, therefore the
print statement will be executed.

Changing that to
if (!$a || (anything at all here)) {print "Here"}
will print "Here" since !$a is true when $a == "0".

So, perl is doing exactly what you've told it to do when it prints "0Here".
It is detecting zero in both cases and printing both strings.
-Joe

 

[knobo]

When I run on the below statements, I do not understand why it zero
and here. The $a contains value zero. Why can't perl skip the print
statement?

if (0) => never
if (!0) => allways

#!/usr/intel/bin/perl
$count = "17.0";
if ($count =~ /\./)
{
($before_decimal, $a) = split(/\./, $count);

$a == 0

if (!$before_decimal || $before_decimal =~ /\D/)
{
exit;
}

if (!$a)

Remember if (!0) => allways

 

[Hon Seng Phuah]

Do you mean why it prints "0Here"? That's because you coded it to do so.

Maybe you should try to explain in English what it is you are *trying*
to do. I'm not able to figure it out from the code.

Oops. I missed some words on my sentense.

Refer to the code below,

#!/usr/intel/bin/perl
$count = "17.0";
if ($count =~ /\./)
{
($before_decimal, $a) = split(/\./, $count);
if (!$before_decimal || $before_decimal =~ /\D/)
{
exit;
}

if (!$a)
{
print "$a";
}
}

I do not understand why the above code print 0. The $a contains value,
"0". I thought perl should skip the print statement since the $a has
been initialized.

 

[Gunnar Hjalmarsson]

Refer to the code below,

if (!$a)
{
print "$a";
}
}

I do not understand why the above code print 0. The $a contains
value, "0". I thought perl should skip the print statement since
the $a has been initialized.

Okay. As others have explained, both 0 (the number) and '0' (the
string) return a false value, just like the nullstring or an undefined
value. Accordingly, the statements

print "False\n" unless 0;
print "False\n" unless '0';
print "False\n" unless '';
print "False\n" unless undef;

all print "False\n".

Since $a in your example contains '0', $a is false, and the opposite
!$a is true.

Maybe this is what you want to do:

unless (!$a or $a eq '0')
{
print "$a";
}

HTH