On 12 Jun 2001 12:38:48 -0400, John Pimentel wrote:
> Forgive me if I'm missing something, but I've used the following:
> $varnew = "$varold" ;
>
> $varnew will return the same value as $varold did/does.
>
> Just like if you wanted to initialize a bunch of counters: $ctr1 = $ctr2 =
> $ctr3 = $ctr4 = 1; .
>
> I don't know that it is proper form, but it seems to work.
<snip />
This is double plus ungood. $varnew = "$varold"; says that the scalar
value in $varold make it into a string and then assign that string to
$varnew. First off the double quotes are unessessary for the simple
cases:
my $varold = 5;
my $varnew = "$varold"; #this is equal to $varnew = "5",
#how is this better than $varnew = $varold;?
Second it is downright terrible in the complex cases;
my $old = new Object;
my $new = "$old"; #$new is now equal to "Object=HASH(0x80f8a20)"
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