Re: my $x = <expr>; vs my $x; $x = <expr>;

No, they are equivalent as you have written them. I prefer to initialize at the same time I declare lexical, as in your first example. That helps enforce locality of the variable. I generally split them only when $x is set within a smaller scope, but $x is needed in a wider one. An example,

my $x;
{
    open my $fh, '<', '/path/to/foo.file' or die $!
    $x = <$fh>;
    close $fh;
}
[download]

$x is available outside the braces, but the apparatus needed to set it is hidden from the rest of the program.

After Compline,
Zaxo

Comment on Re: my $x = <expr>; vs my $x; $x = <expr>; Download Code

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Re^2: my $x = <expr>; vs my $x; $x = <expr>; by Aristotle (Chancellor) on Jun 05, 2004 at 21:32 UTC
I'd write that as `my $x = do { open my $fh, '<', '/path/to/foo.file' or die $! <$fh>; };` [download] Note that `$fh` is auto-closed due to falling out of scope. Update: the last line was not meant to be `$x = <$fh>` of course. Makeshifts last the longest.	[reply] [d/l]
Re^3: my $x = <expr>; vs my $x; $x = <expr>; by Roy Johnson (Monsignor) on Jun 06, 2004 at 03:41 UTC
You can't use $x in the same statement as you declare it. Your second use of $x is global. The last line in the block should be just `scalar <$fh>` The PerlMonk `tr///` Advocate	[reply] [d/l]
Re^4: my $x = <expr>; vs my $x; $x = <expr>; by Aristotle (Chancellor) on Jun 07, 2004 at 15:20 UTC
That was just a mistake. Thanks for catching it. scalar is not necessary here, btw, because `my $x =` already provides scalar context. Makeshifts last the longest.	[reply]