Re: $foo = "Foo::Bar"; $foo->new() works?

$ perl -MO=Deparse,-x7 -e'Foo::Bar->new()'
'Foo::Bar'->new;
-e syntax OK
[download]

Pay attention to the quotes: as far as Perl is concerned, class methods are always invoked on a string containing the package name. Whether that's a literal string or one stored in a variable matters naught.

Makeshifts last the longest.

Comment on Re: $foo = "Foo::Bar"; $foo->new() works? Download Code

Replies are listed 'Best First'.
Re^2: $foo = "Foo::Bar"; $foo->new() works? by gaal (Parson) on Jan 23, 2005 at 16:32 UTC
It even turns out that there is no speed difference. (I expected some advantage to the inline string, because there'd be an extra indirection. But it's lost to the method call overhead.) `use Benchmark; package Bar; sub new {}; package main; my $obj; my $class = "Bar"; # "our" is slower, though. timethese -1, { bareword => sub { $x = Bar->new() }, var => sub { $x = $class->new() }, }; ######################################### Benchmark: running bareword, var for at least 1 CPU seconds... bareword: 1 wallclock secs ( 1.06 usr + 0.00 sys = 1.06 CPU) @ 59 +0160.38/s (n=625570) var: 1 wallclock secs ( 1.06 usr + 0.00 sys = 1.06 CPU) @ 59 +0160.38/s (n=625570)` [download]	[reply] [d/l]

Replies are listed 'Best First'.

Re^2: $foo = "Foo::Bar"; $foo->new() works?
by gaal (Parson) on Jan 23, 2005 at 16:32 UTC

use Benchmark;

package Bar;

sub new {};

package main;

my $obj;
my $class = "Bar";  # "our" *is* slower, though.

timethese -1, {
    bareword => sub { $x = Bar->new()    },
    var      => sub { $x = $class->new() },
};

#########################################

Benchmark: running bareword, var for at least 1 CPU seconds...
  bareword:  1 wallclock secs ( 1.06 usr +  0.00 sys =  1.06 CPU) @ 59
+0160.38/s (n=625570)
       var:  1 wallclock secs ( 1.06 usr +  0.00 sys =  1.06 CPU) @ 59
+0160.38/s (n=625570)
[download]

[reply]
[d/l]