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


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

by Aristotle (Chancellor)

on Jan 23, 2005 at 16:00 UTC ( [id://424400]=note: print w/replies, xml )

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in reply to $foo = "Foo::Bar"; $foo->new() works?

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

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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)
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