Perl dynamic function reference-CodePudding

Given a package name $package and a function name from within that package $function, I can create a reference to that function:

my $ref = eval( "\\&${package}::${function}" );

As a complete example:

#!/usr/bin/perl -w

use strict;

package foo;

sub func()
{
    print "foo::func\n";
}

package main;

my $package = "foo";
my $function = "func";

my $ref = eval( "\\&${package}::$function" );
$ref->();

I do not particularly like that eval() in there and wonder if the same result could be achieved without eval()?

CodePudding user response：

This can be made by accessing the function via the symbol table. Note that this needs no strict 'refs', which can be restricted to a block though:

my $package = "foo";
my $function = "func";
my $ref = do {
   no strict 'refs';
   *{"${package}::${function}"};
};
$ref->();

Example:

getref("Foo","bar")->(); # Foo::bar
getref("Bar","foo")->(); # Bar::foo

sub getref {
    my ($pkg,$sub) = @_;
    no strict 'refs';
    return *{"${pkg}::${sub}"};
}


package Foo;
sub bar { print "Foo::bar\n" }

package Bar;
sub foo { print "Bar::foo\n" }

CodePudding user response：

Using strings to refer to subroutines is seldom the right way. If you are doing it this way, you are most likely doing something wrong. You can use a code reference to store a reference to a function, in a scalar or hash. The hash allows you to correctly look up code references via string inputs.

use strict;
use warnings;

package foo;
sub func() {
    print "inside foo::func (@_)\n";
}

package main;

# use a direct reference, in a scalar variable
my $ref = \&foo::func;
$ref->(1,2,3);

# ...or in a hash
my $pack = "foo";
my $func = "func";
my %table = ( foo => { func => \&foo::func } );
$table{$pack}{$func}->(qw(a b c));

Output:

inside foo::func (1 2 3)
inside foo::func (a b c)