默认情况下，穆斯非常宽容。您可以拥有一个名为Cucumber的类，并将未声明的属性（如wheels）传递给构造函数。 Moose默认不会抱怨。但我可能更喜欢穆斯而不是接受未申报的属性die。我怎样才能做到这一点？我似乎记得读过它是可能的，但在文档中找不到它所说的地方。
package Gurke;
use Moose;
has color => is => 'rw', default => 'green';
no Moose;
__PACKAGE__->meta->make_immutable;
package main; # small test for the above package
use strict;
use warnings;
use Test::More;
use Test::Exception;
my $gu = Gurke->new( color => 'yellow' );
ok $gu->color, 'green';
if ( 1 ) {
my $g2 = Gurke->new( wheels => 55 );
ok ! exists $g2->{wheels}, 'Gurke has not accepted wheels :-)';
# But the caller might not be aware of such obstinate behaviour.
diag explain $g2;
}
else {
# This might be preferable:
dies_ok { Gurke->new( wheels => 55 ) } q(Gurken can't have wheels.);
}
done_testing;
好的，这是说明解决方案的更新测试：
package Gurke;
use Moose;
# By default, the constructor is liberal.
has color => is => 'rw', default => 'green';
no Moose;
__PACKAGE__->meta->make_immutable;
package Tomate;
use Moose;
# Have the Moose constructor die on being passed undeclared attributes:
use MooseX::StrictConstructor;
has color => is => 'rw', default => 'red';
no Moose;
__PACKAGE__->meta->make_immutable;
package main; # small test for the above packages
use strict;
use warnings;
use Test::More;
use Test::Exception;
my $gu = Gurke->new( color => 'yellow' );
ok $gu->color, 'green';
my $g2 = Gurke->new( wheels => 55 );
ok ! exists $g2->{wheels}, 'Gurke has not accepted wheels :-)';
diag 'But the caller might not be aware of such obstinate behaviour.';
diag explain $g2;
diag q(Now let's see the strict constructor in action.);
my $to = Tomate->new( color => 'blue' );
diag explain $to;
dies_ok { Tomate->new( wheels => 55 ) } q(Tomaten can't have wheels.);
done_testing;