如何在构造对象时不执行触发器

Question

我有这个代码:

package Foo;
use Moo;

has attr => ( is => "rw", trigger => 1 );

sub _trigger_attr 
    { print "trigger! value:". shift->attr ."\n" }


package main;
use Foo;

my $foo = Foo->new( attr => 1 );
$foo->attr( 2 );

它返回:

$ perl test.pl
trigger! value:1
trigger! value:2

这是Moo中触发器的默认记录行为.

如果通过构造函数设置属性,如何禁用触发器执行？

我当然可以这样做:

package Foo;
use Moo;

has attr        => ( is => "rw", trigger => 1 );
has useTriggers => ( is => "rw", default => 0 );

sub _trigger_attr 
{ 
    my $self = shift;
    print "trigger! value:". $self->attr ."\n" if $self->useTriggers 
}

package main;
use Foo;

my $foo = Foo->new( attr => 1 );
$foo->useTriggers( 1 );
$foo->attr( 2 );

得到:

$ perl testt.pl
trigger! value:2

所以它有效,但......感觉不对;).

Answer 1

我不太了解Moo,但是Moose你可以在构造函数之后实现自己的代码.如果你可以做这样的事情Moo会给你想要的效果.

sub BUILD {
    my $self = shift;

    # Sets "useTriggers" AFTER the object is already constructed.
    $self->useTriggers(1);
};

这将导致useTriggers在构造之后刚刚设置,因此触发器将在构造对象之后激活,而不是在构造之前激活.

所以你应该写:

my $foo->new(attr => 1);
$foo->attr(2);

并获得相同的输出.

Answer 2

package Foo;
use Moo;

has attr => ( accessor => '_attr' );

sub attr { 
    my $self = shift;
    my $rv = $self->_attr(@_);
    print "trigger! value: ", $rv if @_;
    return $rv;
}


package main;
use Foo;

my $foo = Foo->new( attr => 1 );
$foo->attr( 2 );