$obj->SUPER::promote(); $obj->SUPER->promote();
谁知道他们是否一样？
解决方法
不. – > operator表示调用引用(在本例中为对象引用),它将查找SUPER方法,而不是超级基类.
#!/usr/bin/perl -w
package MyOBJ;
use strict;
use warnings;
use Data::Dumper;
sub new {
my ($class) = @_;
my $self = {};
bless $self,$class;
return $self;
}
sub promote {
my ($self) = @_;
print Dumper($self);
}
1;
package MyOBJ::Sub;
use strict;
use warnings;
use base 'MyOBJ';
1;
use strict;
use warnings;
my $obj = MyOBJ::Sub->new();
$obj->SUPER::promote();
运行它,你会得到：
$VAR1 = bless( {},'MyOBJ::Sub' );
当您更改要使用的最后一行时 – >而不是::你得到：
Can't locate object method "SUPER" via package "MyOBJ" at test.pl line 45.
来自“perldoc perlop”手册
The Arrow Operator
If the right side is either a “[…]”,“{…}”,or a “(…)” subscript,then the left side must be either a hard or symbolic reference to an array,a hash,or a subroutine respectively.
Otherwise,the right side is a method name or a simple scalar variable containing either the method name or a subroutine reference,and the left side must be either an object (a blessed reference) or a class name (that is,a package name)
由于左侧既不是对象引用也不是类名(SUPER是多态性的语言定义的裸字),因此它被视为一种方法,它不存在,因此也就是错误.