perl - 这两个语句在 perl 中总是相同的吗？

Question

$obj->SUPER::promote();

$obj->SUPER->promote();

有谁知道他们是一样的吗？

Answer 1

号 -> 运算符意味着调用一个引用(在这种情况下，一个对象引用)，它会寻找 super 方法，而不是 super 基类。
这是显示它的代码:

#!/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' );

当您将最后一行更改为使用 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 是一种为多态定义的裸字语言)，因此它被视为不存在的方法，因此会出现错误。