为什么perl会解析$x | $y | $z为$z | ($x | $y)?

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我正在尝试使用perl的重载功能来构建一个简单的解析树.
我不需要太多 – 实际上,我只需要一个左关联的运算符.
但是,perl解析$x op $y与更长时间的方式似乎存在不一致
像$x op $y op $z op ….

这就是我所拥有的:

package foo;

use overload '|' => \&pipe,"**" => \&pipe,">>" => \&pipe;

sub pipe { [ $_[0],$_[1] ] }

package main;

my $x = bless ["x"],"foo";
my $y = bless ["y"],"foo";
my $z = bless ["z"],"foo";
my $w = bless ["w"],"foo";

                               # how perl parses it:
my $p2 = $x | $y;              # Cons x y
my $p3 = $x | $y | $z;         # Cons z (Cons x y)
my $p4 = $x | $y | $z | $w;    # Cons w (Cons z (Cons x y))
my $p5 = $z | ($x | $y);       # same as p3???

my $s2 = $x ** $y;             # Cons x y
my $s3 = $x ** $y ** $z;       # Cons x (Cons y z)
my $s4 = $x ** $y ** $z ** $w; # Cons x (Cons y (Cons z w))

sub d { Dumper(\@_) }

say "p2 = ".d($p2);
say "p3 = ".d($p3);
say "p4 = ".d($p4);
say "p5 = ".d($p5);

say "s2 = ".d($s2);
say "s3 = ".d($s3);
say "s4 = ".d($s4);

输出类似于:

p2 = [bless( ['x'],'foo' ),bless( ['y'],'foo' )]
p3 = [bless( ['z'],[bless( ['x'],'foo' )]]
p4 = [bless( ['w'],[bless( ['z'],'foo' )]]]
p5 = [bless( ['z'],'foo' )]]

s2 = [bless( ['x'],'foo' )]
s3 = [bless( ['x'],[bless( ['y'],bless( ['z'],'foo' )]]
s4 = [bless( ['x'],bless( ['w'],'foo' )]]]

p2不应该反转x和y与其他情况一致吗?
请注意,p3和p5产生相同的输出 – 所以我如何分辨它们?

我没有看到右关联运算符**的相同问题.

有没有解决这个问题?

解决方法

use feature ":5.14";
use warnings FATAL => qw(all);
use strict;
use Data::Dump qw(dump pp);

sub foo() 
 {package foo;

  use overload '|' => \&p;

  sub p {bless [@{$_[0]},@{$_[1]}]}
 }

my $x = bless ["x"],"foo";

my $p = $x | $y | $z;

pp($p)

生产:

bless(["x","y","z"],"foo")
原文链接:https://www.f2er.com/Perl/241721.html

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