替换中的空白替换会将匹配的字符串更改为空

Question

我正在尝试以下一段代码。根据我的理解，基本上这段代码不应该改变 $data 。我错过了什么吗？我正在使用 5.22.3（草莓 perl）。

use strict;
use warnings;
use 5.010;
my $data = 'we are here';
$data =~ s///g;
print "DATA1: $data\n";
$data =~ s{(we)}{
    my $x1 = ;
    $x1 =~ s///g;
    print "x1: ^^$x1^^\n";
    "$x1"
}e;
print "DATA2: $data\n";

O/P-

DATA1: we are here
x1: ^^^^
DATA2:  are here

Answer 1

除非被 split 使用，否则空模式告诉 match/substitute 运算符使用最后一个模式来成功匹配。

例如，

$ perl -e'$_ = "abba"; s//c/g if /a/ || /b/; CORE::say;'
cbbc

这意味着

$x1 =~ s///g;

等同于

$x1 =~ s/(we)//g;

要绕过该异常，您可以使用

$x1 =~ s/(?:)//g;

说你的真实情况使用

$re = ...;
s/$re//g;

你可以使用

$re = ...;
s/(?:$re)//g;

或

$re = ...;
$re = qr/$re/;
s/$re//g;

引用 perlop,

The empty pattern //

If the PATTERN evaluates to the empty string, the last successfully matched regular expression is used instead. In this case, only the g and c flags on the empty pattern are honored; the other flags are taken from the original pattern. If no match has previously succeeded, this will (silently) act instead as a genuine empty pattern (which will always match).

替换中的空白替换会将匹配的字符串更改为空

blank substitute inside a substitution changes matched string to empty

regex

perl

substitution