Perl 从字符串中捕获的数字总是 1

Question

开始学习的路上perl...

我遇到了这种有趣的行为，似乎无法理解它。

my $username = "bandit0";
my $command = "";

sub next_command {
    my ($user) = @_;
    my $num = $user =~ /(\d+)/;
    print "$num\n";
    if ($num == 0){
        return "cat readme";
    }
    elsif ($num == 1){
        return "ls";
    }
}

sub next_level {
    my ($user) = @_;
    my $num = $user =~ /(\d+)/;
    $user =~ s/\d//g;
    $user = $user . $num++;
    return $user;
}

$command = next_command($username);
print $command . "\n";
$username = next_level($username);
$command = next_command($username);
print $command . "\n";

在运行这个小的 perl 程序之后，我希望是这样的：

0
cat readme
1
ls

但是我得到了这样的东西...

1
ls
1
ls

我不明白为什么，一开始字符串 bandit0 $num 会是 1 而不是 0

Answer 1

您可以尝试一下，看看是否有效。如果有帮助，我会解释一切。我现在无法自己测试。

use strict;   # always use strict
use warnings; # always use warnings

my $username = "bandit0"; #try bandit1 to ensure it prints 1 and ls instead
my $command = "";

sub next_command {
    my ($user) = @_;
    my ($num) = $user =~ /(\d+)/;
    print "$num\n";
    if ($num == 0){
       return "cat readme";
    }

    elsif ($num == 1){
        return "ls";
    }
}

sub next_level {
    my ($user) = @_;
    my ($num) = $user =~ /(\d+)/;
    $user =~ s/\d//g;
    $user = $user . ++$num;
     return $user;
}

$command = next_command($username);
print $command . "\n";
$username = next_level($username);
$command = next_command($username);
print $command . "\n";

Answer 2

标量上下文中的匹配运算符的计算结果为布尔值，指示匹配是否成功。

my $success = $user =~ /(\d+)/;

列表上下文中的匹配运算符 returns 成功时捕获的字符串（或 1 如果没有捕获）和错误时的空列表。

my ($num) = $user =~ /(\d+)/;

你用的是前者，但你想要后者。这为您提供了以下内容（在其他一些小修复之后）：

sub next_level {
    my ($user) = @_;
    my ($num) = $user =~ /(\d+)\z/;
    $user =~ s/\d+\z//g;
    $user .= ++$num;
    return $user;
}

但这种方法复杂且效率低下。更简单的解决方案：

sub next_level {
    my ($user) = @_;
    $user =~ s/(\d+)\z/  + 1 /e;
    return $user;
}

Perl 从字符串中捕获的数字总是 1

Perl captured digits from string are always 1

perl

digits