Perl的：从子程序返回哈希 [英] Perl: Return hash from subroutine

问题描述

我一直在尝试了几个小时的例子，但我似乎无法把握怎么做我想做的事情。

I have been trying examples for hours but I can't seem to grasp how to do what I want to do.

我想从一个子程序返回哈希，和我猜想的参考是最好的选择。这里就是它变得有点棘手。我要引用如$哈希{$ X}哈希值。我仍然在Perl中的小白：/

I want to return a hash from a subroutine, and I figured a reference was the best option. Here's where it gets a bit tricky. I want to reference a hash like $hash{$x}. I am still a noob at perl :/

1。首先的问题，我用的例子似乎表明它是确定使用$ hashTable中{$登录}，我应该使用％hashTable中{$登录}或者它没有关系？下面是code：

1.First question, the examples I use seem to show it is ok to use $hashTable{$login}, should I be using %hashTable{$login} or does it not matter? Below is the code:

sub authUser  {
    $LocalPath = "/root/UserData";
    open(DATAFILE, "< $LocalPath");
    while( $linebuf = <DATAFILE> ) {
        chomp($linebuf);
        my @arr = split(/:/, $linebuf);
        my $login = $arr[1];        # arr[1] contains the user login names
        my $hashTable{ $login } = "$arr[0]";        #$arr[0] is account number
    }
    close DATAFILE;
    return \$hashTable{ $login };
}

然后我要测试这个数据，看看是否登录为present，这里是我的测试方法

I then want to test this data to see if a login is present, here is my test method

# test login Dr. Brule which is present in UserData
my $test = "Dr. Brule";
my $authHash = &authUser();

if ( $authHash{ $test } )  {
    print "Match for user $test";
}
else  {
    print "No Match for user $test";
}

2.Should我的$ authHash真的$ authHash {$}的东西，我对这个很困惑

2.Should my $authHash be really $authHash{ $something }, I am so confused on this

编辑：经过一些阅读技巧，仍在试图但没有骰子，任何帮助将大大AP preciated

After some reading tips, still attempting but no dice, any help would be greatly appreciated

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编辑2：任何人都可以修改我的code，这样我可以理解的答案更好？对不起，我似乎无法得到这在所有的工作，我一直在努力了几个小时，我真的想知道正确的方式做到这一点，我可以张贴我的各种尝试，但我觉得这将是一种浪费房产。

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Edit 2: Can anyone modify my code so that I can understand the answers better? I'm sorry I can't seem to get this to work at all, I have been trying for hours and I really want to know the correct way to do this, I can post my various tries but I feel that will be a waste of real estate.

推荐答案

首先，如评论所说mpapec，使用严格的;使用警告; 。这将赶上最常见的错误，其中最萎靡不振的你问这里的问题（通常提供有关你应该做的，而不是提示）。

First off, as mpapec mentioned in comments, use strict; use warnings;. That will catch most common mistakes, including flagging most of the problems you're asking about here (and usually providing hints about what you should do instead).

现在来回答问题1和问题2：

Now to answer questions 1 and 2:

％哈希是哈希作为一个整体。完整的数据结构。

%hash is the hash as a whole. The complete data structure.

$ {哈希键} 是哈希内的单个元素。

$hash{key} is a single element within the hash.

因此​​， \\％哈希是％哈希，参考也就是说，整个哈希，这似乎是你打算在这种情况下返回的内容。 \\ $ {哈希键} 是单个元素的引用。

Therefore, \%hash is a reference to %hash, i.e., the whole hash, which appears to be what you intend to return in this case. \$hash{key} is a reference to a single element.

如果它在你的第二个问题变得复杂的是，引用总是标量，不管它们是指什么。

Where it gets tricky in your second question is that references are always scalars, regardless of what they refer to.

$ hash_ref = \\％哈希

要出去，你必须参考哈希的元素，你需要取消对它的引用第一。这通常是用完成 - ＆GT; 运营商，像这样：

To get an element out of a hash that you have a reference to, you need to dereference it first. This is usually done with the -> operator, like so:

$ hash_ref-＆GT; {键}

请注意，您使用 - ＆GT; 当您从引用（ $ hash_ref-＆GT启动; {键} ），而不是当你从一个实际的哈希启动（ $ {哈希键} ）。

Note that you use -> when you start from a reference ($hash_ref->{key}), but not when you start from an actual hash ($hash{key}).

（如在问题2一个侧面说明，不要用＆放preFIX次通话; - 只需使用 AUTHUSER（），而不是＆放大器; AUTHUSER（）的＆放大器; 不再需要在Perl 5 +并有副作用，你通常不希望，所以你不应该在使用它的地方它没有必要的习惯得到的。）

(As a side note on question 2, don't prefix sub calls with & - just use authUser() instead of &authUser(). The & is no longer needed in Perl 5+ and has side-effects that you usually don't want, so you shouldn't get in the habit of using it where it's not needed.)

有关问题3，如果你只打算数组，你可能也只是循环过来检查一次，检查每一个元素：

For question 3, if you're only going to check once, you may as well just loop over the array and check each element:

my $valid;
for my $username (@list_of_users) {
  if ($login eq $username) {
    $valid = 1;
    last; # end the loop since we found what we're looking for
  }
}

if ($valid) {
  print "Found valid username $login\n";
} else {
  print "Invalid user! $login does not exist!\n";
}

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