blocks|key|1689612|text|你根本不需要对它进行转换：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1689613|%25+perl+-e+'print+"5.45"+%2B+0.1;'
5.55|code-block|syntax|javascript|1689614|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

You don't need to convert it at all:

<pre><code>% perl -e 'print "5.45" + 0.1;'
5.55
</code></pre>

blocks|key|1475028|text|这是一个简单的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1475029|示例1|1475030|my+$var1+=+"123abc";
print+$var1+%2B+0;|code-block|syntax|javascript|1475031|结果|1475032|123|1475033|示例2|1475034|my+$var2+=+"abc123";
print+$var2+%2B+0;|1475035|1475036|0|1475037|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|W|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|X|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Y|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|Z|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|10|8|@]|9|@]|A|$]]|$1|O|3|P|5|F|7|11|8|@]|9|@]|A|$G|H]]|$1|Q|3|J|5|6|7|12|8|@]|9|@]|A|$]]|$1|R|3|S|5|F|7|13|8|@]|9|@]|A|$G|H]]|$1|T|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|U|$]]

This is a simple solution:

Example 1

<pre><code>my $var1 = "123abc";
print $var1 + 0;
</code></pre>

Result

<pre><code>123
</code></pre>

Example 2

<pre><code>my $var2 = "abc123";
print $var2 + 0;
</code></pre>

Result

<pre><code>0
</code></pre>

blocks|key|1689792|text|Perl是一种基于上下文的语言。它不会根据你给它的数据来做它的工作。相反，它会根据您使用的运算符和使用这些运算符的上下文来确定如何处理数据。如果你做数字排序的事情，你会得到数字：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1689793|#+numeric+addition+with+strings:
my+$sum+=+'5.45'+%2B+'0.01';+#+5.46|code-block|syntax|javascript|1689794|如果你做字符串排序的事情，你会得到字符串：|1689795|#+string+replication+with+numbers:
my+$string+=+(+45/2+)+x+4;+#+"22.522.522.522.5"|1689796|Perl主要知道要做什么，而且大部分都是正确的。另一种说法是Perl更关心动词而不是名词。|1689797|你是否正在尝试做一些事情，但它不起作用？|1689798|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Perl is a context-based language. It doesn't do its work according to the data you give it. Instead, it figures out how to treat the data based on the operators you use and the context in which you use them. If you do numbers sorts of things, you get numbers:

<pre><code># numeric addition with strings:
my $sum = '5.45' + '0.01'; # 5.46
</code></pre>

If you do strings sorts of things, you get strings:

<pre><code># string replication with numbers:
my $string = ( 45/2 ) x 4; # "22.522.522.522.5"
</code></pre>

Perl mostly figures out what to do and it's mostly right. Another way of saying the same thing is that Perl cares more about the verbs than it does the nouns.

Are you trying to do something and it isn't working?

blocks|key|1689875|text|谷歌把我带到这里，同时搜索phill提出的相同问题(排序浮点数)，所以我认为尽管这个帖子有点老了，但还是值得把答案贴出来。我是perl的新手，还在为它着迷，但brian+d+foy的说法是："Perl更关心动词而不是名词。“上面的话确实一针见血。在应用排序之前，您不需要将字符串转换为浮点数。您需要告诉排序将值排序为数字，而不是字符串。即|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1689876|my+@foo+=+('1.2',+'3.4',+'2.1',+'4.6');
my+@foo_sort+=+sort+{$a+<=>+$b}+@foo;|code-block|syntax|javascript|1689877|有关排序的更多详细信息，请参见http://perldoc.perl.org/functions/sort.html|offset|length|1689878|entityMap|0|LINK|mutability|MUTABLE|url|http://perldoc.perl.org/functions/sort.html^0|0|0|F|17|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

Google lead me here while searching on the same question phill asked (sorting floats) so I figured it would be worth posting the answer despite the thread being kind of old. I'm new to perl and am still getting my head wrapped around it but brian d foy's statement "Perl cares more about the verbs than it does the nouns." above really hits the nail on the head. You don't need to convert the strings to floats before applying the sort. You need to tell the sort to sort the values as numbers and not strings.
i.e.

<pre><code>my @foo = ('1.2', '3.4', '2.1', '4.6');
my @foo_sort = sort {$a &lt;=&gt; $b} @foo;
</code></pre>

See <a href="http://perldoc.perl.org/functions/sort.html" rel="noreferrer">http://perldoc.perl.org/functions/sort.html</a> for more details on sort

blocks|key|1475003|text|据我所知，int()并不是一个用于指定数据类型的“强制转换”函数，它只是(Ab)用于将上下文定义为算术类型。我过去(Ab)使用(0%2B$val)来确保$val被当作一个数字来处理。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1475004|entityMap|0|LINK|mutability|MUTABLE|url|http://perldoc.perl.org/functions/int.html^0|5|5|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

As I understand it <a href="http://perldoc.perl.org/functions/int.html" rel="noreferrer">int()</a> is not intended as a 'cast' function for designating data type it's simply being (ab)used here to define the context as an arithmetic one. I've (ab)used (0+$val) in the past to ensure that $val is treated as a number.

blocks|key|1474659|text|Perl实际上只有三种类型:标量、数组和散列。即使是这种区别也是有争议的。;)每个变量的处理方式取决于您如何处理它：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1474660|%25+perl+-e+"print+5.4+.+3.4;"
5.43.4


%25+perl+-e+"print+'5.4'+%2B+'3.4';"
8.8|code-block|syntax|javascript|1474661|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Perl really only has three types: scalars, arrays, and hashes. And even that distinction is arguable. ;) The way each variable is treated depends on what you do with it:

<pre><code>% perl -e "print 5.4 . 3.4;"
5.43.4


% perl -e "print '5.4' + '3.4';"
8.8
</code></pre>

blocks|key|1475105|text|相比之下，如果标量是一个字符串的一个数字，就会有所不同。而且这并不总是可以决定的。我可以报告一个例子，perl以“科学”表示法检索了一个浮点数，并在下面的几行代码中进行了比较：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1475106|use+strict;
....
next+unless+$line+=~+/and+your+result+is:\s*(.*)/;
my+$val+=+$1;
if+($val+<+0.001)+{
+++print+"this+is+small\n";
}|code-block|syntax|javascript|1475107|在这里，$val不被解释为数字，例如从$line检索的"2e-77"。添加0(对于优秀的ole+C程序员来说是0.0+)会有所帮助。|offset|length|style|CODE|1475108|entityMap^0|0|0|4|4|J|5|R|7|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]|$I|T|J|U|K|L]|$I|V|J|W|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|N|$]]

In comparisons it makes a difference if a scalar is a number of a string. And it is not always decidable. I can report a case where perl retrieved a float in "scientific" notation and used that same a few lines below in a comparison:

<pre><code>use strict;
....
next unless $line =~ /and your result is:\s*(.*)/;
my $val = $1;
if ($val &lt; 0.001) {
 print "this is small\n";
}
</code></pre>

And here <code>$val</code> was not interpreted as numeric for e.g. <code>"2e-77"</code> retrieved from <code>$line</code>. Adding 0 (or 0.0 for good ole C programmers) helped.

blocks|key|1475171|text|Perl是弱类型的，并且基于上下文。许多标量可以被视为字符串和数字，这取决于您使用的运算符。$a+=+7*6;+$b+=+7x6;+print+"$a+$b\n";|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1475172|你得到了42+777777。|1475173|然而，这里有一个细微的区别。当您将数字数据从文本文件读取到数据结构中，然后使用Data::Dumper查看它时，您会注意到您的数字被引用了。Perl在内部将它们视为字符串。|1475174|阅读：$my_hash{$1}+=+$2+if+/(.%2B)=(.%2B)\n/;。|1475175|转储：'foo'+=>+'42'|1475176|如果您想要转储中未加引号的数字：|1475177|阅读：$my_hash{$1}+=+$2%2B0+if+/(.%2B)=(.%2B)\n/;。|1475178|转储：'foo'+=>+42|1475179|在$2%2B0注意到您已将$2视为一个数字之后，因为您使用了一个数字运算符。|1475180|在尝试用Data::Dumper比较两个散列时，我注意到了这一点。|1475181|entityMap^0|1A|10|0|4|9|0|13|C|0|3|Z|0|3|D|0|0|3|11|0|3|B|0|1|4|0|4|C|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|N|3|O|5|6|7|1E|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|1F|8|@$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|R|3|S|5|6|7|1I|8|@$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|T|3|U|5|6|7|1L|8|@$9|1M|A|1N|B|C]]|D|@]|E|$]]|$1|V|3|W|5|6|7|1O|8|@$9|1P|A|1Q|B|C]]|D|@]|E|$]]|$1|X|3|-4|5|6|7|1R|8|@]|D|@]|E|$]]]|Y|$]]

Perl is weakly typed and context based. Many scalars can be treated both as strings and numbers, depending on the operators you use.
<code>$a = 7*6; $b = 7x6; print "$a $b\n";</code>
 You get <code>42 777777</code>.

There is a subtle difference, however. When you read numeric data from a text file into a data structure, and then view it with <code>Data::Dumper</code>, you'll notice that your numbers are quoted. Perl treats them internally as strings.
 Read:<code>$my_hash{$1} = $2 if /(.+)=(.+)\n/;</code>.
 Dump:<code>'foo' => '42'</code>

If you want unquoted numbers in the dump:
 Read:<code>$my_hash{$1} = $2+0 if /(.+)=(.+)\n/;</code>.
 Dump:<code>'foo' => 42</code>

After <code>$2+0</code> Perl notices that you've treated $2 as a number, because you used a numeric operator. 

I noticed this whilst trying to compare two hashes with <code>Data::Dumper</code>.

I have a string which holds a decimal value in it and I need to convert that string into a floating point variable. So an example of the string I have is "5.45" and I want a floating point equivalent so I can add .1 to it. I have searched around the internet, but I only see how to convert a string to an integer.

How can I convert a string to a number in Perl?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个字符串，其中包含一个十进制值，我需要将该字符串转换为浮点变量。因此，我有一个字符串的例子是"5.45“，我想要一个浮点等效值，这样我就可以向它添加.1。我在互联网上搜索过，但我只看到如何将字符串转换为整数。

问如何在Perl中将字符串转换为数字？
EN

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问如何在Perl中将字符串转换为数字？
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