如何在 Perl 中将字符串转换为数字？

我有一个字符串，其中包含一个十进制值，我需要转换成一个浮点变量的字符串。所以我有一个字符串的例子是“5.45”，我想要一个浮点数，这样我就可以把.1加到它上面。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。

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最佳答案

You don't need to convert it at all:

% perl -e 'print "5.45" + 0.1;'
5.55
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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:

% perl -e "print 5.4 . 3.4;"
5.43.4




% perl -e "print '5.4' + '3.4';"
8.8
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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:

# numeric addition with strings:
my $sum = '5.45' + '0.01'; # 5.46

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

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

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?

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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.

my @foo = ('1.2', '3.4', '2.1', '4.6');
my @foo_sort = sort {$a <=> $b} @foo;

See http://perldoc.perl.org/functions/sort.html for more details on sort

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As I understand it int() 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.

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This is a simple solution:

Example 1

my $var1 = "123abc";
print $var1 + 0;

Result

123

Example 2

my $var2 = "abc123";
print $var2 + 0;

Result

0
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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:

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

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

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$var += 0

probably what you want. Be warned however, if $var is string could not be converted to numeric, you'll get the error, and $var will be reset to 0:

my $var = 'abc123';
print "var = $var\n";
$var += 0;
print "var = $var\n";

logs

var = abc123
Argument "abc123" isn't numeric in addition (+) at test.pl line 7.
var = 0
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Perl is weakly typed and context based. Many scalars can be treated both as strings and numbers, depending on the operators you use. $a = 7*6; $b = 7x6; print "$a $b\n";
You get 42 777777.

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

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

After $2+0 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 Data::Dumper.


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