问题
Is there a different way to concatenate variables in Perl?
I accidentally wrote the following line of code:
print "$linenumber is: \n" . $linenumber;
And that resulted in output like:
22 is:
22
I was expecting:
$linenumber is:
22
So then I wondered. It must be interpreting the $linenumber in the double quotes as a reference to the variable (how cool!).
What are the caveats to using this method and how does this work?
回答1:
Variable interpolation occurs when you use double quotes. So, special characters need to be escaped. In this case, you need to escape the $:
print "\$linenumber is: \n" . $linenumber;
It can be rewritten as:
print "\$linenumber is: \n$linenumber";
To avoid string interpolation, use single quotes:
print '$linenumber is: ' . "\n$linenumber"; # No need to escape `$`
回答2:
I like .= operator method:
#!/usr/bin/perl
use strict;
use warnings;
my $text .= "... contents ..."; # Append contents to the end of variable $text.
$text .= $text; # Append variable $text contents to variable $text contents.
print $text; # Prints "... contents ...... contents ..."
回答3:
In Perl any string that is built with double quotes will be interpolated, so any variable will be replaced by its value. Like many other languages if you need to print a $, you will have to escape it.
print "\$linenumber is:\n$linenumber";
OR
print "\$linenumber is:\n" . $linenumber;
OR
printf "\$linenumber is:\n%s", $linenumber;
Scalar Interpolation
回答4:
If you change your code from
print "$linenumber is: \n" . $linenumber;
to
print '$linenumber is:' . "\n" . $linenumber;
or
print '$linenumber is:' . "\n$linenumber";
it will print
$linenumber is:
22
What I find useful when wanting to print a variable name is to use single quotes so that the variables within will not be translated into their value making the code easier to read.
回答5:
When formulating this response, I found this webpage which explains the following information:
###################################################
#Note that when you have double quoted strings, you don't always need to concatenate. Observe this sample:
#!/usr/bin/perl
$a='Big ';
$b='Macs';
print 'I like to eat ' . $a . $b;
#This prints out:
# I like to eat Big Macs
###################################################
#If we had used double quotes, we could have accomplished the same thing like this:
#!/usr/bin/perl
$a='Big ';
$b='Macs';
print "I like to eat $a $b";
#Printing this:
# I like to eat Big Macs
#without having to use the concatenating operator (.).
###################################################
#Remember that single quotes do not interpret, so had you tried that method with single quotes, like this:
#!/usr/bin/perl
$a='Big ';
$b='Macs';
print 'I like to eat $a $b';
#Your result would have been:
# I like to eat $a $b
#Which don't taste anywhere near as good.
I thought this would be helpful to the community so I'm asking this and answering my own question. Other helpful answers are more than welcome!
回答6:
You can backslash the $ to print it literally:
print "\$linenumber is: \n" . $linenumber;
That prints what you were expecting. You can also use single quotes if you don't want Perl to interpolate variable names, but then the "\n" will be interpolated literally.
来源：https://stackoverflow.com/questions/11782435/how-to-concatenate-variables-in-perl