Perl: file test operator without condition

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I have this simple code from Perl Cookbook which prints all directories and files recursively:

use File::Find;

@ARGV = qw(.) unless @ARGV;
find sub { print $File::Find::name, -d && '/', "\n" }, @ARGV;

I do not understand the grammar of print $File::Find::name, -d. How is this to interpret? If -d tests if $File::Find::nameis a directory so -dis a parameter of the function print? Or does Perl explicitly interpret a standalone -d as if -d?

perl testing operators
1 Answers

No, the -d is a stand alone statement, it tests $_. So it is in essence identical to

-d $_ && '/'

Which says "if file is a directory, return a slash character (to print)". The sub code block is used by the find function from File::Find, where $_ contains the file name of the current file.

The commas , separate a list of statements that return strings for the print statement:

print $File::Find::name,   # print the files name
-d && '/',                 # if it is a dir, print /
"\n"                       # print a newline

In the documentation for -d (contained in perldoc for -X where all the file tests are listed) states:

If the argument is omitted, tests $_ ...

This applies to all file tests under -X.

The reason && can be used this way is that it has a higher precedence than the comma operator ,. This is documented in perldoc perlop

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