New user, confusion over print and lists

[jktstance]

Hi all, I'm learning Perl (my only other programming experience is Tcl) and I'm having some trouble. I'm confused over lists, so I looked up some help and found this thread at Stack Overflow. I looked up the perldoc of reverse (reverse) and I on the page, there is some code, which I've slightly modified:

#!/usr/bin/perl
use strict;
use warnings;

print join(", ", reverse "world", "One"), "\n"; # One, world
print scalar reverse ("dlrow ,", "owT"),"\n";    # Two, world
 
$_ = "dlrow ,eerhT";
print reverse;                         # No output, list context
 
$_ = "dlrow ,ruoF";
print scalar reverse;                  # Four, world

I get the following output:

One, world
Two, world
Four, world

My first question is why isn't the third line printed? Reverse supposedly returns a list and the perldoc of print says that a list is one of it's arguments. Why doesn't this work? My second question is that if I explicitly state the $_ after reverse in the third print statement, then it prints out the string, but it is NOT reversed:

#!/usr/bin/perl
use strict;
use warnings;

print join(", ", reverse "world", "One"), "\n"; # One, world
print scalar reverse ("dlrow ,", "owT"),"\n";    # Two, world
 
$_ = "dlrow ,eerhT";
print reverse $_;                         # No output, list context
 
$_ = "dlrow ,ruoF";
print scalar reverse $_;                  # Four, world

I get the following output:

One, world
Two, world
dlrow, eerhTFour, world

What is going on here? Why does listing $_ explicitly as an argument to reverse cause print to work, but reverse() is not performed?

[tobyink]

From the documentation for reverse:

Used without arguments in scalar context, reverse() reverses $_.

That is, $_ is only "special" to reverse when:

• reverse is called with no arguments; and
• reverse is called in scalar context.

In the case of print reverse;, it is not in scalar context, so $_ is not special. You're calling reverse with the empty list as an argument. It reverses that list and return the result (which is still the empty list).

[choroba]

When you specify $_ as an argument for reverse in list context, it is taken as the single member of the list to reverse. Equivalently, no arguments represent an empty list.

[jktstance]

OK, I understand why the string is being printed back as is (1-length list). I'm still confused on why the behavior is different when I explicitly state $_ and when I do not.

When reverse has no arguments, doesn't it just take $_? Is writing reverse; the same as reverse $_;? I'm imagining this to be the behavior, because print; and print $_; seem identical when I test them.

EDIT: From the doc, I seems that reverse will ONLY work on $_ if it is in scalar context, which print is not. So I give it no arguments in list context, does reverse ignore $_, which the doc seems to imply it does. If so, is it just returning an empty list which is then passed to print?

[LanX]

This comment is wrong

$_ = "dlrow ,eerhT";
print reverse $_;                         # No output, list context

you get the one-element list ($_) reversed, which is the same list. Harder to spot since you forgot the newline this time.

reverse does exactly like documented and explained by choroba

• In list context, returns a list value consisting of the elements of LIST in the opposite order.

• In scalar context, concatenates the elements of LIST and returns a string value with all characters in the opposite order.

I think you are misunderstanding how context works, it's the LHS which decides how the RHS of reverse is interpreted and print always induces list context.