2. One way to do this would be with join:

print join " ",(reverse @list);
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# In real use you'd probably want this in a block
local $, = ' ';
print (reverse @list);
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1. Concatenating the list with " " causes it to evaluate in a scalar context. In a scalar context, reverse returns the reverse of the concatenation of all elements of the list. The space will be concatenated to the end of that.
   
   
2. The following code should do what you want to do:

   #!/usr/bin/perl - w
   
   use strict;
   
   my @list = qw (first    second    third    fourth    fifth    sixth   
   + seventh    eigth    ninth    tenth);
   print join ' ', reverse (@list);
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   And here's another way:

   #!/usr/bin/perl - w
   
   use strict;
   
   my @list = qw (first    second    third    fourth    fifth    sixth   
   + seventh    eigth    ninth    tenth);
   my @reverse = reverse (@list);
   print "@reverse";
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Others have answered both your questions well, but I'm going to be a bit more verbose on this one (I recently did a bunch of messing around with reverse, so it's fresh in my mind)

reverse works on a list, and returns either a list or a scalar, depending on context. Your original (Camel) code: print (reverse (@list)); It's in a list context (checking perldoc -f print we see that print operates on a list, so that's what context you're in). (Note also that the outer ()'s are syntactic sugar, they have NOTHING to do with putting this into list context, print reverse @list works the same. See a thread where merlyn gives the quick explanation here.)

Now your modified version: print (reverse (@list) . " ") concatenates the results of reverse with a string, meaning that the reverse is used in a scalar context.

With me so far? Okay, now there are times when you WANT this kind of behavior. For example, when learning how to commify a number, you can't just say print reverse "10000000"; Because you'll get just "10000000" as output. (I.e. the reverse order of a list of one element). To see what it really looks like reversed, try print scalar reverse "10000000"; Of course, if you assign the result to a scalar variable rather than simply printing the result, it will be taken in a scalar context without the scalar.

reverse is an easily underestimated function. Many regexes and the like are much easier to do on the reverse of a string rather than the string itself.

1. From the Perl docs:

   # perldoc -f reverse
   =item reverse LIST
   
   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.
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2. Just try this: print "@list";

Christian Lemburg
Brainbench MVP for Perl
http://www.brainbench.com/

(remembering the note about testing code before posting it...) That won't do the job...printing a space between every listitem will work the way kudra pointed out (using the join)

Jouke Visser, Perl 'Adept'

This code:

#!/usr/bin/perl -w

use strict;

my @list = qw (first    second    third    fourth
fifth    sixth   seventh    eigth    ninth    tenth);

print "@list";
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Does the following on my machine:

Mi 04.10.2000 15:41:50,56
D:\tmp
>perl try.pl
first second third fourth fifth sixth seventh eigth ninth tenth
Mi 04.10.2000 15:43:47,85
D:\tmp
>
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Christian Lemburg
Brainbench MVP for Perl
http://www.brainbench.com

Second, the title was meant to say, 'A simple, "why?"' but the HTML overlooked the quotes when rendered. Oops -- is there an HTMLmonks.org? Haha...

Lastly, two more questions!

1. Why then, don't I see a space on the print out with my second version? My guess is that it is there - albeit at the end and therefor not seen?
2. Why did my second version reverse the element, i.e. THIRD => DRIHT, and also reverse the list? Almost as if I used two reverse statements; first one in a scalar context, then a second on the list itself.

Again, thanks everyone for the help.

-- Just another Perl slacker
Steampunk.

In answer to number 1, the space is still there. You can easily confirm your theory by adding a line like:

print "-\n";
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after you print the list. The space should be quite obvious then.

As for your second question, remember: "concatenates all the elements of LIST together and then returns the reverse of that, character by character." So first your list becomes "firstsecondthird" (a short version of it), then that is reversed to "driht...".