the setenv example should not have an = for csh.

Instead of:

% setenv CHARACTER=Fred

It should be:

% setenv CHARACTER Fred

Note from the Author or Editor:
Change as noted

"Scalar variables in Perl are always referenced with the leading $&."
The "&" must be removed. It's correct in the 'Safari books online' version

Note from the Author or Editor:
Change as noted

page 30, the small example.
If $meat has never recieved a variable, then $barney in
$barney = "fred ate a $meat";
has the value "fred ate a " and not "fred ate a", so the trailing " "
of the strings should remain?

Note from the Author or Editor:
On the second code second of page 30, the code comment reads

$barney is now "fred ate a"

it should read

$barney is now "fred ate a "

with a trailing space

``use 5.6'' should read: ``use 5.006''

Note from the Author or Editor:
change as noted

``http://www.stonehenge.com'' is not a hostname, ``www.stonehenge.com'' is...

Note from the Author or Editor:
change http://www.stonehenge.com to www.stonehenge.com as noted

A copy/paste error from the 1st snippet on the same page:

The pattern should be /(.)(.)\2\1/, instead of /y(.)(.)\2\1/, and the output message should be something like "It matched a palindrome!\n".

Note from the Author or Editor:
The error isn't in the regex, it's the print statement. In the second example, change the print statement to:

print "It matched a after the y"

The pattern /(.)(.)\g{-1}11/ will not be matched from the string "aa11bb".

However, if the string contained, for example, "aaa11bb", the pattern would be matched.

The existing pattern could be modified to /(.)\g{-1}11/ to achieve a match with the existing string: "aa11bb".

Note from the Author or Editor:
This is the last code snippet before the "Alternatives" section in chapter 7. Change the code to read:
------
use 5.010;
$_ = "daa11bb";
if (/(.)(.)\g{-1}11/) {
print "It matched!\n";
}
-----

The change is the addition of the letter "d" at the beginning of "aa11bb".

When discussing fallthroughs in given/when statements, it appears the bottom example on p227 does not behave as expected. e.g. if the program is run as
$ ./fallthroughs.pl Frederick

Then we get the output:

Name has fred in it
Name starts with Fred
I don't see a Fred

Which seems rather self contradictory.

Note that this was also a problem in perlsyn.pod (http://perldoc.perl.org/perlsyn.html - see fallthrough example) and has now been patched in the latest perl 5 development tree (see thread on http://www.xray.mpe.mpg.de/mailing-lists/perl5-porters/2008-03/msg00375.html)

Regards

Martin

Note from the Author or Editor:
This was indeed a documentation problem. I figured the docs would be right and perl would change, but they ended up changing the docs.

This isn't just a problem, then, with the last example on page 227.

The example can also be:

use 5.010;

given( $ARGV[0] ) {
when( 'Fred' ) { say 'Name is Fred'; continue }
when( /^Fred/ ) { say 'Name starts with Fred'; continue }
when( /fred/i ) { say 'Name has fred in it'; }
default { say "I don't see a Fred" }
}

As Michael already stated in his unconfirmed errata, running the bottom example with Frederick as argument does a fall-through right down to the default case. This happens since the argument begins with Fred and the check is continued after matching to both fred and ^Fred, Frederick isn't equal to Fred so we continue to the default.

A change of logic might be good in this case, the best case would be to find Fred, if we can't we'd settle for Frederick, and if that's not possible just having fred would be OK.

use 5.010;
given( $ARGV[0] ) {
when( $_ ~~ 'Fred' ) { say 'Name is Fred'; continue }
when( $_ ~~ /^Fred/ ) { say 'Name starts with Fred';continue }
when( $_ ~~ /fred/i ) { say 'Name has fred in it'; }
default { say "I don't see a Fred" }
}

Staying with the logic of your example, one could also "unless /fred/i" at the end of the default's say.

Note from the Author or Editor:
This is basically how I'm rewriting all the Fred examples.

The books says

when( ! /fred/i } { #dumb
~ => this should be a ,,)'' and not a ,,}''

It is a syntax error.

Note from the Author or Editor:
Change as noted