Why does `$v = () = split` return 1? - perl

perldoc says "a list assignment in scalar context returns the number of elements on the right-hand side of the list assignment" but when I try this code:
perl -e '$_="aaaaa";print $v=(()=split //)'
The output is 1 which makes me confused. (The answer I expect is 5.)
Can anybody explain this?

According to split documentation:
When assigning to a list, if LIMIT is omitted, or zero, Perl supplies
a LIMIT one larger than the number of variables in the list <...>
Since you specify empty list, split only returns 1 result and this number of results is exactly what ends in your variable.

split has some kind of crazy ultra-magic in it that allows it to know when it is on the right hand side of an assignment that has a list on the left hand side, and adjusts its behavior according to the number of items in that list.
This is described in perlfunc as being done "to avoid unnecessary work", but you've found an observable difference in behavior caused by that optimization.
To see some evidence of what happened, run your script through Deparse like this:
perl -MO=Deparse -e '$_="aaaaa";print $v=(()=split //)'
Update: I went looking for the code that implements this, and it's not where I expected it to be. Actually the optimization is performed by the assignment operator (op.c:Perl_newASSIGNOP) . split doesn't know that much about its context.

Why are you assigning to an empty array? the ()=(split //) bit. That's going to end up with - um, well, a mess. Or, in your case, an array with a size of one with not much in it.
Also, that's excessively obscure. perl has a sad reputation for being write-only, and all that modifying $_ and using it doesn't help others - or you - understand what is going on.
Try something like
perl -e '$v = (split //, "aaaaa"); print "$v\n"'
or, if you wish to replicate the behavior of your test:
perl -e '$v = () = (split //, "aaaaa"); print "$v\n"'

Yes, but :
perl -e '$_="aaaaa";print $v=(split //)'
gives 5, as well as
perl -e '$_="aaaaa";print $v=(#x=split //)'
Maybe your left-value () is dropping additional array elements ?
edit : by the way :
perl -e '$_="aaaaa";print $v=(($x,$y)=split //)'
returns 3, because the right sight of the $v=... command gives :
( $x , $y , ( a , a , a ) )
So in your original case, ()=split // returns ( ( a , a , a , a , a ) ) (which has only one element)
Edit : bad array notation, and result was wrong because of a last minute changed of my test-case

Related

Why are ##, #!, #, etc. not interpolated in strings?

First, please note that I ask this question out of curiosity, and I'm aware that using variable names like ## is probably not a good idea.
When using doubles quotes (or qq operator), scalars and arrays are interpolated :
$v = 5;
say "$v"; # prints: 5
$# = 6;
say "$#"; # prints: 6
#a = (1,2);
say "#a"; # prints: 1 2
Yet, with array names of the form #+special char like ##, #!, #,, #%, #; etc, the array isn't interpolated :
#; = (1,2);
say "#;"; # prints nothing
say #; ; # prints: 1 2
So here is my question : does anyone knows why such arrays aren't interpolated? Is it documented anywhere?
I couldn't find any information or documentation about that. There are too many articles/posts on google (or SO) about the basics of interpolation, so maybe the answer was just hidden in one of them, or at the 10th page of results..
If you wonder why I could need variable names like those :
The -n (and -p for that matter) flag adds a semicolon ; at the end of the code (I'm not sure it works on every version of perl though). So I can make this program perl -nE 'push#a,1;say"#a"}{say#a' shorter by doing instead perl -nE 'push#;,1;say"#;"}{say#', because that last ; convert say# to say#;. Well, actually I can't do that because #; isn't interpolated in double quotes. It won't be useful every day of course, but in some golfing challenges, why not!
It can be useful to obfuscate some code. (whether obfuscation is useful or not is another debate!)
Unfortunately I can't tell you why, but this restriction comes from code in toke.c that goes back to perl 5.000 (1994!). My best guess is that it's because Perl doesn't use any built-in array punctuation variables (except for #- and #+, added in 5.6 (2000)).
The code in S_scan_const only interprets # as the start of an array if the following character is
a word character (e.g. #x, #_, #1), or
a : (e.g. #::foo), or
a ' (e.g. #'foo (this is the old syntax for ::)), or
a { (e.g. #{foo}), or
a $ (e.g. #$foo), or
a + or - (the arrays #+ and #-), but not in regexes.
As you can see, the only punctuation arrays that are supported are #- and #+, and even then not inside a regex. Initially no punctuation arrays were supported; #- and #+ were special-cased in 2000. (The exception in regex patterns was added to make /[\c#-\c_]/ work; it used to interpolate #- first.)
There is a workaround: Because #{ is treated as the start of an array variable, the syntax "#{;}" works (but that doesn't help your golf code because it makes the code longer).
Perl's documentation says that the result is "not strictly predictable".
The following, from perldoc perlop (Perl 5.22.1), refers to interpolation of scalars. I presume it applies equally to arrays.
Note also that the interpolation code needs to make a decision on
where the interpolated scalar ends. For instance, whether
"a $x -> {c}" really means:
"a " . $x . " -> {c}";
or:
"a " . $x -> {c};
Most of the time, the longest possible text that does not include
spaces between components and which contains matching braces or
brackets. because the outcome may be determined by voting based on
heuristic estimators, the result is not strictly predictable.
Fortunately, it's usually correct for ambiguous cases.
Some things are just because "Larry coded it that way". Or as I used to say in class, "It works the way you think, provided you think like Larry thinks", sometimes adding "and it's my job to teach you how Larry thinks."

How does this Perl one-liner actually work?

So, I happened to notice that last.fm is hiring in my area, and since I've known a few people who worked there, I though of applying.
But I thought I'd better take a look at the current staff first.
Everyone on that page has a cute/clever/dumb strapline, like "Is life not a thousand times too short for us to bore ourselves?". In fact, it was quite amusing, until I got to this:
perl -e'print+pack+q,c*,,map$.+=$_,74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34'
Which I couldn't resist pasting into my terminal (kind of a stupid thing to do, maybe), but it printed:
Just another Last.fm hacker,
I thought it would be relatively easy to figure out how that Perl one-liner works. But I couldn't really make sense of the documentation, and I don't know Perl, so I wasn't even sure I was reading the relevant documentation.
So I tried modifying the numbers, which got me nowhere. So I decided it was genuinely interesting and worth figuring out.
So, 'how does it work' being a bit vague, my question is mainly,
What are those numbers? Why are there negative numbers and positive numbers, and does the negativity or positivity matter?
What does the combination of operators +=$_ do?
What's pack+q,c*,, doing?
This is a variant on “Just another Perl hacker”, a Perl meme. As JAPHs go, this one is relatively tame.
The first thing you need to do is figure out how to parse the perl program. It lacks parentheses around function calls and uses the + and quote-like operators in interesting ways. The original program is this:
print+pack+q,c*,,map$.+=$_,74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34
pack is a function, whereas print and map are list operators. Either way, a function or non-nullary operator name immediately followed by a plus sign can't be using + as a binary operator, so both + signs at the beginning are unary operators. This oddity is described in the manual.
If we add parentheses, use the block syntax for map, and add a bit of whitespace, we get:
print(+pack(+q,c*,,
map{$.+=$_} (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21,
18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34)))
The next tricky bit is that q here is the q quote-like operator. It's more commonly written with single quotes:
print(+pack(+'c*',
map{$.+=$_} (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21,
18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34)))
Remember that the unary plus is a no-op (apart from forcing a scalar context), so things should now be looking more familiar. This is a call to the pack function, with a format of c*, meaning “any number of characters, specified by their number in the current character set”. An alternate way to write this is
print(join("", map {chr($.+=$_)} (74, …, -34)))
The map function applies the supplied block to the elements of the argument list in order. For each element, $_ is set to the element value, and the result of the map call is the list of values returned by executing the block on the successive elements. A longer way to write this program would be
#list_accumulator = ();
for $n in (74, …, -34) {
$. += $n;
push #list_accumulator, chr($.)
}
print(join("", #list_accumulator))
The $. variable contains a running total of the numbers. The numbers are chosen so that the running total is the ASCII codes of the characters the author wants to print: 74=J, 74+43=117=u, 74+43-2=115=s, etc. They are negative or positive depending on whether each character is before or after the previous one in ASCII order.
For your next task, explain this JAPH (produced by EyesDrop).
''=~('(?{'.('-)#.)#_*([]#!#/)(#)#-#),#(##+#)'
^'][)#]`}`]()`#.#]#%[`}%[#`#!##%[').',"})')
Don't use any of this in production code.
The basic idea behind this is quite simple. You have an array containing the ASCII values of the characters. To make things a little bit more complicated you don't use absolute values, but relative ones except for the first one. So the idea is to add the specific value to the previous one, for example:
74 -> J
74 + 43 -> u
74 + 42 + (-2 ) -> s
Even though $. is a special variable in Perl it does not mean anything special in this case. It is just used to save the previous value and add the current element:
map($.+=$_, ARRAY)
Basically it means add the current list element ($_) to the variable $.. This will return a new array with the correct ASCII values for the new sentence.
The q function in Perl is used for single quoted, literal strings. E.g. you can use something like
q/Literal $1 String/
q!Another literal String!
q,Third literal string,
This means that pack+q,c*,, is basically pack 'c*', ARRAY. The c* modifier in pack interprets the value as characters. For example, it will use the value and interpret it as a character.
It basically boils down to this:
#!/usr/bin/perl
use strict;
use warnings;
my $prev_value = 0;
my #relative = (74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34);
my #absolute = map($prev_value += $_, #relative);
print pack("c*", #absolute);

(3 lines) from bash to perl?

I have these three lines in bash that work really nicely. I want to add them to some existing perl script but I have never used perl before ....
could somebody rewrite them for me? I tried to use them as they are and it didn't work
note that $SSH_CLIENT is a run-time parameter you get if you type set in bash (linux)
users[210]=radek #where 210 is tha last octet from my mac's IP
octet=($SSH_CLIENT) # split the value on spaces
somevariable=$users[${octet[0]##*.}] # extract the last octet from the ip address
These might work for you. I noted my assumptions with each line.
my %users = ( 210 => 'radek' );
I assume that you wanted a sparse array. Hashes are the standard implementation of sparse arrays in Perl.
my #octet = split ' ', $ENV{SSH_CLIENT}; # split the value on spaces
I assume that you still wanted to use the environment variable SSH_CLIENT
my ( $some_var ) = $octet[0] =~ /\.(\d+)$/;
You want the last set of digits from the '.' to the end.
The parens around the variable put the assignment into list context.
In list context, a match creates a list of all the "captured" sequences.
Assigning to a scalar in a list context, means that only the number of scalars in the expression are assigned from the list.
As for your question in the comments, you can get the variable out of the hash, by:
$db = $users{ $some_var };
# OR--this one's kind of clunky...
$db = $users{ [ $octet[0] =~ /\.(\d+)$/ ]->[0] };
Say you have already gotten your IP in a string,
$macip = "10.10.10.123";
#s = split /\./ , $macip;
print $s[-1]; #get last octet
If you don't know Perl and you are required to use it for work, you will have to learn it. Surely you are not going to come to SO and ask every time you need it in Perl right?

When is Perl's scalar comma operator useful?

Is there any reason to use a scalar comma operator anywhere other than in a for loop?
Since the Perl scalar comma is a "port" of the C comma operator, these comments are probably apropos:
Once in a while, you find yourself in
a situation in which C expects a
single expression, but you have two
things you want to say. The most
common (and in fact the only common)
example is in a for loop, specifically
the first and third controlling
expressions. What if (for example) you
want to have a loop in which i counts
up from 0 to 10 at the same time that
j is counting down from 10 to 0?
So, your instinct that it's mainly useful in for loops is a good one, I think.
I occasionally use it in the conditional (sometimes erroneously called "the ternary") operator, if the code is easier to read than breaking it out into a real if/else:
my $blah = condition() ? do_this(), do_that() : do_the_other_thing();
It could also be used in some expression where the last result is important, such as in a grep expression, but in this case it's just the same as if a semicolon was used:
my #results = grep { setup(), condition() } #list;

How does this Perl one liner to check if a directory is empty work?

I got this strange line of code today, it tells me 'empty' or 'not empty' depending on whether the CWD has any items (other than . and ..) in it.
I want to know how it works because it makes no sense to me.
perl -le 'print+(q=not =)[2==(()=<.* *>)].empty'
The bit I am interested in is <.* *>. I don't understand how it gets the names of all the files in the directory.
It's a golfed one-liner. The -e flag means to execute the rest of the command line as the program. The -l enables automatic line-end processing.
The <.* *> portion is a glob containing two patterns to expand: .* and *.
This portion
(q=not =)
is a list containing a single value -- the string "not". The q=...= is an alternate string delimiter, apparently used because the single-quote is being used to quote the one-liner.
The [...] portion is the subscript into that list. The value of the subscript will be either 0 (the value "not ") or 1 (nothing, which prints as the empty string) depending on the result of this comparison:
2 == (()=<.* *>)
There's a lot happening here. The comparison tests whether or not the glob returned a list of exactly two items (assumed to be . and ..) but how it does that is tricky. The inner parentheses denote an empty list. Assigning to this list puts the glob in list context so that it returns all the files in the directory. (In scalar context it would behave like an iterator and return only one at a time.) The assignment itself is evaluated in scalar context (being on the right hand side of the comparison) and therefore returns the number of elements assigned.
The leading + is to prevent Perl from parsing the list as arguments to print. The trailing .empty concatenates the string "empty" to whatever came out of the list (i.e. either "not " or the empty string).
<.* *>
is a glob consisting of two patterns: .* are all file names that start with . and * corresponds to all files (this is different than the usual DOS/Windows conventions).
(()=<.* *>)
evaluates the glob in list context, returning all the file names that match.
Then, the comparison with 2 puts it into scalar context so 2 is compared to the number of files returned. If that number is 2, then the only directory entries are . and .., period. ;-)
<.* *> means (glob(".*"), glob("*")). glob expands file patterns the same way the shell does.
I find that the B::Deparse module helps quite a bit in deciphering some stuff that throws off most programmers' eyes, such as the q=...= construct:
$ perl -MO=Deparse,-p,-q,-sC 2>/dev/null << EOF
> print+(q=not =)[2==(()=<.* *>)].empty
> EOF
use File::Glob ();
print((('not ')[(2 == (() = glob('.* *')))] . 'empty'));
Of course, this doesn't instantly produce "readable" code, but it surely converts some of the stumbling blocks.
The documentation for that feature is here. (Scroll near the end of the section)