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 " . $x -> {c};
Most of the time, the longest possible text that does not include
spaces between components and which contains matching braces or
brackets. because the outcome may be determined by voting based on
heuristic estimators, the result is not strictly predictable.
Fortunately, it's usually correct for ambiguous cases.
Some things are just because "Larry coded it that way". Or as I used to say in class, "It works the way you think, provided you think like Larry thinks", sometimes adding "and it's my job to teach you how Larry thinks."
This question already has answers here:
Why aren't newlines being printed in this Perl code?
(3 answers)
Closed 8 years ago.
Let's take the following minimalistic script:
#!/usr/bin/perl
#
# conditional_operator.pl
#
use strict;
print ( 1 ? "true" : "false" )." statement\n";
exit;
I expect the output always to be "true statement". But when I execute this snippet, I see ...
deviolog#home:~/test$ perl conditional_operator.pl
true
The " statement\n" concatenation seems to be ignored.
My perl version is v5.14.2. I read the perlop manual about the conditional operator and think, a string concatenation should be possible.
Can somebody explain this behaviour?
Always include use warnings; at the top of every script.
To get your desired behavior, just add parenthesis so print is called with the entire argument instead of just the first part:
print(( 1 ? "true" : "false" )." statement\n");
If you'd had warnings turned on, you would've gotten this alert:
Useless use of concatenation (.) or string in void context
You can also avoid the undesired behavior by leading with a blank concatenation, or you could put a plus sign before the parenthesis:
print +( 1 ? "true" : "false" )." statement\n";
print ''.( 1 ? "true" : "false" )." statement\n";
Adding use warnings to your code gives this:
print (...) interpreted as function at ./cond line 10.
Useless use of concatenation (.) or string in void context at ./cond line 10.
Even better, add use diagnostics and you get this:
print (...) interpreted as function at ./cond line 11 (#1)
(W syntax) You've run afoul of the rule that says that any list operator
followed by parentheses turns into a function, with all the list
operators arguments found inside the parentheses. See
"Terms and List Operators (Leftward)" in perlop.
Useless use of concatenation (.) or string in void context at ./cond line 11 (#2)
(W void) You did something without a side effect in a context that does
nothing with the return value, such as a statement that doesn't return a
value from a block, or the left side of a scalar comma operator. Very
often this points not to stupidity on your part, but a failure of Perl
to parse your program the way you thought it would. For example, you'd
get this if you mixed up your C precedence with Python precedence and
said
$one, $two = 1, 2;
when you meant to say
($one, $two) = (1, 2);
Another common error is to use ordinary parentheses to construct a list
reference when you should be using square or curly brackets, for
example, if you say
$array = (1,2);
when you should have said
$array = [1,2];
The square brackets explicitly turn a list value into a scalar value,
while parentheses do not. So when a parenthesized list is evaluated in
a scalar context, the comma is treated like C's comma operator, which
throws away the left argument, which is not what you want. See
perlref for more on this.
This warning will not be issued for numerical constants equal to 0 or 1
since they are often used in statements like
1 while sub_with_side_effects();
String constants that would normally evaluate to 0 or 1 are warned
about.
Perl wants to explain the problems to you. You just need to ask it what you're doing wrong :-)
So, I happened to notice that last.fm is hiring in my area, and since I've known a few people who worked there, I though of applying.
But I thought I'd better take a look at the current staff first.
Everyone on that page has a cute/clever/dumb strapline, like "Is life not a thousand times too short for us to bore ourselves?". In fact, it was quite amusing, until I got to this:
perl -e'print+pack+q,c*,,map$.+=$_,74,43,-2,1,-84, 65,13,1,5,-12,-3, 13,-82,44,21, 18,1,-70,56, 7,-77,72,-7,2, 8,-6,13,-70,-34'
Which I couldn't resist pasting into my terminal (kind of a stupid thing to do, maybe), but it printed:
Just another Last.fm hacker,
I thought it would be relatively easy to figure out how that Perl one-liner works. But I couldn't really make sense of the documentation, and I don't know Perl, so I wasn't even sure I was reading the relevant documentation.
So I tried modifying the numbers, which got me nowhere. So I decided it was genuinely interesting and worth figuring out.
So, 'how does it work' being a bit vague, my question is mainly,
What are those numbers? Why are there negative numbers and positive numbers, and does the negativity or positivity matter?
What does the combination of operators +=$_ 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);
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;
I am just learning Perl's comparison operators. I tried the below code :-
$foo=291;
$bar=30;
if ($foo < $bar) {
print "$foo is less than $bar (first)\n";
}
if ($foo lt $bar) {
print "$foo is less than $bar (second)\n";
}
The output is 291 is less than 30 (second). Does this mean the lt operator always converts the variables to string and then compare? What is the rationale for Perl making lt operator behave differently from the < operator?
Thanks,
Your guess is right. The alphabetic operators like lt compare the variables as strings whereas the symbolic ones like < compare them as numbers. You can read the perlop man page for more details.
The rationale is that scalars in Perl are not typed, so without you telling it Perl would not know how to compare two variables. If it did guess then it would sometimes getting it wrong, which would lead to having to do things like ' ' + $a < ' ' + $b to force string comparsion which is probably worse than lt.
That said this is a horrible gotcha which probably catches out everyone new to Perl and still catches me out when coming back to Perl after some time using a less post-modern language.
Since Perl is loosely typed, and values can silently convert between strings and integers at any moment, Perl needs two different types of comparison operators to distinguish between integer comparison (<) and string comparison (lt). If you only had one operator, how would you tell the difference?
Rationale? It's a string operator. From "perldoc perlop":
Binary "lt" returns true if the left argument is stringwise less than the right argument.
If that's not what you want, don't use it.
lt compares values lexically (i.e. in ASCII/UNICODE or locale order) and < compares values numerically. Perl has both operators for the same reason "10" + 5 is 15 rather than a type error: it is weakly typed. You must always tell the computer something unambiguous. Languages that are strongly typed tend to use casting to resolve ambiguity, whereas, weakly typed languages tend to use lots of operators. The Python (a strongly typed language) equivalent to "10" + 5 is float("10") + 5.
Does this mean the 'lt' operator
always converts the variables to
string and then compare?
Yes, see perlop
What is the rationale for Perl making
'lt' operator behave differently from
'<' operator?
Because having a numeric comparison operator and a string comparison operator makes a lot more sense then having a mumble mumble operator and another, identical mumble mumble operator.