I try to get the first element of
$object->method()
that is returning a list.
My first though was to try:
$object->method()[0]
But I get this error:
syntax error at script.pl line 42, near ")["
Execution of script.pl aborted due to compilation errors.
so I tried:
print ($object->method())[0];
but Perl 'eat' the ( ) to use with print, and still have the error.
what I need is to do:
print((object->method())[0]);
Is there a simpler way to do this?
It's Perl, so there are many special tricks to do this.
print [$object->method()]->[0]
for another one.
There's a special trick to do this:
print +($object->method())[0]
from perldoc perlfunc:
Any function in the list below may be used either with or without
parentheses around its arguments. (The syntax descriptions omit the
parentheses.) If you use parentheses, the simple but occasionally
surprising rule is this: It looks like a function, therefore it is a function, and precedence doesn't matter. Otherwise it's a list operator
or unary operator, and precedence does matter. Whitespace between the
function and left parenthesis doesn't count, so sometimes you need to be
careful:
print 1+2+4; # Prints 7.
print(1+2) + 4; # Prints 3.
print (1+2)+4; # Also prints 3!
print +(1+2)+4; # Prints 7.
print ((1+2)+4); # Prints 7.
From #perl#libera IRC:
Related
My RAKU Code:
sub comments {
if ($DEBUG) { say "<filtering comments>\n"; }
my #filteredtitles = ();
# This loops through each track
for #tracks -> $title {
##########################
# LAB 1 TASK 2 #
##########################
## Add regex substitutions to remove superflous comments and all that follows them
## Assign to $_ with smartmatcher (~~)
##########################
$_ = $title;
if ($_) ~~ s:g:mrx/ .*<?[\(^.*]> / {
# Repeat for the other symbols
########################## End Task 2
# Add the edited $title to the new array of titles
#filteredtitles.push: $_;
}
}
# Updates #tracks
return #filteredtitles;
}
Result when compiling:
Error Compiling! Placeholder variable '#_' may not be used here because the surrounding block doesn't take a signature.
Is there something obvious that I am missing? Any help is appreciated.
So, in contrast with #raiph's answer, here's what I have:
my #tracks = <Foo Ba(r B^az>.map: { S:g / <[\(^]> // };
Just that. Nothing else. Let's dissect it, from the inside out:
This part: / <[\(^]> / is a regular expression that will match one character, as long as it is an open parenthesis (represented by the \() or a caret (^). When they go inside the angle brackets/square brackets combo, it means that is an Enumerated character class.
Then, the: S introduces the non-destructive substitution, i.e., a quoting construct that will make regex-based substitutions over the topic variable $_ but will not modify it, just return its value with the modifications requested. In the code above, S:g brings the adverb :g or :global (see the global adverb in the adverbs section of the documentation) to play, meaning (in the case of the substitution) "please make as many as possible of this substitution" and the final / marks the end of the substitution text, and as it is adjacent to the second /, that means that
S:g / <[\(^]> //
means "please return the contents of $_, but modified in such a way that all its characters matching the regex <[\(^]> are deleted (substituted for the empty string)"
At this point, I should emphasize that regular expressions in Raku are really powerful, and that reading the entire page (and probably the best practices and gotchas page too) is a good idea.
Next, the: .map method, documented here, will be applied to any Iterable (List, Array and all their alikes) and will return a sequence based on each element of the Iterable, altered by a Code passed to it. So, something like:
#x.map({ S:g / foo /bar/ })
essencially means "please return a Sequence of every item on #x, modified by substituting any appearance of the substring foo for bar" (nothing will be altered on #x). A nice place to start to learn about sequences and iterables would be here.
Finally, my one-liner
my #tracks = <Foo Ba(r B^az>.map: { S:g / <[\(^]> // };
can be translated as:
I have a List with three string elements
Foo
Ba(r
B^az
(This would be a placeholder for your "list of titles"). Take that list and generate a second one, that contains every element on it, but with all instances of the chars "open parenthesis" and "caret" removed.
Ah, and store the result in the variable #tracks (that has my scope)
Here's what I ended up with:
my #tracks = <Foo Ba(r B^az>;
sub comments {
my #filteredtitles;
for #tracks -> $_ is copy {
s:g / <[\(^]> //;
#filteredtitles.push: $_;
}
return #filteredtitles;
}
The is copy ensures the variable set up by the for loop is mutable.
The s:g/...//; is all that's needed to strip the unwanted characters.
One thing no one can help you with is the error you reported. I currently think you just got confused.
Here's an example of code that generates that error:
do { #_ }
But there is no way the code you've shared could generate that error because it requires that there is an #_ variable in your code, and there isn't one.
One way I can help in relation to future problems you may report on StackOverflow is to encourage you to read and apply the guidance in Minimal Reproducible Example.
While your code did not generate the error you reported, it will perhaps help you if you know about some of the other compile time and run time errors there were in the code you shared.
Compile-time errors:
You wrote s:g:mrx. That's invalid: Adverb mrx not allowed on substitution.
You missed out the third slash of the s///. That causes mayhem (see below).
There were several run-time errors, once I got past the compile-time errors. I'll discuss just one, the regex:
.*<?[...]> will match any sub-string with a final character that's one of the ones listed in the [...], and will then capture that sub-string except without the final character. In the context of an s:g/...// substitution this will strip ordinary characters (captured by the .*) but leave the special characters.
This makes no sense.
So I dropped the .*, and also the ? from the special character pattern, changing it from <?[...]> (which just tries to match against the character, but does not capture it if it succeeds) to just <[...]> (which also tries to match against the character, but, if it succeeds, does capture it as well).
A final comment is about an error you made that may well have seriously confused you.
In a nutshell, the s/// construct must have three slashes.
In your question you had code of the form s/.../ (or s:g/.../ etc), without the final slash. If you try to compile such code the parser gets utterly confused because it will think you're just writing a long replacement string.
For example, if you wrote this code:
if s/foo/ { say 'foo' }
if m/bar/ { say 'bar' }
it'd be as if you'd written:
if s/foo/ { say 'foo' }\nif m/...
which in turn would mean you'd get the compile-time error:
Missing block
------> if m/⏏bar/ { ... }
expecting any of:
block or pointy block
...
because Raku(do) would have interpreted the part between the second and third /s as the replacement double quoted string of what it interpreted as an s/.../.../ construct, leading it to barf when it encountered bar.
So, to recap, the s/// construct requires three slashes, not two.
(I'm ignoring syntactic variants of the construct such as, say, s [...] = '...'.)
I have some doubt about the outcome of a binding operator expression in perl. I mean expression like
string =~ /pattern/
I have done some simple test
$ss="a1b2c3";
say $ss=~/a/; # 1
say $ss=~/[a-z]/g; # abc
#aa=$ss=~/[a-z]/g;say #aa; # abc
$aa=#aa;say $aa; # 3
$aa=$ss=~/[a-z]/g;say $aa; # 1
note the comment part above is the running result.
So here comes the question, what on earth is returned by $ss=~/[a-z]/g, it seems that it returned an array according to code line 3,4,5. But what about the last line, why it gives 1 instead of 3 which is the length of array?
The return of the match operator depends on the context: in list context it returns all captured matches, in scalar context the true/false. The say imposes list context, but in the first example nothing is captured in the regex so you only get "success."
Next, the behavior of /g modifier also differs across contexts. In list context, with it the string keeps being scanned with the given pattern until all matches are found, and a list with them is returned. These are your second and third examples.
But in scalar context its behavior is a bit specific: with it the search will continue from the position of the last match, the next time round. One typical use is in the loop condition
while (/(\w+)/g) { ... }
This is a bit of a tokenizer: after the body of the loop runs the next word is found, etc.
Then the last example doesn't really make sense; you are getting the "normal" scalar-context matching success/fail, and /g doesn't do anything -- until you match on $ss the next time
perl -wE'
$s=shift||q(abc);
for (1..2) { $m = $s=~/(.)/g; say "$m: $1"; }
'
prints lines 1:a and then 1:b.
Outside of iterative structures (like while condition) the /g in scalar context is usually an error, pointless at best or a quiet bug.
See "Global matching" under "Using regular expressions" in perlretut for /g.
See regex operators in perlop in general, and about /g as well. A useful tool to explore /g workings is pos.
Iam trying to remove a test segment frequently occurs in a file. I have a matching keyword let suppose
FIND_WORD(---------
-----------
-----------);
is terminating after 2 or 3 lines at " );" but i need to remove next few lines also till ");"
I am using
if ($line =~ m/FIND_WORD/) {return 0;}
which removes one line
I need to remove till ");"
Sounds like this is exactly what the flip-flop operator is for.
while (<>) {
print unless /FIND_WORD\(/ .. /\);/;
}
From perlop:
In scalar context, ".." returns a boolean value. The operator is
bistable, like a flip-flop, and emulates the line-range (comma)
operator of sed, awk, and various editors. Each ".." operator
maintains its own boolean state, even across calls to a subroutine
that contains it. It is false as long as its left operand is false.
Once the left operand is true, the range operator stays true until the
right operand is true, AFTER which the range operator becomes false
again.
See this recent GeekUni blog post for more details.
You need to use multi-line matching. You will need to slurp the whole file into a variable and then substitute that.
$all_lines =~ s/FIND_WORD.*?\)\;//sg;
This matches from FIND_WORD to the nearest ");" and deletes it.
I am pretty new to Perl. I have the following code fragment that works just fine, but I don't fully understand it:
for ($i = 1; $i <= $pop->Count(); $i++) {
foreach ( $pop->Head( $i ) ) {
/^(From|Subject):\s+/i and print $_, "\n";
}
}
$pop->Head is a string or an array of strings returned by the function Mail::POP3Client, and it is the headers of a bunch of emails. Line 3 is some kind of regular expression that extracts the FROM and the SUBJECT from the header.
My question is how does the print function only print the From and the Subject without all the other stuff in the header? What does "and" mean - this surely can't be a boolean and can it? Most important, I want to put the From string into its own variable (my $fromline). How do I do this?
I am hoping that this will be easy for some Perl professional, it has got me baffled!
Thanks in advance.
ARGHHH... The question was edited while I was typing the answer. OK, throwing out the part of my answer that's no longer relevant, and focusing on the specific questions:
The outer loop iterates over all the messages in the mailbox.
The inner loop doesn't specify a loop variable, so the special variable $_ is used.
In each iteration through the inner loop, $_ is one header line from message number $i.
/^(From|Subject):\s+/i and print $_, "\n";
The first part of this line, up to the and is a pattern. We didn't specify what to do with the pattern, so it's implicitly matched against $_. (That's one of the things that makes $_ special.) This gives us a yes/no test: does the pattern match the header line or not?
The pattern tests whether that item begins with (<) either of the words "From" or "Subject", followed immediately by a colon and one or more whitespace characters. (This not the correct pattern to match an RFC 822 header. Whitespace is optional on both sides of the colon. The pattern should more properly be /^(From|Subject)\s*:\s*/i. But that's a separate issue.) the i at the end of the pattern says to ignore case, so from or SUBJECT would be OK.
The and says to continue evaluating (i.e., executing) the expression if there is a match. If there's no match, whatever follows and is ignored.
The rest of the expression prints the header line ($_) and a newline ("\n").
In perl, and and or are boolean operators. They're synonyms for && and ||, except that they have much lower precedence, making it easier to write short-ciruit expressions without clutter from lots of parentheses.
The smallest change that captures the From line into a separate variable would be to add the following line to the inner loop:
/^From\s*:\s*(.*)$/i and $fromline = $1;
You should probably also put
$fromline = undef
before the loop so you can test, after the loop, whether there was a From: line.
There are other ways to do it. In fact, that's one of the mantras of perl: "There's more than one way to do it." I've stripped out the "From: " from the beginning of the line before storing the balance in $fromline, but I don't know your needs.
It's a logical and with short-circuiting. If the left side evaluates to true -- say, if that regular expression matches -- it'll evaluate the right side, the print.
If the expression on the left is false, it doesn't need to evaluate the right hand side, because the net result would still be false, so it skips it.
See also: perldoc perlop
I am trying to understand syntax of
attributeMap[tuple[0]] = tuple[1]
from
I have Python code for detecting input parameter - How to do similar in Powershell
It doesn't look correct because the brackets are uneven, but the program is interpreted without error. On the other hand if I change it to
attributeMap[tuple[0] = tuple[1]]
I get the error
File "lookup.py", line 15
attributeMap[tuple[0] = tuple[1]]
The brackets are not "uneven" at all:
attributeMap[tuple[0]] = tuple[1]
We have three expressions here:
tuple[0] # first element of tuple
tuple[1] # second element of tuple
attributeMap[tuple[0]] # value in attributeMap which has the key matching first element of tuple
As you can see, the third expression makes use of the first, and at the end all we do is assign the second to the third. The brackets are in the right places.