I am trying to call a multiple variable from a subroutine but when i try to print, it just print only one variable.
In python the script will be f=mode()[0] and b=mode()[1] and it works.
subroutinefile a.pl
sub mode() {
my ($f, $b);
$f=41;
$b=2;
return ($f,$b);
}
And another file that calls the a.pl
use strict;
use warnings;
require 'a.pl';
my ($f,$b);
$f= mode(0);
$b= mode(1);
print "$f\n";
print "$b\n";
The problem is it only prints 2 for both f and b.
You have passed arguments to subroutine: $f = mode(0); $b = mode(1);.
Try the following:
my ($f,$b) = mode();
Your subroutine returns a fixed length list. You may assign values in the list to your variables.
The problem is the mismatch between the definition of the subroutine (no parameters) and the way you are trying to call it (one argument).
Change those calls to this and it works:
$f= (mode())[0];
$b= (mode())[1];
Think there is a syntactical difference between getting an array index FROM the return set versus sending an argument; function(xxx) sends the argument, (function()[])is using an array slice of the return.
Related
I need to control the print method using a variable
My code is below
#!/usr/bin/perl
# test_assign_func.pl
use strict;
use warnings;
sub echo {
my ($string) = #_;
print "from echo: $string\n\n";
}
my $myprint = \&echo;
$myprint->("hello");
$myprint = \&print;
$myprint->("world");
when I ran, I got the following error for the assignment of print function
$ test_assign_func.pl
from echo: hello
Undefined subroutine &main::print called at test_assign_func.pl line 17.
Looks like I need to prefix a namespace to print function but I cannot find the name space. Thank you for any advice!
print is an operator, not a sub.
perlfunc:
The functions in this section can serve as terms in an expression. They fall into two major categories: list operators and named unary operators.
Perl provides a sub for named operators that can be duplicated by a sub with a prototype. A reference to these can be obtained using \&CORE::name.
my $f = \&CORE::length;
say $f->("abc"); # 3
But print isn't such an operator (because of the way it accepts a file handle). For these, you'll need to create a sub with a more limited calling convention.
my $f = sub { print #_ };
$f->("abc\n");
Related:
What are Perl built-in operators/functions?
As mentioned in CORE, some functions can't be called as subroutines, only as barewords. print is one of them.
Here is a Perl program:
use strict;
use warnings;
use Data::Dumper;
sub f {
foreach (()) { }
}
print Dumper(f());
This outputs:
$VAR1 = '';
Since no value is explicitly returned from f, and no expressions are evaluated inside it, shouldn't the result be undef? Where did the empty string come from?
It hasn't quite returned the empty string; it has returned "false", an internal Perl value (called PL_no). This false value is numerically zero but stringily empty. Data::Dumper can't represent it directly as PL_no and so chooses a representation which will work.
Other ways you can generate it:
$ perl -MData::Dumper -E 'say Dumper(1 eq 2)'
$VAR1 = '';
Since no value is explicitly returned from f, and no expressions are evaluated inside it, shouldn't the result be undef?
Nope. perldoc perlsub says the return value is unspecified:
If no return is found and if the last statement is an expression, its value is returned. If the last statement is a loop control structure like a foreach or a while, the returned value is unspecified.
"Unspecified" is short for "we're not going to document the exact behavior because we could change it at any time and you shouldn't rely on it." Right now, it returns PL_no as LeoNerd explained; in a future Perl version, it could return undef, or something else altogether.
I have a subroutine called grepText, which simply greps a text from another variable. I am trying to split the output. Is it possible to pass the output of grepText as an argument to split directly? without putting the value of grepText in a variable first ? grepText returns a string.
What i am trying to do is:
$output = (split ":", grepText("findThis", $Alltext))[1];
grepText is as follows
sub grepText(){
my #text = split "\n", $_[1];
my $output = grep /$_[0]/, #text;
return $output;
}
it doesn't work. Error is
Too many arguments for main::grepText at a line 115, near "$Alltext)"
It is very much possible to pass the input of a subroutine to any perl function directly without using a perl variable.
I think the issue might be with your "grepText" subroutine. To debug the issue in detail, much more information is required.
I did try your routine and I was able to get the required output:
#!/usr/bin/perl
sub grepText
{
return "hello:world"; # returns a test string
}
my $output = (split ":", grepText($textToFind, $Alltext))[1];
print "$output";
Output:
world
Sure it is. But as you've written it grepText is getting some strange parameters. In
(split ":", grepText(/$textToFind/, $Alltext))[1];
you're calling grepText(/$textToFind/, $Alltext) which is searching for the value of $textToFind in the global variable $_ and, in list context, is inserting either an empty list () or a list containing 1 (1) into the parameters
So you're calling grepText($Alltext) or grepText(1, $Alltext) depending on whether $_ contains the regex pattern in $textToFind
I'm pretty certain that's not what you want to do, so some more information would be nice!
However, whatever grepText returns will be split on colons : and (split ":", grepText(...))[1] will give you the second colon-separated field, which seems to be what you're asking
I have perl function I dont what does it do?
my what does min in perl?
#ARVG what does mean?
sub getArgs
{
my $argCnt=0;
my %argH;
for my $arg (#ARGV)
{
if ($arg =~ /^-/) # insert this entry and the next in the hash table
{
$argH{$ARGV[$argCnt]} = $ARGV[$argCnt+1];
}
$argCnt++;
}
return %argH;}
Code like that makes David sad...
Here's a reformatted version of the code doing the indentations correctly. That makes it so much easier to read. I can easily tell where my if and loops start and end:
sub getArgs {
my $argCnt = 0;
my %argH;
for my $arg ( #ARGV ) {
if ( $arg =~ /^-/ ) { # insert this entry and the next in the hash table
$argH{ $ARGV[$argCnt] } = $ARGV[$argCnt+1];
}
$argCnt++;
}
return %argH;
}
The #ARGV is what is passed to the program. It is an array of all the arguments passed. For example, I have a program foo.pl, and I call it like this:
foo.pl one two three four five
In this case, $ARGV is set to the list of values ("one", "two", "three", "four", "five"). The name comes from a similar variable found in the C programming language.
The author is attempting to parse these arguments. For example:
foo.pl -this that -the other
would result in:
$arg{"-this"} = "that";
$arg{"-the"} = "other";
I don't see min. Do you mean my?
This is a wee bit of a complex discussion which would normally involve package variables vs. lexically scoped variables, and how Perl stores variables. To make things easier, I'm going to give you a sort-of incorrect, but technically wrong answer: If you use the (strict) pragma, and you should, you have to declare your variables with my before they can be used. For example, here's a simple two line program that's wrong. Can you see the error?
$name = "Bob";
print "Hello $Name, how are you?\n";
Note that when I set $name to "Bob", $name is with a lowercase n. But, I used $Name (upper case N) in my print statement. As it stands, now. Perl will print out "Hello, how are you?" without a care that I've used the wrong variable name. If it's hard to spot an error like this in a two line program, imagine what it would be like in a 1000 line program.
By using strict and forcing me to declare variables with my, Perl can catch that error:
use strict;
use warnings; # Another Pragma that should always be used
my $name = "Bob";
print "Hello $Name, how are you doing\n";
Now, when I run the program, I get the following error:
Global symbol "$Name" requires explicit package name at (line # of print statement)
This means that $Name isn't defined, and Perl points to where that error is.
When you define variables like this, they are in scope with in the block where it's defined. A block could be the code contained in a set of curly braces or a while, if, or for statement. If you define a variable with my outside of these, it's defined to the end of the file.
Thus, by using my, the variables are only defined inside this subroutine. And, the $arg variable is only defined in the for loop.
One more thing:
The person who wrote this should have used the Getopt::Long module. There's a major bug in their code:
For example:
foo.pl -this that -one -two
In this case, my hash looks like this:
$args{'-this'} = "that";
$args{'-one'} = "-two";
$args{'-two'} = undef;
If I did this:
if ( defined $args{'-two'} ) {
...
}
I would not execute the if statement.
Also:
foo.pl -this=that -one -two
would also fail.
#ARGV is a special variable (refer to perldoc perlvar):
#ARGV
The array #ARGV contains the command-line arguments intended for the
script. $#ARGV is generally the number of arguments minus one, because
$ARGV[0] is the first argument, not the program's command name itself.
See $0 for the command name.
Perl documentation is also available from your command line:
perldoc -v #ARGV
I have the following test code:
sub one_argument {
($a) = #_;
print "In one argument: \$a = $a\n";
return "one_argument";
}
sub mul_arguments {
(#a) = #_;
return "mul_argument";
}
print &one_argument &mul_arguments "something", "\n";
My goal is to be able to understand a bit better how perl decides which arguments to go into each function, and to possibly clear up any misunderstandings that I might have. I would've expected the above code to output:
In one argument: mul_argument
one_argument
However, the below is output:
Use of uninitialized value $a in concatenation (.) or string at ./test.pl line 5.
In one argument: $a =
mdd_argument
I don't understand where 'mdd_argument' comes from (Is it a sort of reference to a function?), and why one_argument receives no arguments.
I would appreciate any insight as to how perl parses arguments into functions when they are called in a similar fashion to above.
Please note that this is purely a learning exercise, I don't need the above code to perform as I expected, and in my own code I wouldn't call a function in such a way.
perldoc perlsub:
If a subroutine is called using the & form, the argument list is optional, and if omitted, no #_ array is set up for the subroutine: the #_ array at the time of the call is visible to subroutine instead. This is an efficiency mechanism that new users may wish to avoid.
In other words, in normal usage, if you use the &, you must use parentheses. Otherwise, the subroutine will be passed the caller's #_.
The mysterious "mdd" is caused because &one_argument doesn't have any arguments and perl is expecting an operator to follow it, not an expression. So the & of &mul_arguments is actually interpreted as the stringwise bit and operator:
$ perl -MO=Deparse,-p -e 'sub mul_arguments; print &one_argument &mul_arguments "something", "\n"'
print((&one_argument & mul_arguments('something', "\n")));
and "one_argument" & "mul_arguments" produces "mdd_argument".