Perl, how to print evaluation of sum of numbers using "print" - perl

Look at the following code sample.
$ct=5;
print "hey $ct+1";
When evaluated, this returns:
hey 5+1
However, I'm trying to get the code to return:
hey 6
Does anybody know if there is some sort of evaluation call that I can make to have this occur correctly, or a way to change the syntax a tad? I know I could simply do:
$dum=$ct+1;
print "hey $dum";
But what I'm showing you is a simple version of something more complicated. If I could get this small example to work correctly, my larger problem is solved. Thanks.

You want to execute a Perl addition operator, but the code only features a Perl string literal. You want
print "hey ", $ct+1;
If you did have Perl source code in a variable and you wanted to execute it, you'd have to invoke the Perl parser and compiler: eval EXPR.

Another alternative is formatted printing.
my $ct = 5;
printf "hey %s", $ct+1;

This is kind of intermediate Perl, but you can put your expression into a reference and dereference it inside double quotes:
print "hey #{[$ct+1]}";
print "hey ${\($ct+1)}";
Code like this is harder to read and gives Perl the reputation of being unfriendly to newbies, so I don't recommend it over extracting the expressions out of the quotes:
print "hey ", ($ct+1);
print "hey " . ($ct+1);

Related

Why is the perl print filehandle syntax the way it is?

I am wondering why the perl creators chose an unusual syntax for printing to a filehandle:
print filehandle list
with no comma after filehandle. I see that it's to distinguish between "print list" and "print filehandle list", but why was the ad-hoc syntax preferred over creating two functions - one to print to stdout and one to print to given filehandle?
In my searches, I came across the explanation that this is an indirect object syntax, but didn't the print function exist in perl 4 and before, whereas the object-oriented features came into perl relatively late? Is anyone familiar with the history of print in perl?
Since the comma is already used as the list constructor, you can't use it to separate semantically different arguments to print.
open my $fh, ...;
print $fh, $foo, $bar
would just look like you were trying to print the values of 3 variables. There's no way for the parser, which operates at compile time, to tell that $fh is going to refer to a file handle at run time. So you need a different character to syntactically (not semantically) distinguish between the optional file handle and the values to actually print to that file handle.
At this point, it's no more work for the parser to recognize that the first argument is separated from the second argument by blank space than it would be if it were separated by any other character.
If Perl had used the comma to make print look more like a function, the filehandle would always have to be included if you are including anything to print besides $_. That is the way functions work: If you pass in a second parameter, the first parameter must also be included. There isn't one function I can think of in Perl where the first parameter is optional when the second parameter exists. Take a look at split. It can be written using zero to four parameters. However, if you want to specify a <limit>, you have to specify the first three parameters too.
If you look at other languages, they all include two different ways ways to print: One if you want STDOUT, and another if you're printing to something besides STDOUT. Thus, Python has both print and write. C has both printf and fprintf. However, Perl can do this with just a single statement.
Let's look at the print statement a bit more closely -- thinking back to 1987 when Perl was first written.
You can think of the print syntax as really being:
print <filehandle> <list_to_print>
To print to OUTFILE, you would say:
To print to this file, you would say:
print OUTFILE "This is being printed to myfile.txt\n";
The syntax is almost English like (PRINT to OUTFILE the string "This is being printed to myfile.txt\n"
You can also do the same with thing with STDOUT:
print STDOUT "This is being printed to your console";
print STDOUT " unless you redirected the output.\n";
As a shortcut, if the filehandle was not given, it would print to STDOUT or whatever filehandle the select was set to.
print "This is being printed to your console";
print " unless you redirected the output.\n";
select OUTFILE;
print "This is being printed to whatever the filehandle OUTFILE is pointing to\n";
Now, we see the thinking behind this syntax.
Imagine I have a program that normally prints to the console. However, my boss now wants some of that output printed to various files when required instead of STDOUT. In Perl, I could easily add a few select statements, and my problems will be solved. In Python, Java, or C, I would have to modify each of my print statements, and either have some logic to use a file write to STDOUT (which may involve some conniptions in file opening and dupping to STDOUT.
Remember that Perl wasn't written to be a full fledge language. It was written to do the quick and dirty job of parsing text files more easily and flexibly than awk did. Over the years, people used it because of its flexibility and new concepts were added on top of the old ones. For example, before Perl 5, there was no such things as references which meant there was no such thing as object oriented programming. If we, back in the days of Perl 3 or Perl 4 needed something more complex than the simple list, hash, scalar variable, we had to munge it ourselves. It's not like complex data structures were unheard of. C had struct since its initial beginnings. Heck, even Pascal had the concept with records back in 1969 when people thought bellbottoms were cool. (We plead insanity. We were all on drugs.) However, since neither Bourne shell nor awk had complex data structures, so why would Perl need them?
Answer to "why" is probably subjective and something close to "Larry liked it".
Do note however, that indirect object notation is not a feature of print, but a general notation that can be used with any object or class and method. For example with LWP::UserAgent.
use strict;
use warnings;
use LWP::UserAgent;
my $ua = new LWP::UserAgent;
my $response = get $ua "http://www.google.com";
my $response_content = decoded_content $response;
print $response_content;
Any time you write method object, it means exactly the same as object->method. Note also that parser seems to only reliably work as long as you don't nest such notations or do not use complex expressions to get object, so unless you want to have lots of fun with brackets and quoting, I'd recommend against using it anywhere except common cases of print, close and rest of IO methods.
Why not? it's concise and it works, in perl's DWIM spirit.
Most likely it's that way because Larry Wall liked it that way.

perl: printing object properties

I'm playing a bit with the Net::Amazon::EC2 libraries, and can't find out a simple way to print object properties:
This works:
my $snaps = $ec2->describe_snapshots();
foreach my $snap ( #$snaps ) {
print $snap->snapshot_id . " " . $snap->volume_id . "\n";
}
But if I try:
print "$snap->snapshot_id $snap->volume_id \n";
I get
Net::Amazon::EC2::Snapshot=HASH(0x4c1be90)->snapshot_id
Is there a simple way to print the value of the property inside a print?
$snap->volume_id is not a property, it is a method call. While you could interpolate a method call inside a string, it is exceedingly ugly.
To get all the properties of an object you can use the module Data::Dumper, included with core perl:
use Data::Dumper;
print Dumper($object);
Not in the way you want to do it. In fact, what you're doing with $snap->snapshot_id is calling a method (as in sub). Perl cannot do that inside a double-quoted string. It will interpolate your variable $snap. That becomes something like HASH(0x1234567) because that is what it is: a blessed reference of a hash.
The interpolation only works with scalars (and arrays, but I'll omit that). You can go:
print "$foo $bar"; # scalar
print "$hash->{key}"; # scalar inside a hashref
print "$hash->{key}->{moreKeys}->[0]"; # scalar in an array ref in a hashref...
There is one way to do it, though: You can reference and dereference it inside the quoted string, like I do here:
use DateTime;
my $dt = DateTime->now();
print "${\$dt->epoch }"; # both these
print "#{[$dt->epoch]}"; # examples work
But that looks rather ugly, so I would not recommend it. Use your first approach instead!
If you're still interested in how it works, you might also want to look at these Perl FAQs:
What's wrong with always quoting "$vars"?
How do I expand function calls in a string?
From perlref:
Here's a trick for interpolating a subroutine call into a string:
print "My sub returned #{[mysub(1,2,3)]} that time.\n";
The way it works is that when the #{...} is seen in the double-quoted
string, it's evaluated as a block. The block creates a reference to an
anonymous array containing the results of the call to mysub(1,2,3) .
So the whole block returns a reference to an array, which is then
dereferenced by #{...} and stuck into the double-quoted string. This
chicanery is also useful for arbitrary expressions:
print "That yields #{[$n + 5]} widgets\n";
Similarly, an expression that returns a reference to a scalar can be
dereferenced via ${...} . Thus, the above expression may be written
as:
print "That yields ${\($n + 5)} widgets\n";
Stick with the first sample you showed. It looks cleaner and is easier to read.
I'm answering this because it took me a long time to find this and I feel like other people may benefit as well.
For nicer printing of objects use Data::Printer and p():
use DateTime;
use Data::Printer;
my $dt = DateTime->from_epoch( epoch => time );
p($dt);
The PERL translator has limited depth perception within quotes. Removing them should solve the problem. Or just load the real values into a simple variable that you can print within the quotes. Might need to do that if you have objects which contain pointers to other objects:
SwissArmyChainSaw =/= PureMagic:
print("xxx".$this->{whatever}."rest of string\n");
The problem is that $snap is being interpolated inside the string, but $snap is a reference. As perldoc perlref tells us: "Using a reference as a string produces both its referent's type, including any package blessing as described in perlobj, as well as the numeric address expressed in hex."
In other words, within a string, you can't dereference $snap. Your first try was the correct way to do it.
I agree with most comment, stick to concatenation for easy reading. You can use
say
instead of print to spare of using the "\n".

Print the return of a method

I know in php I can do something like this
echo "{$this->method}";
and I swear there was a way to do it in perl
Update:
What I am trying to do is print a scalar that the method returns. I was kind of hoping of doing within the string like in php, just because I'm lazy :P.
Are you just trying to evaluate an arbitrary expression inside a double quoted string? Then maybe you're thinking of
print "#{[$this->method]}";
There is also a trick to call the method in scalar context, but the syntax is a little less clean.
print "${\($this->method)}";
Well, if $this->method outputs a string or a number (like PHP, Perl can automatically convert numbers to strings when required), then you can do print $this->method . "\n";.
If $this->method outputs a data structure (eg an array reference or a hash reference), you can use Data::Dumper to look at the structure of the data. Basically, print Dumper($foo) is the Perl equivalent of PHP's var_dump($foo).
What are you trying to do, exactly?
If $this->method is returning a string, you can do this:
print $this->method . "\n";
without quotes. That will print your string. Sometimes, that can lead to a clumsy looking statement:
print "And we have " . $this->method . " and " . $that->method . " and " . $there->method . "\n";
In that case you can use a little programming trick of:
print "And we have #{[$this->method]} and #{[that->method]} and #{[$their->method]}\n";
Surrounding a function with #{[]} prints out the function's value. Someone explained this to me once, but I can't remember why it works.

Why does this base64 string comparison in Perl fail?

I am trying to compare an encode_base64('test') to the string variable containing the base64 string of 'test'. The problem is it never validates!
use MIMI::Base64 qw(encode_base64);
if (encode_base64("test") eq "dGVzdA==")
{
print "true";
}
Am I forgetting anything?
Here's a link to a Perlmonks page which says "Beware of the newline at the end of the encode_base64() encoded strings".
So the simple 'eq' may fail.
To suppress the newline, say encode_base64("test", "") instead.
When you do a string comparison and it fails unexpectedly, print the strings to see what is actually in them. I put brackets around the value to see any extra whitespace:
use MIME::Base64;
$b64 = encode_base64("test");
print "b64 is [$b64]\n";
if ($b64 eq "dGVzdA==") {
print "true";
}
This is a basic debugging technique using the best debugger ever invented. Get used to using it a lot. :)
Also, sometimes you need to read the documentation for things a couple time to catch the important parts. In this case, MIME::Base64 tells you that encode_base64 takes two arguments. The second argument is the line ending and defaults to a newline. If you don't want a newline on the end of the string you need to give it another line ending, such as the empty string:
encode_base64("test", "")
Here's an interesting tip: use Perl's wonderful and well-loved testing modules for debugging. Not only will that give you a head start on testing, but sometimes they'll make your debugging output a lot faster. For example:
#!/usr/bin/perl
use strict;
use warnings;
use Test::More 0.88;
BEGIN { use_ok 'MIME::Base64' => qw(encode_base64) }
is( encode_base64("test", "dGVzdA==", q{"test" encodes okay} );
done_testing;
Run that script, with perl or with prove, and it won't just tell you that it didn't match, it will say:
# Failed test '"test" encodes okay'
# at testbase64.pl line 6.
# got: 'gGVzdA==
# '
# expected: 'dGVzdA=='
and sharp-eyed readers will notice that the difference between the two is indeed the newline. :)

How can I eval environment variables in Perl?

I would like to evaluate an environment variable and set the result to a variable:
$x=eval($ENV{EDITOR});
print $x;
outputs:
/bin/vi
works fine.
If I set an environment variable QUOTE to \' and try the same thing:
$x=eval($ENV{QUOTE});
print $x;
outputs:
(nothing)
$# set to: "Can't find a string terminator anywhere before ..."
I do not wish to simply set $x=$ENV{QUOTE}; as the eval is also used to call a script and return its last value (very handy), so I would like to stick with the eval(); Note that all of the Environment variables eval'ed in this manner are set by me in a different place so I am not concerned with malicious access to the environment variables eval-ed in this way.
Suggestions?
Well, of course it does nothing.
If your ENV varaible contains text which is half code, but isn't and you give the resulting string to something that evaluates that code as Perl, of course it's not going to work.
You only have 3 options:
Programmatically process the string so it doesn't have invalid syntax in it
Manually make sure your ENV variables are not rubbish
Find a solution not involving eval but gives the right result.
You may as well complain that
$x = '
Is not valid code, because that's essentially what's occurring.
Samples of Fixing the value of 'QUOTE' to work
# Bad.
QUOTE="'" perl -wWe 'print eval $ENV{QUOTE}; print "$#"'
# Can't find string terminator "'" anywhere before EOF at (eval 1) line 1.
# Bad.
QUOTE="\'" perl -wWe 'print eval $ENV{QUOTE}; print "$#"'
# Can't find string terminator "'" anywhere before EOF at (eval 1) line 1.
# Bad.
QUOTE="\\'" perl -wWe 'print eval $ENV{QUOTE}; print "$#"'
# Can't find string terminator "'" anywhere before EOF at (eval 1) line 1.
# Good
QUOTE="'\''" perl -wWe 'print eval $ENV{QUOTE}; print "$#"'
# '
Why are you eval'ing in the first place? Should you just say
my $x = $ENV{QUOTE};
print "$x\n";
The eval is executing the string in $ENV{QUOTE} as if it were Perl code, which I certainly hope it isn't. That is why \ disappears. If you were to check the $# variable you would find an error message like
syntax error at (eval 1) line 2, at EOF
If you environment variables are going to contain code that Perl should be executing then you should look into the Safe module. It allows you to control what sort of code can execute in an eval so you don't accidentally wind up executing something like "use File::Find; find sub{unlink $File::Find::file}, '.'"
Evaluating an environment value is very dangerous, and would generate errors if running under taint mode.
# purposely broken
QUOTE='`rm system`'
$x=eval($ENV{QUOTE});
print $x;
Now just imagine if this script was running with root access, and was changed to actually delete the file system.
Kent's answer, while technically correct, misses the point. The solution is not to use eval better, but to not use eval at all!
The crux of this problem seems to be in understanding what eval STRING does (there is eval BLOCK which is completely different despite having the same name). It takes a string and runs it as Perl code. 99.99% this is unnecessary and dangerous and results in spaghetti code and you absolutely should not be using it so early in your Perl programming career. You have found the gun in your dad's sock drawer. Discovering that it can blow holes in things you are now trying to use it to hang a poster. It's better to forget it exists, your code will be so much better for it.
$x = eval($ENV{EDITOR}); does not do what you think it does. I don't even have to know what you think it does, that you even used it there means you don't know. I also know that you're running with warnings off because Perl would have screamed at you for that. Why? Let's assume that EDITOR is set to /bin/vi. The above is equivalent to $x = /bin/vi which isn't even valid Perl code.
$ EDITOR=/bin/vi perl -we '$x=eval($ENV{EDITOR}); print $x'
Bareword found where operator expected at (eval 1) line 1, near "/bin/vi"
(Missing operator before vi?)
Unquoted string "vi" may clash with future reserved word at (eval 1) line 2.
Use of uninitialized value $x in print at -e line 1.
I'm not sure how you got it to work in the first place. I suspect you left something out of your example. Maybe tweaking EDITOR until it worked?
You don't have to do anything magical to read an environment variable. Just $x = $ENV{EDITOR}. Done. $x is now /bin/vi as you wanted. It's just the same as $x = $y. Same thing with QUOTE.
$ QUOTE=\' perl -wle '$x=$ENV{QUOTE}; print $x'
'
Done.
Now, I suspect what you really want to do is run that editor and use that quote in some shell command. Am I right?
Well, you could double-escape the QUOTE's value, I guess, since you know that it's going to be evaled.
Maybe what you want is not Perl's eval but to evaluate the environment variable as the shell would. For this, you want to use backticks.
$x = `$ENV{QUOTE}`