Perl else error - perl

Hello I am new to programming in perl I am trying to make a number adder (math) but it gives me 1 error here's my code:
sub main {
print("First: ");
$num1 = <STDIN>;
print("Second: ");
$num2 = <STDIN>;
$answer = $num1 + $num2;
print("$answer")
} else {
print("You have entered invalid arguments.")
}
main;
now obviously its not done but I get an error on ONLY else here is the error:
C:\Users\aries\Desktop>"Second perl.pl"
syntax error at C:\Users\aries\Desktop\Second perl.pl line 9, near "} else"
Execution of C:\Users\aries\Desktop\Second perl.pl aborted due to compilation errors.
please help (also I tried googling stuff still error)

Since you're new to Perl, I recommend you add strict and warnings at the top of your scripts. This will help identify common problems and potentially dangerous code.
The main problem with your code is that you've appended the else statement to your subroutine. Here is an example of your code as I think you intended it to be:
use strict;
use warnings;
use Scalar::Util qw(looks_like_number);
sub main {
print 'First :';
my $num1 = <STDIN>;
print 'Second: ';
my $num2 = <STDIN>;
if( looks_like_number($num1) && looks_like_number($num2) ) {
my $answer = $num1 + $num2;
print "$answer\n";
}
else {
die 'You have entered invalid arguments.';
}
}
main();
There are a few things I should note here about the differences between Perl and Python that I think will help you understand Perl better.
Unlike Python, Perl doesn't care about white space. It uses curly braces and semi-colons to indicate the end of blocks and statements and will happily ignore any white space that isn't in quotes.
The else statement appended to the subroutine won't work because Perl isn't designed to evaluate code blocks that way. A subroutine is simply a named code block that can be called at any other point in the script. Any error handling will need to be done inside of the subroutine rather than to it.
Perl is very context-sensitive and doesn't make a solid distinction between strings and integers in variables. If you write $var_a = 1, Perl will read it as an integer. $var_b = '1', it will read it as a string. But, you can still add them together: $var_c = ($var_a + $var_b), and Perl will make $var_c = 2.
This is another reason the else statement would not work as you've written it. Python would throw an error if you try to add non-integers, but Perl will just figure out how to combine them, and give you the result. If you try to add a letter and a number, Perl still won't fail, but it will warn you if you put "use warnings;" at the top of your script.
In the example, and as Dada mentioned, you can use the looks_like_number() method from the Scalar::Utils module to evaluate the variables as you had intended.
Apart from the syntax, if/else statements work the same way in Perl as in Python. The else-if is slightly different as is has an extra s:
if (condition) {
...
}
elsif (other condition) {
...
}
else {
...
}
In Perl, it's good practice to assign lexical scope to variables using the my function. Since Perl is so context-sensitive, this can help prevent unexpected behavior when moving between different scopes.
Single- and double-quotes have different uses in Perl. Single-quotes are read literally and double-quotes are interpolated, so if you want to combine variables and text together, you can skip concatenating the strings and just do: print "Got variable: $var\n";
Lastly, note that I added parentheses after the main subroutine call. This is another best practice to make it clearer that you are calling a subroutine as opposed to it being a bare word or a bad variable name.

Related

How do I compare two strings using Perl

I have googled until I ran out of options. I must be doing something wrong but I cannot figure out what. (obviously I am new to Perl and inherited this project).
I am just trying to see if a variable is equal to 'Y'. As you can see towards the bottom of the code below, I have tried 'eq', '=' and '=='. Other than setting a new variable equal to the result of an 'eq' operation, all the attempts result in the page blowing up. The result of the string test ($stringtest) appears to be empty and I would not know how to test it if it wasn't.
#!/usr/bin/perl -w
#######################################################################
# Test script to show how we can pass in the needed information
# to a web page and use it to run a program on the webserver and
# redirect its output to a text file.
#######################################################################
require 5; #Perl 5 or greater
require "jwshttputil.pl";
require "jwsdbutil.pl";
# Parse QUERY_STRING
&HTTPGet;
# Parse POST_STRING - NOTE: This CLEARS the post data
&HTTPPost;
print STDOUT "Content-type: text/html\n\n";
print STDOUT "<HTML><BODY>";
$canrun = "Y";
$is4ge = $Data{'is4ge'};
$sYes = "Y";
#Step 1: Check for needed values
if ( !($Data{'user'}) )
{
print "user was not found<br/>";
$canrun = "N";
}
if ( !($Data{'pwd'}) )
{
print "pwd was not found<br/>";
$canrun = "N";
}
if ( !($Data{'is4ge'}) )
{
print "is4ge was not found<br/>";
$canrun = "N";
}
print "$Data{'is4ge'}"; #prints Y
print $is4ge; #prints Y
if ( !($Data{'db'}) )
{
print "db was not found<br/>";
#if (!($is4ge = "Y")) #dies
# $canrun = "N";
#if (!($is4ge eq "Y")) #dies
# $canrun = "N";
#$stringtest = ($is4ge eq $sYes);
#print $stringtest; #displays nothing
#if (($is4ge == $sYes)) #dies
# $canrun = "N";
}
print STDOUT "</BODY></HTML>";
You have a good answer to your question, but it might be useful to point out how you could have investigated this yourself.
Firstly, your original title for this question was "How do I compare two strings in a CGI page under Apache?" That title has been corrected because this problem has nothing to do with CGI or Apache, it's simply a misunderstanding about Perl syntax. I know you couldn't have known that initially, but a good way to investigate strange errors like this is to eliminate as many of the complications as possible. So forget about CGI and web servers - just write a simple Perl program.
#!/usr/bin/perl
$canrun = 'N';
$is4ge = 'Y';
if (!$is4ge eq 'Y')
$canrun = 'Y';
print $canrun;
Running this, we get:
$ perl testif
Scalar found where operator expected at testif line 7, near ")
$canrun"
(Missing operator before $canrun?)
syntax error at testif line 7, near ")
$canrun "
Execution of testif aborted due to compilation errors.
That makes it clear that the problem is the syntax around lines 6 and 7. That might be enough to send you off to the Perl syntax manual page, where you would learn that condition statements in Perl always require braces around the block. If not, you could add the diagnostics pragma to get more information.
So as you've already been told, the correct format for your test is
if ($is4ge ne 'Y') {
$canrun = 'Y';
}
But there are a few other things in this program that will make it hard to maintain in the future. Basically, it is rather falling behind with Perl best practices.
You use -w when that was replaced with use warnings in Perl 5.6.0 (released in 2000)
You don't have use strict in the code (that will point out many bad practices in your code which you'll want to fix - in particular, you'll need to declare your variables).
The two libraries that are required at the top of the program might well be better rewritten as proper Perl modules.
I worry that jwshttpdutil.pl might be a home-baked CGI parser that should be replaced with CGI.pm.
The embedded HTML in the code should be replaced with a templating engine of some kind.
Also, I recommend reading CGI::Alternatives to bring yourself up to date with modern Perl methods for writing web applications. You are using techniques that were out of date before the end of the last millennium.
Curlies around the body of flow control statements are not optional in Perl.
eq is indeed the string comparison operator in Perl.
So,
if (!($is4ge eq "Y")) {
$canrun = "N";
}
Or just
if ($is4ge ne "Y") {
$canrun = "N";
}
You really should be using 1 and 0 instead of 'Y' and 'N', since they are true and a false values that can more easily be tested.

Definition of empty {} after If() statement in eval{} statement

I am currently attempting to document a Perl script in preparation for converting it to .NET. I have no prior experience in Perl before now, however I was managing to get through it with a lot of Google-fu. I have run into a single line of code that has stopped me as I am unsure of what it does. I've been able to figure out most of it, but I'm missing a piece and I don't know if it's really that important. Here is the line of code:
eval { if(defined $timeEnd && defined $timeStart){}; 1 } or next;
I know that defined is checking the variables $timeEnd and $timeStart to see if they are null/nothing/undef. I also believe that the eval block is being used as a Try/Catch block to trap any exceptions. The line of code is in a foreach loop so I believe the next keyword will continue on with the next iteration of the foreach loop.
The part I'm having difficulty deciphering is the {};1 bit. My understanding is that the ; is a statement separator in Perl and since it's not escaped with a backslash, I have no idea what it is doing there. I also don't know what the {} means. I presume it has something to do with an array, but it would be an empty array and I don't know if it means something special when it is directly after an if() block. Lastly, I no idea what a single integer of 1 means and is doing there at the end of an eval block.
If someone could break that line of code down into individual parts and their definitions, I would greatly appreciate it.
Bonus: If you can give me a .NET conversion, and how each Perl bit relates to it, I will most certainly give you my internet respects. Here's how I would convert it to VB.NET with what I know now:
For each element in xmlList 'This isn't in the Perl code I posted, but it's the `foreach` loop that the code resides in.
Try
If Not IsNothing(timeEnd) AND Not IsNothing(timeStart) then
End If
Catch ex as Exception
Continue For
End Try
Next
Ignoring elsif and else clasuses, the syntax of an if statement is the following:
if (EXPR) BLOCK
The block is executed if the EXPR evaluates to something true. A block consists of a list of statements in curly braces. The {} in your code is the block of the if statement.
It's perfectly valid for blocks to be empty (to contain a list of zero statements). For example,
while (s/\s//) { }
is an inefficient way of doing
s/\s//g;
The thing is, the condition in the following has no side-effects, so it's quite useless:
if(defined $timeEnd && defined $timeStart){}
It can't even throw an exception![1] So
eval { if(defined $timeEnd && defined $timeStart){}; 1 } or next;
is equivalent to
eval { 1 } or next;
which is equivalent to[2]
1 or next;
which is equivalent to
# Nothing to see here...
Technically, it can if the variables are magical.
$ perl -MTie::Scalar -e'
our #ISA = "Tie::StdScalar";
tie(my $x, __PACKAGE__);
sub FETCH { die }
defined($x)
'
Died at -e line 4.
I doubt the intent is to check for this.
Technically, it also clears $#.
eval{} returns result of last expresion (1 in your example) or undef if there was an exception. You can write same code as,
my $ok = eval {
if (defined $timeEnd && defined $timeStart){};
1
};
$ok or next;
From perldoc -f eval
.. the value returned is the value of the last expression evaluated inside the mini-program; a return statement may be also used, just as with subroutines.
If there is a syntax error or runtime error, or a die statement is executed, eval returns undef in scalar context or an empty list in list context, and $# is set to the error message

2 Sub references as arguments in perl

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

Perl script for creating two arrays

Input: A list of numbers on command line
Output: Two lists of numbers ,one with input numbers that are greater than zero and one with those that are less than zero (Ignoring zero valued numbers)
here is my code
#!/usr/bin/perl
$i++ = 0;
$j++ = 0;
while ($number = <>)
{
if($number<0)
$first[$i++]=$number;
else
$second[$j++]=$number;
}
print "The numbers with value less than zero are\n";
foreach $number (#first)
print $number;
print "The numbers with value greater than zero are\n"
foreach $number(#second)
print $number;
I am getting the following silly errors which i am not able to rectify.The errors are
divide.pl: 2: ++: not found
divide.pl: 3: ++: not found
divide.pl: 5: Syntax error: ")" unexpected
Can anybody help me out with rectifying these errors please? I am new to perl script
Curly braces on compound statements are not optional in Perl.
Your statements:
$i++=0;
$j++=0;
don't make sense; you probably just want to delete the "++".
You're missing a semicolon on one of your print statements.
Once you've got those problems fixed, you should add
use strict;
use warnings;
after the #! line. This will introduce more error messages; you'll need to fix those as well. For example, you'll need to declare your variables using my().
The code you present will hardly compile. Loops should have {} around the main block, arrays are better created with push (or unshift), you should use strict and warnings, and you can't do increments at the same time as assignments (e.g. $i++ = 0).
use v5.10;
use strict;
use warnings;
my (#first, #second);
while (<STDIN>) { # <STDIN> clearer than <> in this case
chomp;
if ($_ < 0) {
push #first, $_;
} elsif ($_ > 0) {
push #second, $_;
}
}
say "Numbers less than zero:";
say "#first";
say "Numbers greater than zero:";
say "#second";
I don't know what $i++ = 0 is supposed to mean, but change that to $i = 0 to initialize the variables.
Also, the first thing yuu should do in the while loop is call chomp($number) to remove spurious newlines - 5\n is not a number and treating it as one will confuse perl.
Once you've fixed that, post any new errors that show up - I don't see any other problems though.
How are you executing this perl script? Beyond the errors mentioned about the code itself. It looks like you are attempting to evaluate the code using dash instead of perl.
The errors you should be seeing if you were executing it with Perl would be like:
Can't modify postincrement (++) in scalar assignment at /tmp/foo.pl
line 2, near "0;"
But instead, your errors are more in line with what dash outputs:
$ dash /tmp/foo.pl
/tmp/foo.pl: 2: ++: not found
/tmp/foo.pl: 3: ++: not found
Once you've verified that you are running your perl script properly you can start working through the other problems people have mentioned your code. The easiest way to do this is to run it via perl divide.pl instead of whatever you are doing.

Perl while loop / reading file

I'm new to Perl and I have a problem when reading a file line by line. I started with a tutorial which suggested using a while loop. That worked fine, however I wanted to have an option to break out of the loop in case of an error. I know I can use the "last" keyword, but I don't like that very much, I would like to include it in the while loop's boolean expression so it is easier to see on what conditions it stops. So I did
$error=0;
while ( (!$error) && (<MYFILE>) ) {
print $_;
...
}
Unfortunately, this doesn't work because $_ seems to contain a "uninitialized value". For some reason, when I change (!$error) to (!0), it works again. I would understand that it was not working had I used || because of lazy evaluation, but with && in this case both sides need to be evaluated so I don't understand why it doesn't initialize the line variable.
The magical assignment to $_ only occurs if there is nothing else in the loop:
while (<MYFILE>) {
print $_;
...
}
If you want the error test too - which is not recommended - then you need to do the assignment:
my $error = 0;
while (!$error && ($_ = <MYFILE>)) {
print;
$error = 1 if m/quit/i;
}
Don't forget to add:
use warnings;
use strict;
The reason why the version with '!0' worked is probably because Perl optimizes and recognizes that '!0' is always true and removes it from the loop, leaving just the I/O operator, which then assigns to $_.
NB: it is far better to use idiomatic Perl and the last statement:
while (<MYFILE>) {
print;
last if m/quit/i;
}
First, as Jonathan points out you should always write languages idiomatically unless you have a very good reason and here you don't.
Your larger problem is misunderstanding the while loop. The automatic assignment of the result of the angle operator to $_ only occurs if it is the only thing in the while conditional test. If there are multiple items in the conditional test they are evaluated and then discarded (Programming Perl pp. 80-81 and I/O Operators at Perldoc).