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).
Related
I have a script in Perl that is reading a file. At some point, the code utilize the following if statement inside a for loop:
for (my $i = 0; $i<10 ; $i++ ) {
$_ = <INPUT>;
if (!$_) {last;}
...
I am new in Perl, so I would like to know the meaning of !$_. In this example, $_ is a line of my file. So, what content the line should have to the if statement be true.
The if condition, what is inside (), is evaluated in a boolean scalar context to be tested for "truthiness." So if it's undef or '' (empty string) or 0 (or string "0") it's false.
That ! negates what follows it, so if (!$_) is true if $_ is false (undef or '' or 0 or "0"). However, in this case that $_ is assigned from <> operator so it'll always have a linefeed at the end -- unless the source for <> was exhausted in which case <> returns undef.
So, in this case, that if (!$_) tests for whether there is nothing more to read from INPUT, and exits the for loop with last if that is the case.
A few comments on the shown code.
That C-style for loop can also be written as for my $i (0..9), what is considered far nicer and more readable.† See foreach, and really follow links for flow-control key-words
The piece of code
$_=<INPUT>
if (!$_) { last; }
...
reads from INPUT filehandle and exits its loop (see last) once there is end-of-file. (That need not be an actual file but any resource readable via a filehandle.)
This is clumsy, to say the least; a common way of doing it is
while (<INPUT>) {
...
}
† So much so that even hard-core compiled languages now have it. The C++11 introduced the range-based for loop
for (auto var: container) ... // (really, const auto&), or auto&, or auto&&
and the standard reference linked above says
Used as a more readable equivalent to the traditional for loop [...]
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.
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
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.
Here is my code:
#!/usr/bin/env perl
sub start
{
...
}
sub stop
{
...
}
if (defined $ARGV[0])
{
if ($ARGV[0]=='start') {start}
elsif ($ARGV[0]=='stop') {stop}
else {die "Unrecognized command: $ARGV[0]"}
}
Whatever I do though, it always executes &start. Am I doing something wrong?
I'm using Linux Mint 10 and Perl 5.10.1.
You're using numeric comparison to check string equality, which converts the arguments to numbers first before comparing them.
In this case, 'start' is not a number, so it gets converted to 0; the value in $ARGV[0] (which is expected to be a word here) is also converted, resulting in another 0 and a final condition of if (0 == 0), which is always true.
You want to use the eq string-wise comparison operator instead:
if ($ARGV[0] eq 'start') { start }
See the docs for more information on the different comparison operators.
Note that (as has been pointed out in the comments) including use warnings; at the top of your script would have caused perl to warn you about this from the start. Always including use warnings; and use strict; is generally best practice, as it helps catch these kinds of errors early.
The bug is that your program neglected to start with the standard boilerplate for Perl code.