I am making a concatenating do-while loop wherein each of the entered strings will be added to the main string $food. Here is the code:
do {
print("\nEnter your order(-1 to end): ");
$order = <>;
chop($order);
if ($order != -1) {
$food .= " ".$order;
print($food);
}
} while ( $order != -1)
print ($food); #PROBLEM HERE!
The problem is that whenever I put print ($food) outside the loop, a syntax error for this line appears plus an Execution of file.pl aborted due to compilation errors message.
The code works once I put print ($food) inside the loop but I am curious to why this error is happening.
Your line before last print has syntax error - it does not end with ;.
Just add it and it should work.
A thing worth mentioning in addition to the syntax error: I see that you you start the string with a newline. While this is fine, it might be considered "better" to end strings with the newline before printing, as printing a newline would result in flushing the STDOUT buffer right away. In most cases, this is the behavior you want, as it prevents delayed output and text flickering (especially if printing a lot).
Related
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
This is a extract from my code :
foreach my $nouveau_code (#tableau_codes_utilises) {
my $code_found = 0;
foreach my $ligne_code_prospect (#tableau_liste_codes_prospects) {
//SOME CODE
}
print "HERE IS $nouveau_code\n";
if ( $code_found == 0 ) {
print "le code $nouveau_code n'a pas été trouvé\n";
STDOUT->autoflush;
}
}
And the result of an iteration where $code_found is equal to 0 is (code is hidden here, but is alphanumeric) :
HERE IS PAR****
n'a pas été trouvé
I do not understand why the second print (in the if sequence) does not print the whole line. I guess there is a buffering problem but I cannot find what is going on.
Any help would be appreciated.
At a guess, $nouveau_code has come from a file that originated on a Windows system and you are processing it on Linux?
If you use chomp in such a circumstance then it will remove the trailing newline, but not the carriage return, and printing the value will reset the screen cursor to the beginning of the line before printing anything else
The simple fix is to replace chomp with s/\R//, which will remove the CR as well as the LF on the end of the line
I suspect what is happening here is that your second data item contains backspaces (BS) or a carriage return (CR). So
your first item runs and prints out the first time, but since code_found=1 it doesn't go into the if.
your second item runs and the CR or BS in the data obscure the fact that the HERE IS was printed
your second item does not have a code found so the second print happens and the first part of it is overwritten by the same CR or BS sequence in your variable
The easy way to confirm this is to redirect your output and look at it like so:
script > check.log
vim check.log
and you should see the extra data in there. less would work fine in place of vim also.
Here is a bit of code from a program I'm writing, the syntax error lays in the 3rd line.
if($header gt $word{
$wordlist{$word} = $header;
$header = $word;
return;
}
In short - you're missing a closing parenthesis on the first line
It's quite funny actually because you'd expect Perl to point you to the right location with its error message. However, due to a stroke of bad luck it seems just like the beginning of a perfectly valid code that just happens to do something else than what you intended.
Perl actually thinks you look up a hash called %word (using $word{...} with the value of the assignment evaluated as key).
So, this would have been a perfectly valid code if you'd have done this:
if ($header gt $word{
$wordlist{$word} = $header # removed the ;
}) { # closed the condition
$header = $word;
return;
}
Perl is only confused once it reaches the end of the second line and sees the ;
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).