I have the silly question
given($num){
...
when($num >= 13 and $num <= 99) return "$num_in_eng{$num}teen" =~ s/tt/t/;
...
}
but it gave an error
"Can't modify the string in substitution near s/tt/t/g;
$_ = "$num_in_eng{$num}teen";
s/tt/t/;
This work fine
You can't assign to a string literal. It wouldn't make sense to say "foo" = "bar" either.
If you try to use the match operator =~ with a substitution (for example, a regex term starting with s, like in s/foo/bar/), the result of the substitution will be assigned to the left hand side. If that is a literal, you will get this error.
It does not matter that you actually used string interpolation to create the literal.
Using a local variable and substituting its contents like you have shown the right solution. You can also use a named variable instead of the "default" $_:
$foo = whatever();
$foo =~ s/tt/t/;
return $foo;
Related
I'm trying to truncate a string in a select input option using perl if it is longer than a set value, though i can't get it to work correctly.
my $value = defined $option->{value} ? $option->{value} : '';
my $maxValueLength = 50;
if ($value.length > $maxValueLength) {
$value = substr $value, 0, $maxValueLength + '...';
}
Another option is regex
$string =~ s/.{$maxLength}\K.*/.../;
It matches any character (.) given number of times ({N}, here $maxLength), what is the first $maxLength characters in $string; then \K makes it "forget" all previous matches so those won't get replaced later. The rest of the string that is matched is then replaced by ...
See Lookaround assertions in perlre for \K.
This does start the regex engine for a simple task but it doesn't need any conditionals -- if the string is shorter than the maximum length the regex won't match and nothing happens.
Your code has several syntax errors. Turn on use strict and use warnings if you don't have it, and then read the error messages it tells you about. This is a bit tricky because of Perl's very complex syntax (see also Damian Conway's keynote from the 2020 Perl and Raku Conference), but it boils down to these:
Use of uninitialized value in concatenation (.) or string at line 7
Argument "..." isn't numeric in addition (+) at line 8
I've used the following adaption of your code to produce these
use strict;
use warnings;
my $value = '1234567890' x 10;
my $maxValueLength = 50;
if ( $value.length > $maxValueLength ) {
$value = substr $value, 0, $maxValueLength + '...';
}
print $value;
Now let's see what they mean.
The . operator in Perl is a concatenation. You cannot use it to call methods, and length is not a method on a string. Perl thinks you are using the built-in length (a function, not a method) without an argument, which makes it default to $_. Most built-ins do this, to make one-liners shorter. But $_ is not defined. Now the . tries to concatenate the length of undef to $value. And using undef in a string operation leads to this warning.
The correct way of doing this is length $value (or with parentheses if you prefer them, length($value)).
The + operator is not concatenation (we just learned that the . is). It's a numerical addition. Perl is pretty good at converting between strings and numbers as there aren't really any types, so saying 1 + "5" would give you 6 without problems, but it cannot do that for a couple of dots in a string. Hence it complains about a non-number value in an addition.
You want the substring with a given length, and then you want to attach the three dots. Because of associativity (or stickyness) of operators you will need to use parentheses () for your substr call.
$value = substr($value, 0, $maxValueLength) . '...';
To find a length of the string use length(STRING)
Here is the code snippet how you can modify the script.
#!/usr/bin/perl
use strict;
use warnings;
use feature qw(say);
my $string = "abcdefghijklmnopqrstuvwxyz abcdefghijklmnopqrstuvwxyz abcdefghijklmnopqrstuvwxyz";
say "length of original string is:".length($string);
my $value = defined $string ? $string : '';
my $maxValueLength = 50;
if (length($value) > $maxValueLength) {
$value = substr $value, 0, $maxValueLength;
say "value:$value";
say "value's length:".length($value);
}
Output:
length of original string is:80
value:abcdefghijklmnopqrstuvwxyz abcdefghijklmnopqrstuvw
value's length:50
I am trying to do delete some characters matching a regex in perl and when I do that it returns an integer value.
I have tried substituting multiple spaces in a string with empty string or basically deleting the space.
#! /usr/intel/bin/perl
my $line = "foo/\\bar car";
print "$line\n";
#$line = ~s/(\\|(\s)+)+//; <--Ultimately need this, where backslash and space needs to be deleted. Tried this, returns integer value
$line = ~s/\s+//; <-- tried this, returns integer value
print "$line\n";
Expected results:
First print: foo/\bar car
Second print: foo/barcar
Actual result:
First print: foo/\\bar car
Second print: 18913234908
The proper solution is
$line =~ s/[\s\\]+//g;
Note:
g flag to substitute all occurrences
no space between = and ~
=~ is a single operator, binding the substitution operator s to the target variable $line.
Inserting a space (as in your code) means s binds to the default target, $_, because there is no explicit target, and then the return value (which is the number of substitutions made) has all its bits inverted (unary ~ is bitwise complement) and is assigned to $line.
In other words,
$line = ~ s/...//
parses as
$line = ~(s/...//)
which is equivalent to
$line = ~($_ =~ s/...//)
If you had enabled use warnings, you would've gotten the following message:
Use of uninitialized value $_ in substitution (s///) at prog.pl line 6.
You've already accepted an answer, but I thought it would be useful to give you a few more details.
As you now know,
$line = ~s/\s+//;
is completely different to:
$line =~ s/\s+//;
You wanted the second, but you typed the first. So what did you end up with?
~ is "bitwise negation operator". That is, it converts its argument to a binary number and then bit-flips that number - all the zeroes become ones and all the ones become zeros.
So you're asking for the bitwise negation of s/\s+//. Which means the bitwise negation works on the value returned by s/\s+//. And the value returned by a substitution is the number of substitutions made.
We can now work out all of the details.
s/\s+// carries out your substitution and returns the number of substitutions made (an integer).
~s/\s+// returns the bitwise negation of the integer returned by the substitution (which is also an integer).
$line = ~s/\s+// takes that second integer and assigns it to the variable $line.
Probably, the first step returns 1 (you don't use /g on your s/.../.../, so only one substitution will be made). It's easy enough to get the bitwise negation of 1.
$ perl -E'say ~1'
18446744073709551614
So that might well be the integer that you're seeing (although it might be different on a 32-bit system).
I'm using Perl 5.16.2 to try to count the number of occurrences of a particular delimiter in the $_ string. The delimiter is passed to my Perl program via the #ARGV array. I verify that it is correct within the program. My instruction to count the number of delimiters in the string is:
$dlm_count = tr/$dlm//;
If I hardcode the delimiter, e.g. $dlm_count = tr/,//; the count comes out correctly. But when I use the variable $dlm, the count is wrong. I modified the instruction to say
$dlm_count = tr/$dlm/\t/;
and realized from how the tabs were inserted in the string that the operation was substituting every instance of any of the four characters "$", "d", "l", or "m" to \t — i.e. any of the four characters that made up my variable name $dlm.
Here is a sample program that illustrates the problem:
$_ = "abcdefghij,klm,nopqrstuvwxyz";
my $dlm = ",";
my $dlm_count = tr/$dlm/\t/;
print "The count is $dlm_count\n";
print "The modified string is $_\n";
There are only two commas in the $_ string, but this program prints the following:
The count is 3
The modified string is abc efghij,k ,nopqrstuvwxyz
Why is the $dlm token being treated as a literal string of four characters instead of as a variable name?
You cannot use tr that way, it doesn't interpolate variables. It runs strictly character by character replacement. So this
$string =~ tr/a$v/123/
is going to replace every a with 1, every $ with 2, and every v with 3. It is not a regex but a transliteration. From perlop
Because the transliteration table is built at compile time, neither the SEARCHLIST nor the REPLACEMENTLIST are subjected to double quote interpolation. That means that if you want to use variables, you must use an eval():
eval "tr/$oldlist/$newlist/";
die $# if $#;
eval "tr/$oldlist/$newlist/, 1" or die $#;
The above example from docs hints how to count. For $dlms in $string
$dlm_count = eval "\$string =~ tr/$dlm//";
The $string is escaped so to not be interpolated before it gets to eval. In your case
$dlm_count = eval "tr/$dlm//";
You can also use tools other than tr (or regex). For example, with string being in $_
my $dlm_count = grep { /$dlm/ } split //;
When split breaks $_ by the pattern that is empty string (//) it returns the list of all characters in it. Then the grep block tests each against $dlm so returning the list of as many $dlm characters as there were in $_. Since this is assigned to a scalar, $dlm_count is set to the length of that list, which is the count of all $dlm.
In the section of the docs on perlop 'Quote Like Operators', it states:
Because the transliteration table is built at compile time, neither
the SEARCHLIST nor the REPLACEMENTLIST are subjected to double quote
interpolation. That means that if you want to use variables, you must
use an eval():
As documented and as you discovered, tr/// doesn't interpolate. The simple solution is to use s/// instead.
my $dlm = ",";
$_ = "abcdefghij,klm,nopqrstuvwxyz";
my $dlm_count = s/\Q$dlm/\t/g;
If the transliteration is being performed in a loop, the following might speed things up noticeably:
my $dlm = ",";
my $tr = eval "sub { tr/\Q$dlm\E/\\t/ }";
for (...) {
my $dlm_count = $tr->();
...
}
Although several answers have hinted at the eval() idiom for tr///, none have the form that covers cases where the string has tr syntax characters in it, e.g.- (hyphen):
$_ = "abcdefghij,klm,nopqrstuvwxyz";
my $dlm = ",";
my $dlm_count = eval sprintf "tr/%s/%s/", map quotemeta, $dlm, "\t";
But as others have noted, there are lots of ways to count characters in Perl that avoid eval(), here's another:
my $dlm_count = () = m/$dlm/go;
$value = $list[1] ~ s/\D//g;
syntax error at try1.pl line 53, near "] ~"
Execution of try1.pl aborted due to compilation errors.
I am trying to extract the digits from the second element of #list, and store it into $value.
You mean =~, not ~. ~ is a unary bitwise negation operator.
A couple of ways to do this:
($value) = $list[1] =~ /(\d+)/;
Both sets of parens are important; only if there are capturing parentheses does the match operation return actual content instead of just an indication of success, and then only in list context (provided by the list-assign operator ()=).
Or the common idiom of copy and then modify:
($value = $list[1]) =~ s/\D//;
maybe you wanted the =~ operator?
P.S. note that $value will not get assigned the resulting string (the string itself is changed in place). $value will get assigned the number of substitutions that were made
You said in a comment that are trying to get rid of non-digits. It looks like you are trying to preserve the old value and get the modified value in a new variable. The Perl idiom for that is:
( my $new = $old ) =~ s/\D//g;
And wanted \digits not non-\Digits. And have a superfluous s/ubstitute operator where a match makes more sense.
if ($list[1] =~ /(\d+)/) {
$value = $1;
}
If I had:
$foo= "12."bar bar bar"|three";
how would I insert in the text ".." after the text 12. in the variable?
Perl allows you to choose your own quote delimiters. If you find you need to use a double quote inside of an interpolating string (i.e. "") or single quote inside of a non-interpolating string (i.e. '') you can use a quote operator to specify a different character to act as the delimiter for the string. Delimiters come in two forms: bracketed and unbracketed. Bracketed delimiters have different beginning and ending characters: [], {}, (), [], and <>. All other characters* are available as unbracketed delimiters.
So your example could be written as
$foo = qq(12."bar bar bar"|three);
Inserting text after "12." can be done many ways (TIMTOWDI). A common solution is to use a substitution to match the text you want to replace.
$foo =~ s/^(12[.])/$1../;
the ^ means match at the start of the sting, the () means capture this text to the variable $1, the 12 just matches the string "12", and the [] mean match any one of the characters inside the brackets. The brackets are being used because . has special meaning in regexes in general, but not inside a character class (the []). Another option to the character class is to escape the special meaning of . with \, but many people find that to be uglier than the character class.
$foo =~ s/^(12\.)/$1../;
Another way to insert text into a string is to assign the value to a call to substr. This highlights one of Perl's fairly unique features: many of its functions can act as lvalues. That is they can be treated like variables.
substr($foo, 3, 0) = "..";
If you did not already know where "12." exists in the string you could use index to find where it starts, length to find out how long "12." is, and then use that information with substr.
Here is a fully functional Perl script that contains the code above.
#!/usr/bin/perl
use strict;
use warnings;
my $foo = my $bar = qq(12."bar bar bar"|three);
$foo =~ s/(12[.])/$1../;
my $i = index($bar, "12.") + length "12.";
substr($bar, $i, 0) = "..";
print "foo is $foo\nbar is $bar\n";
* all characters except whitespace characters (space, tab, carriage return, line feed, vertical tab, and formfeed) that is
If you want to use double quotes in a string in Perl you have two main options:
$foo = "12.\"bar bar bar\"|three";
or:
$foo = '12."bar bar bar"|three';
The first option escapes the quotes inside the string with backslash.
The second option uses single quotes. This means the double quotes are treated as part of the string. However, in single quotes everything is literal so $var or #array isn't treated as a variable. For example:
$myvar = 123;
$mystring = '"$myvar"';
print $mystring;
> "$myvar"
But:
$myvar = 123;
$mystring = "\"$myvar\"";
print $mystring;
> "123"
There are also a large number of other Quote-like Operators you could use instead.
$foo = "12.\"bar bar bar\"|three";
$foo =~s/12\./12\.\.\./;
print $foo; # results in 12...\"bar bar bar\"|three"