Orabîg Orabîg - 1 year ago 94
Perl Question

perl operator s///e not working as expected

I'm trying to figure out why this is working :

s/(\d+)(.)(\d+)/"\$x$2$2"/ee; # This is working, does $x++

print "x=$x \n"; # x=2

while this is not :

s/(\d+)(.)(\d+)/\$x$2$2/e; # This is NOT working

# Error message is :
# Scalar found where operator expected at ./test.pl line 2, near "$x$2"
# (Missing operator before $2?)

I have the guts that
should behave the same, but obviously I'm wrong.

What am I missing ?

Answer Source

You're misunderstanding the expression modifier -- a single /e

It is essentially an alternative to the standard mode, which is to process the replacement string as if it were in double-quotes


my $x = 1;
my $y = '12+34';

$y =~ s/(\d+)(.)(\d+)/\$x$2$2/

produces a replacement equivalent to the string qq{\$x$2$2}, which is $x++

If you add an /e then the replacement is treated as a Perl expression, and you are getting errors because \$x$2$2 isn't valid Perl. You could get the same result as before by using

s/(\d+)(.)(\d+)/'$x' . $2 . $2/e

or, as you have seen, the string expression


But all these do is evaluate the Perl expression. There is no way to execute arbitrary Perl code that is constructed from parts of the target string without adding a second /e modifier which stands for eval

The resulting /ee causes Perl to treat the replacement as an expression (instead of doing double-quote interpolation on it) and then eval the result of that expression

For instance

s/(\d+)(.)(\d+)/'$x' . $2 . $2/ee

first evaluates the expression '$x' . $2 . $2 giving $x++ and then does eval on that string, returning 1 (so the original 12+34 is replaced with 1) and incrementing $x

You can use expression mode with a single /e if you can write a Perl expression that does what you need. Otherwise you need to use /ee to get an eval stage as well

I think it's clearer to use braces if you involve /e at all. That way it looks like Perl code and not a string replacement

    '$x' . $2 . $2
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