Leonid Grinberg wrote:
>> You don't have to escape the colon in a regular expression.
>
> Oh, cool. Didn't know that.
>
>> > my $home = $split_line[5];
>> > my @split_home = split(/\/, $home);
>> ^^^
>> Syntax error. The terminating delimiter is missing.
>
> I am confused. What do you mean, the terminating delimter is missing?
> It seems alright to me...
$ perl -ce'split /\/, "one/two"'
Bareword found where operator expected at -e line 1, near "/\/, "one/two"
(Missing operator before two?)
String found where operator expected at -e line 1, at end of line
(Missing semicolon on previous line?)
syntax error at -e line 1, near "/\/, "one/two"
Can't find string terminator '"' anywhere before EOF at -e line 1.
$ perl -ce'split /\//, "one/two"'
-e syntax OK
The backslash escapes the character following it.
>> "scalar(@split_home) - 1" could also be written as "$#split_home" or
>> just "-1".
>
> I did not know that ``-1'' would work. However, I prefer scalar() to $#.
Using scalar() there is redundant as the expression is already in scalar
context. Also using "scalar(@split_home) - 1" will not work correctly if the
value of $[ has been changed.
John
--
Perl isn't a toolbox, but a small machine shop where you can special-order
certain sorts of tools at low cost and in short order. -- Larry Wall
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