Preceding a number with escape character '\' produces garbage value Ex:
$a = \12;
print $a;
This code gives below output
SCALAR(0x2001ea8)
and the output changes when I again execute the program.
If I take a value(number) from user when the user gives any input starting with zero, I dont want it to interpret as octal number. So I wanted to escape zero if the number starts with zero.
[In a comment, the OP explained he wants numbers input by the user to be treated as decimal even if they have leading zeroes.]
In numerical literals (code that produces a number), a leading zero tells Perl that the number is in octal.
$ perl -E'say 045'
37
But that does not apply to numification (converting a string to a number).
# "045" is the same as reading 045 from handle or @ARGV.
$ perl -E'say 0+"045"'
45
So you don't have to do anything special. A 045 input by the user means forty-five (not thirty-seven) if you use it as a number.
If for some reason you did need to get rid of the leading zero, you could use
$var =~ s/^0+(?!\z)//;
The (?!\z) makes sure "0" doesn't become "".
It's not a garbage value. You are getting what Perl prints out when it prints a reference.
TYPE(ADDRESS)
It's expected functionality. If you want a \ in your string you'll need to escape it.
$str = "\\12";
Or as Ted Hopp pointed out in the comments with a string literal
$str = '\12';
See the Perl doc for more information.
