Question

I have the following small Perl (v5.10) program:

use strict;

my @nums;
my $i = 0;

while ($i < 5) {

    print "Enter number " . $i+1 . ": ";
    $nums[$i] = <STDIN>;
    $i++;
}

foreach (@nums) {

    chomp $_;
    print "$_\t";
}
print "\n";

This is the result of a test run:

1: 2
1: 1
1: 6
1: 3
1: 2
2   1   6   3   2   

The problem, as you can see, is that the print statement prompting the user for input isn’t functioning as expected. Instead of “Enter number 1: ” or “Enter number 3:”, e.t.c., I just get “1:”. I didn’t expect this to work to be honest because I know that the + operator has been overloaded for string concatenation in Perl. How do I get around this problem? And what is the reason for it?

Answer 1

+ is not overloaded. It is a precedence issue. The expression is parsed as

print(((('Enter number ' . $i) + 1) . ': '));

Which is the same as

print((0 + 1) . ': ');

You can use

perl -MO=Deparse,-p -e 'print "Enter number " . $i+1 . ": ";'

to see how Perl parses your scripts.

Adding parentheses sovles the problem.