While learning list operators (v5.26.1), I found this textbook example for grep:

#!/home/localperl/bin/perl

use strict;
use warnings;

my @input_numbers = (1, 2, 4, 8, 16, 32, 64);
my @odd_digit_sum = grep digit_sum_is_odd($_), @input_numbers;
print "@odd_digit_sum\n";

sub digit_sum_is_odd
{
    my $input = shift; # Can use $_ instead of shift
    print "\$input: $input\n";
    my @digits = split //, $input;
    my $sum;
    $sum += $_ for @digits;
    return $sum % 2;
}

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I have tried using $_ instead of shift in the above example and it gives same result. So what I have guessed is:

When "sub digit_sum_is_odd" is called the parameter that is passed for each element of @input_numbers is interpreted as an array of single element, if shift is used.
When "sub digit_sum_is_odd" is called the parameter that is passed for each element of @input_numbers is interpreted as a scalar, if $_ is used.

Is this guess correct or am I mistaken? I would like to know what is the reason it gives same result for both shift and $_?

In reply to [Answered]: Grep: What is the reason following example code gives same result for both shift and $_? by Perl300

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