I do not get to understand how the Perl read($buf) function is able to modify the content of the $buf variable. $buf is not a reference, so the parameter is given by copy (from my c/c++ knowledge). So how come the $buf variable is modified in the caller ?

Is it a tie variable or something ? The C documentation about setbuf is also quite elusive and unclear to me

# Example 1
$buf=''; # It is a scalar, not a ref
$bytes = $fh->read($buf);
print $buf; # $buf was modified, what is the magic ?

# Example 2
sub read_it {
    my $buf = shift;
    return $fh->read($buf);
}
my $buf;
$bytes = read_it($buf);
print $buf; # As expected, this scope $buf was not modified