<html><head></head><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space; "><div>A simpler example of how to use a scalar reference is:</div><div><br></div><div><blockquote type="cite"><font class="Apple-style-span" color="#000000">perl -le 'my $y = 0; my $x = \$y; $$x++; print $$x'</font></blockquote><br></div><div>Basically, the error tells you that the code is actually trying to modify the constant string "0", and you're not allowed to do that. The example here might help you.</div><div><br></div><div>All the best</div><div>Stuart</div><br><div><div>On 2010-04-22, at 9:39 PM, Shaun Fryer wrote:</div><br class="Apple-interchange-newline"><blockquote type="cite"><div>I'm curious why a scalar reference is considered a read-only value?<br><br>$ perl -le 'my $x = \"0"; $$x++; print $$x'<br>Modification of a read-only value attempted at -e line 1.<br><br>$ perl -le 'my $x = {x=> 0}; $x->{x}++; print $x->{x}'<br>1<br><br>Is there an easy way to update a simple scalar ref without having to<br>use an array or hash ref instead?<br><br>Fyi... This is perl, v5.10.0 built for i486-linux-gnu-thread-multi<br>--<br>Shaun Fryer<br>_______________________________________________<br>toronto-pm mailing list<br><a href="mailto:toronto-pm@pm.org">toronto-pm@pm.org</a><br>http://mail.pm.org/mailman/listinfo/toronto-pm<br></div></blockquote></div><br></body></html>