&& is not an lvalue
- From: xhoster@xxxxxxxxx
- Date: 31 Jul 2008 19:14:06 GMT
It seems like almost everything in Perl is an lvalue. So why isn't the
result of && an lvalue?
I wanted to do this:
($h{$big}{$nasty}[$dereferencing]{operation($x)}{$done}[here(y)]and die)=6;
Obviously, that just isn't the way perl was implemented. But is there a
reason that "?:" yields an lvalue but && doesn't?
Thanks,
Xho
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