As I read your question, it looks like you're trying to add $var1 if it's defined or $var2 if it isn't, but you're getting tripped up by testing whether $var1 is true instead.

If my interpretation is correct, then test for definedness with defined: 

$value += defined $var1 ? $var1 : $var2;

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In reply to Re: When is 'eval' too inefficient? by dsheroh
in thread When is 'eval' too inefficient? by argv

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