If you replace your last line with 

printf "two + 2 = %d\n",&two + 2;

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(note the ampersand) you get "two + 2 = 4". So it seems the parser treats your "two" as a bareword and converts it to 0 in numeric context (therefore the result).

I wonder why this does not procuce a warning though...

In reply to Re: two + 2 not equal 4! by morgon
in thread two + 2 not equal 4! by kp2a

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