I know that in the case of a 16 bit word, if we have x - 115 in decimal, the smallest x that would cause overflow would be (32767 + 115 + 1)= 32883 because it would represent a number that is larger than the largest positive integer we could represent in a 16 bit word. However, I was wondering what would the largest value of x be? would it be (2^16)? which would cause 2 carries after the addition. Or would the largest x be a number that would only result in 1 carry alone?
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You get overflow if the result is larger than the maximal integer, in this case 32767. You get underflow if it is smaller than the minimal one, in this case -32768. (In the context of floating-point operations, underflow happens when a non-zero number is rounded to zero.)
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$\begingroup$ If you look at my question, that's exactly what I have explained. My question is not really that though. I would like to know what would the largest value of x be(not the largest integer value as the result) that will cause overflow? is it going to be (2^16 - 1), since this is the largest number the a 16 bit word could represent? $\endgroup$O.A.– O.A.2014-09-28 22:23:28 +00:00Commented Sep 28, 2014 at 22:23
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$\begingroup$ The largest value of $x$ that will cause overflow cannot be 65535, since a 16 bit signed integer cannot hold that number. $\endgroup$Yuval Filmus– Yuval Filmus2014-09-28 23:00:13 +00:00Commented Sep 28, 2014 at 23:00