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OverflowOverflowOne caveat with signed binary numbers is that of overflow, where the answer to an addition or subtraction problem exceeds the magnitude which can be represented with the alloted number of bits. Remember that the place of the sign bit is fixed from the beginning of the problem. With the last example problem, we used five binary bits to represent the magnitude of the number, and the leftmost (sixth) bit as the negativeweight, or sign, bit. With five bits to represent magnitude, we have a representation range of 2^{5}, or thirtytwo integer steps from 0 to maximum. This means that we can represent a number as high as +31_{10} (011111_{2}), or as low as 32_{10} (100000_{2}). If we set up an addition problem with two binary numbers, the sixth bit used for sign, and the result either exceeds +31_{10} or is less than 32_{10}, our answer will be incorrect. Let's try adding 17_{10} and 19_{10} to see how this overflow condition works for excessive positive numbers:
. 17_{10} = 10001_{2} 19_{10} = 10011_{2} . . 1 11 < Carry bits . (Showing sign bits) 010001 . + 010011 .  . 100100
The answer (100100_{2}), interpreted with the sixth bit as the 32_{10} place, is actually equal to 28_{10}, not +36_{10} as we should get with +17_{10} and +19_{10} added together! Obviously, this is not correct. What went wrong? The answer lies in the restrictions of the sixbit number field within which we're working Since the magnitude of the true and proper sum (36_{10}) exceeds the allowable limit for our designated bit field, we have an overflow error. Simply put, six places doesn't give enough bits to represent the correct sum, so whatever figure we obtain using the strategy of discarding the leftmost "carry" bit will be incorrect. A similar error will occur if we add two negative numbers together to produce a sum that is too low for our sixbit binary field. Let's try adding 17_{10} and 19_{10} together to see how this works (or doesn't work, as the case may be!):
. 17_{10} = 101111_{2} 19_{10} = 101101_{2} . . 1 1111 < Carry bits . (Showing sign bits) 101111 . + 101101 .  . 1011100 .  . Discard extra bit . FINAL ANSWER: 011100_{2} = +28_{10}
The (incorrect) answer is a positive twentyeight. The fact that the real sum of negative seventeen and negative nineteen was too low to be properly represented with a five bit magnitude field and a sixth sign bit is the root cause of this difficulty. Let's try these two problems again, except this time using the seventh bit for a sign bit, and allowing the use of 6 bits for representing the magnitude:
. 17_{10} + 19_{10} (17_{10}) + (19_{10}) . . 1 11 11 1111 . 0010001 1101111 . + 0010011 + 1101101 .   . 0100100_{2} 11011100_{2} .  . Discard extra bit . . ANSWERS: 0100100_{2} = +36_{10} . 1011100_{2} = 36_{10}
By using bit fields sufficiently large to handle the magnitude of the sums, we arrive at the correct answers. In these sample problems we've been able to detect overflow errors by performing the addition problems in decimal form and comparing the results with the binary answers. For example, when adding +17_{10} and +19_{10} together, we knew that the answer was supposed to be +36_{10}, so when the binary sum checked out to be 28_{10}, we knew that something had to be wrong. Although this is a valid way of detecting overflow, it is not very efficient. After all, the whole idea of complementation is to be able to reliably add binary numbers together and not have to doublecheck the result by adding the same numbers together in decimal form! This is especially true for the purpose of building electronic circuits to add binary quantities together: the circuit has to be able to check itself for overflow without the supervision of a human being who already knows what the correct answer is. What we need is a simple errordetection method that doesn't require any additional arithmetic. Perhaps the most elegant solution is to check for the sign of the sum and compare it against the signs of the numbers added. Obviously, two positive numbers added together should give a positive result, and two negative numbers added together should give a negative result. Notice that whenever we had a condition of overflow in the example problems, the sign of the sum was always opposite of the two added numbers: +17_{10} plus +19_{10} giving 28_{10}, or 17_{10} plus 19_{10} giving +28_{10}. By checking the signs alone we are able to tell that something is wrong. But what about cases where a positive number is added to a negative number? What sign should the sum be in order to be correct. Or, more precisely, what sign of sum would necessarily indicate an overflow error? The answer to this is equally elegant: there will never be an overflow error when two numbers of opposite signs are added together! The reason for this is apparent when the nature of overflow is considered. Overflow occurs when the magnitude of a number exceeds the range allowed by the size of the bit field. The sum of two identicallysigned numbers may very well exceed the range of the bit field of those two numbers, and so in this case overflow is a possibility. However, if a positive number is added to a negative number, the sum will always be closer to zero than either of the two added numbers: its magnitude must be less than the magnitude of either original number, and so overflow is impossible. Fortunately, this technique of overflow detection is easily implemented in electronic circuitry, and it is a standard feature in digital adder circuits: a subject for a later chapter.


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