Im doing this question and some clarification would be super helpful. What exactly would an overflow entail? If when converting to decimal notation an extra bit would be needed? Fro part 3 "consider the bits as two's complement numbers" does he mean find the 2's complement? Thanks a bunch.
For number 3 he does not mean find the 2's complement. He is telling you to treat the values as signed numbers using 2's complement notation. That would mean the first value in a) is positive and the other three are negative.
For overflow it is different for 2 and 3. For 2, unsigned numbers, overflow occurs if there is a carry out of the high bit. For 3, 2's complement signed numbers, overflow occurs if the sign of the result is not correct. For example, if you add two positive numbers and the result is negative, there was overflow.
If you add x and y and get a result that is less than x or less than y, then the addition has overflowed (wrapped-around).
An overflow would be if the resulting sum is a larger number than can be expressed in an 8 bit system. I believe that would be any number greater than 255 (1 << 8).
Your assumption "an extra bit" is mostly correct. In an 8 bit system, all numbers are stored in 8 bits. Any operation that results in a number greater than the maximum that can be represented will be an overflow. This doesn't happen when you convert to decimal, but when you actually perform the sum with the binary values. If all numbers are 8 bits, you can't just add an additional bit when you need to store a larger number.
Yes, "two's complement" is the same as "2's complement". I'm not aware of any distinction between whether you spell it out or use the numeral.
Related
I just learnt the concept of two's complement. It is useful in the subtraction of binary numbers. But in my college booklet i have been asked to solve 48+23 and -48-23 using 2's complement. What is the logic behind this and how to proceed with this question?
The key insight is that negative numbers are not merely conveniently represented by two's complement, but the scheme actually preserves the properties of addition and subtraction naturally as a result of the encoding itself. That is:
Bit patterns end up identical when you subtract positive numbers or add equivalent negative numbers; and when you add positive numbers or subtract equivalent negative ones
The signum rolls over automatically in either direction: that is, negative numbers extend 'naturally' to 0 and thence to positive numbers by counting up
These properties make working (i.e. arithmetic) using this encoding a lot easier than, say one's complement.
Finally, for a fixed length two's complement bit string, the number of available negative numbers is always exactly one larger than the number of available positive numbers.
I'm reading this article about exponent bias in floating point numbers and it says the following:
n IEEE 754 floating point numbers, the exponent is biased in the
engineering sense of the word – the value stored is offset from the
actual value by the exponent bias. Biasing is done because exponents
have to be signed values in order to be able to represent both tiny
and huge values, but two's complement, the usual representation for
signed values, would make comparison harder. To solve this problem the
exponent is biased before being stored, by adjusting its value to put
it within an unsigned range suitable for comparison. By arranging the
fields so that the sign bit is in the most significant bit position,
the biased exponent in the middle, then the mantissa in the least
significant bits, the resulting value will be ordered properly,
whether it's interpreted as a floating point or integer value. This
allows high speed comparisons of floating point numbers using fixed
point hardware.
I've also found this explanation from wikipedia's article about offset binary:
This has the consequence that the "zero" value is represented by a 1
in the most significant bit and zero in all other bits, and in general
the effect is conveniently the same as using two's complement except
that the most significant bit is inverted. It also has the consequence
that in a logical comparison operation, one gets the same result as
with a two's complement numerical comparison operation, whereas, in
two's complement notation a logical comparison will agree with two's
complement numerical comparison operation if and only if the numbers
being compared have the same sign. Otherwise the sense of the
comparison will be inverted, with all negative values being taken as
being larger than all positive values.
I don't really understand what kind of comparison they are talking about here. Can someone please explain using a simple example?
'Comparison' here refers to the usual comparison of numbers by size: 5 > 4, etc. Suppose floating-point numbers were stored with as
[sign bit] [unbiased exponent] [mantissa]
For example, if the exponent is a 2's complement 3-bit binary number and the mantissa is a 4-bit unsigned binary number, you'd have
1 010 1001 = 4.5
1 110 0111 = 0.21875
You can see that the first is bigger than the second, but to figure this out, the computer would have to calculate 1.001 x 2^2 and 0.111 x 2^(-2) and then compare the resulting floating-point numbers. This is already complex with floating-point hardware, and if there is no such hardware for this computer, then...
So the number is stored as
[sign bit] [biased exponent] [mantissa]
Using the same 3-bit binary number for the exponent (this time biased; see a related question) and unsigned 4-bit mantissa, we have
1 101 1001 = 4.5
1 001 0111 = 0.21875
But now comparison is very easy! You can treat the two numbers as integers 11011001 and 10010111 and see that the first is obviously bigger: obvious even to a computer, as integer comparisons are easy. This is why biased exponents are used.
First of all, this is not a duplicate question of this because it hasn't answered my questions below.
I searched for many resources and ended with no clear perception of how to perform signed number division using 2's complement, specifically for the case where one of the divisors or dividend or both are negative.
I read the Signed Number Division section from chapter 2 of Digital Fundamentals by Floyed and all of its examples and problems. I even read through internet resources that explain division, but they only used positive numbers (as explained below):
, but what about 100/-25 or -100/25 or -100/-25?
What are the rules to follow?
Can anyone give a simple explanation on, for example, how to divide 14 by -7?
The rules states that if the numbers have different sign bits that the quotient and remainder gotten when both are divided(this is after the have both been turned positive) will be negative.So for 14/-7, they will have different sign bits when turned to binary so this means your quotient and remainder will be negative. Now you leave 14 in positive form and turn -7 to it's corresponding positive binary number and carry out the division with the positive forms of both numbers, when you get your quotient and remainder from dividing the positive form of both numbers you turn both the quotient and remainder negative.
I need help in determining if my logic here is right or wrong.
Example Question
"Assuming I have an 8-bit signed decimal value of 200 in two's compliment form..."
My Thought Process
Now because it is 8-bits and is signed, the most significant bit must be reserved for the sign.
Thus, the maximum positive value it can have is:
2^(8-1) - 1 = 127
At first I was confused because I thought, why is the question stating that 200 is able to be 8-bits and signed? Then I thought, that's where the two's compliment statement comes into question.
Because it is two's compliment in reality, this is the case:
8-bit Signed, 2's Compliment, Decimal = 200
Convert to Binary --> 1100 1000
Because it is signed, the actual two's compliment number is ACTUALLY -56 (I would use negating methods to invert the 1's and 0's then + 1, but for the interest of time, I just found a converter online).
So my conclusion is:
8-bit Signed, 2's Compliment, Decimal value of 200 is actually -56.
Ultimate Question
Is my thought process correct with this? If so, I think the most confusing part about this is telling my brain that one number is equal to a completely different number.
Yes, I think your analysis is correct.
To expand a bit more, I think the wording of the question is awkward and would have been better stated as "What is the value of 1100 1000 in base 10, where the number is a two's complemented number?"
The trick here is to think not that 200 == -56, but that the single point of truth is the bits 11001000. These bits of numbers have no meaning by themselves. We have the computer interpret them differently based on the program. So two's complement (with 8 bit numbers) treats that as -56, an unsigned interpretation would treat that as 200, and in ASCII this would be some special character depending on the encoding.
I'm trying to wrap my head around overflow within twos complement for example say I'm trying to take away these two binary numbers:
1111 1000 0100 - 010 111 001 000
I convert the 2nd binary number to it's two complement equivalent and then simply add it but I noticed it resulted in an overflow of 1, do I simply ignore the overflow? or is there a rule I must follow
1111 1000 0100 + 1010 0011 1000 = (1) 1001 1011 1100
Short answer:
if you are performing arithmetic on fixed-width binary numbers, using two's complement representation for negative numbers, then yes, you ignore the one-bit overflow.
Long Answer:
You can consider each ith bit in n-bit two's complement notation have place value 2^i, for 0 <= i < n - 1, with bit n - 1 (the sign bit) having place value -2^(n - 1). That's a negative place value for the sign bit. If you compute the sum of two such numbers as if they were unsigned n-bit binary numbers, these cases are fine:
the sign bit is not set in the either addend or in the result (reinterpreted as being in two's-complement representation),
the sign bit is set in exactly one of the addends, regardless of overflow (which is ignored if it occurs), or
the sign bit is set in both addends (therefore there is an overflow, which is ignored) and in the result.
To understand that, it may be easier to think about the problem as two separate sums: a sum of the sign bits, and a sum of the value (other) bits. An overflow of the value sum yields an overflow bit whose place value is 2^(n-1) -- exactly the inverse of the place value of a sign bit -- therefore such an overflow cancels one sign bit.
The negative + negative case requires such a cancellation for the result to be representable (two sign bits + one value overflow = one sign bit), and the positive + positive case cannot accommodate such a cancellation because there is no sign bit available to be cancelled. In the positive + negative case, there is an overflow of the value-bit sum in exactly those cases where the result is non-negative; you can consider that to cancel the sign bit of the negative addend, wich yields the same result as ignoring the overflow of the overall unsigned sum, and reinterpreting the sum as a two's complement number.
The remaining cases yield mathematical results that cannot be represented in n-bit two's complement format -- either greater than the largest representable number, or less than the smallest. If you ignore overflow then such results can be recognized by an apparent sign flip. What you do with that is a question of error recovery strategy.
From Wikipedia's article on 2's complement in the section on addition at https://en.wikipedia.org/wiki/Two%27s_complement#Addition, my understanding is that carry beyond the given (fixed) bit length (to the left) can be ignored but not overflow as determined when the leftmost two bits of the carry are different. The article shows how to maintain a carry row so as to tell if there was overflow and here is a simple example in the same style:
In 4 bit 2's complement -2 is 1110 and +3 is 0011 so
11110 carry
1110 -2
+0011 +3
----
10001 which is 0001 or simply 1 ignoring the carry in bit 5 and is
safe since the leftmost two bits in the carry row are identical
Although this is a very old question, it comes up frequently. In two's complement addition, a carry out from the leftmost digit is discarded. Why? Although not precisely correct mathematically, it is easiest to think of a two’s complement number as having a sign bit on the left and value bits elsewhere. The only way a carry out of the sign bit can occur is if the sign bits of both addends were one (negative) and there was a carry in to the sign bit. In that case, the sign bit of the result will be one, which is correct. A problem occurs if the carry in to the sign bit is different from the carry out. That causes an incorrect sign bit, which is an overflow condition. That can be detected without referring to the carry out from the sign bit because the sign of the result will be wrong. For example, if two positive numbers are added and the result is negative, something is wrong. The something that’s wrong is that the sum of the value bits has overflowed into the sign bit and the result is in error.
With pen and paper arithmetic, it is usual to discard the carry and check that the sign of the result is correct. In electronic circuits, the easiest way is to compare that carry in to the carry out with an XOR and signal an error if they differ. The carry out is not otherwise used or stored.