So if we have 2 negative numbers and we want to subtract them do the subtraction becomes addition ? Or we subtract them normally
I tried to subtract them but it gave me a wrong number but when I add them I got overflow
Related
I've searched the existing threads and can't figure out where I'm doing it wrong...
I'm trying to get the average of two values and then round the average to the nearest TEN (10).
My expression is:
=ROUND((Fields!Cyl1Stress.Value + Fields!Cyl2Stress.Value) / 2, -1)
The above returns an error.
Rounding to the nearest TENTH works fine, but as soon as I change the 1 to -1, I get the error.
Where Cyl1Stress.Value = 7600 and Cyl2Stress.Value = 7490.
=ROUND((Fields!Cyl1Stress.Value + Fields!Cyl2Stress.Value) / 2, **1**)
The above code returns 7545.
But I need the result to be 7550, so I change my formula to:
=ROUND((Fields!Cyl1Stress.Value + Fields!Cyl2Stress.Value) / 2, **-1**).
This one returns an error.
Can't figure out why this isn't working!
Text Box Properties are Numeric, 0 decimal places.
I don't believe you can use a negative value for the Digits argument of Round. The documentation indicates:
The number of fractional digits in the return value. For Decimal values, it can range from 0 to 28. For Double values, it can range from 0 to 15.
You can accomplish what you are looking to do by dividing the value by 10, round to the nearest whole number, and then multiply by 10. Also note that if you want a value ending in 5 to round up to the next 10, you will need to add a third argument to Round to indicate this, otherwise I believe the default behavior rounds to the nearest even number. This would cause your example to round to 7540 instead of 7550.
This expression should be what you are looking for:
=ROUND(((Fields!Cyl1Stress.Value + Fields!Cyl2Stress.Value) / 2) / 10, 0, MidpointRounding.AwayFromZero) * 10
For your example values, this turns 7545 into 754.5, rounds to the nearest whole number and rounds it away from zero instead of to the nearest even number (755), then multiplies by 10 (7550).
I need a bit advice of the following topics: adding two signed/magnitude and adding two complement binary numbers. I did my calculations on paper and uploaded the picture. Sorry if my picture is sideways (I don't know why the upload does that)
Adding two signed/magnitude
+6 + (-6)
Ignore the carry 1. The sum is 4 and incorrect. No overflow because we have added two numbers with different signed indicators, 0 and 1.
+4 + (+5)
The sum is -1 and incorrect. Overflow because we because we have added two numbers with the same signed indicators 0.
Adding two complement binary numbers
+6 + (-6)
Ignore the carry 1. The sum is 0 and correct. No overflow because we have added two numbers with different signed indicators, 0 and 1.
+4 + (+5)
The sum is -7 and incorrect. Overflow because we because we have added two numbers with the same signed indicators 0.
Did I understand correctly about binary addition overflow and sum correctness though my examples?
From this University course
4.11.4. Addition and Subtraction
Addition and subtraction require attention to the sign bit. If the signs are the same, we simply add the magnitudes as unsigned numbers and watch for overflow. If the signs differ, we subtract the smaller magnitude from the larger, and keep the sign of the larger.
So for sign-magnitude you got the first one wrong.
The signs differ so you subtract the larger from the smaller. As they are the same, it doesn't matter what you pick. The result is 0 and it is correct. +0 or -0. It doesn't matter as they both represent the same thing.
You got the second one wrong too because for the sum +4+5, you should keep the sign + for the result and then add the magnitude. Which is
1 carry (ignore carry to the most sign bit)
0100 +4
0101 +5
--------
0001 +1
The result is +1 and there was an overflow that was detected by the carry out to the most significant bit, which is the sign bit.
Check this for more info http://pages.cs.wisc.edu/~smoler/x86text/lect.notes/arith.int.html
2's complement
Your answers are correct :tada:
I have a calculated field in my table called C. its the result of A-B=C. A & B are number fields (single, fixed). I have having trouble setting up C as a calculated (Decimal Field).
The precision / decimal places seem to work perfectly, I can modify them freely. But no matter what I do to "SCALE". It always seems to return to "0". I need it to be 2 since all my data in my reports are rounding off at the wrong locations giving me hole numbers.
As you can see "scale = 0", no matter what I do to this number. it will always revert to "0". Why is that?
You can’t change the scale in a calculated field, because it takes the values and settings from the calculation.
So the fact of a scale of 0 should not matter. The resulting number if it needs decimal places will (should) have the decimal value. The setting is IGNORED
I mean, if the calculation is:
2 x 3 = 6
Then you get 6.
If you have 4 / 3 = 1.3333
Then, in your case you get:
1.33333333333333
And you WILL get the above EVEN if the scale = 0. So the scale setting is NOT used nor available in a calculated field.
You are certainly free to round, or format the above result. And in fact you could (should) consider using the round() function in the actual calculation. So use something like:
Round([Field1] / [Field2],4)
And you thus get:
1.3333
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.
If you have a binary number say 1010 (which is 10 in base 10), is saying that dividing by two will remove the first digit (making it end up as 010), true?
Basically how do you remove the first digit (i.e. if the binary number is 0 or 1, then it will end up as nothing)? I don't want code or anything, I just want to know like something like you divide or multiply by two.
Also do not consider any of the left most zeroes, of a binary number.
It works the same way as it does in base ten. The number 401, without its first digit, is 1. You've subtracted 400, no? Now, to divide by ten, you would SHIFT the digits to the right. 401 shifted right is 040. 401/10 = 40. Note that the 1 is discarded because we're working with integer division.
So in binary, it's exactly the same, but with powers of 2. Removing the first bit does not DIVIDE by two. It SUBTRACTS the value of its position. So 101b (which is 4+1 = 5), without its largest bit, is 001b, or 1 decimal. It's subtraction: 5 - 4 = 1.
To divide by two, you shift the bits to the right, just like in base 10. So 101b would become 010b, which is 2 decimal. 5/2 == 2 (we're dropping the fractional part since it's integer division)
Make sense? If you're ever confused about binary, just think of how the digits & positions work in base ten, and instead of powers of ten, use powers of two.
If by "first digit" you mean "first significant digit", then what you're looking for is something like number and not (1 shl (int(log number / log 2))), where and and not are the bitwise operations, shl means shift left, and int is rounding down (never up) to integer. log is just a logarithm, in any base (same base for both cases).
If by "first digit" you mean the digit in some nth position (let the rightmost position be 0, counting to the left), then you just do number and not (1 shl position).
Removing a digit is like changing it to 0. Changing 1010 to 0010 is accomplished by subtracting 1000: 1010 - 1000 = 0010.