The documentation of Random.NextDouble():
Returns a random floating-point number that is greater than or equal to 0.0, and less than 1.0.
So, it can be exactly 0. But what are the chances for that?
var random = new Random();
var x = random.NextDouble()
if(x == 0){
// probability for this?
}
It would be easy to calculate the probability for Random.Next() being 0, but I have no idea how to do it in this case...
As mentioned in comments, it depends on internal implementation of NextDouble. In "old" .NET Framework, and in modern .NET up to version 5, it looks like this:
protected virtual double Sample() {
return (InternalSample()*(1.0/MBIG));
}
InternalSample returns integer in 0 to Int32.MaxValue range, 0 included, int.MaxValue excluded. We can assume that the distribution of InternalSample is uniform (in the docs for Next method, which just calls InternalSample, there are clues that it is, and it seems there is no reason to use non-uniform distribution in general-purpose RNG for integers). That means every number is equally likely. Then, we have 2,147,483,647 numbers in distribution, and the probability to draw 0 is 1 / 2,147,483,647.
In modern .NET 6+ there are two implementations. First is used when you provide explicit seed value to Random constructor. This implementation is the same as above, and is used for compatibility reasons - so that old code relying on the seed value to produce deterministic results will not break while moving to the new .NET version.
Second implementation is a new one and is used when you do NOT pass seed into Random constructor. Source code:
public override double NextDouble() =>
// As described in http://prng.di.unimi.it/:
// "A standard double (64-bit) floating-point number in IEEE floating point format has 52 bits of significand,
// plus an implicit bit at the left of the significand. Thus, the representation can actually store numbers with
// 53 significant binary digits. Because of this fact, in C99 a 64-bit unsigned integer x should be converted to
// a 64-bit double using the expression
// (x >> 11) * 0x1.0p-53"
(NextUInt64() >> 11) * (1.0 / (1ul << 53));
We first obtain random 64-bit unsigned integer. Now, we could multiply it by 1 / 2^64 to obtain double in 0..1 range, but that would make the resulting distribution biased. double is represented by 53-bit mantissa (52 bits are explicit and one is implicit) ,exponent and sign. For all integer values exponent is the same, so that leaves us with 53 bits to represent integer values. But we have 64-bit integer here. This means integer values less than 2^53 can be represented exactly by double but bigger integers can not. For example:
ulong l1 = 1ul << 53;
ulong l2 = l1 + 1;
double d1 = l1;
double d2 = l2;
Console.WriteLine(d1 == d2);
Prints "true", so two different integers map to the same double value. That means if we just multiply our 64-bit integer by 1 / 2^64 - we'll get a biased non-uniform distribution, because many integers bigger than 2^53-1 will map to the same values.
So instead, we throw away 11 bits, and multiply the result by 1 / 2^53 to get uniform distibution in 0..1 range. The probability to get 0 is then 1 / 2^53 (1 / 9,007,199,254,740,992). This implementation is better than the old one, because is provides much more different doubles in 0 .. 1 range (2^53 compared to 2^32 in old one).
You also asked in comments:
If one knows how many numbers there are between 0 inclusive and 1
exclusive (according to IEEE 754), it would be possible to answer the
'probability' question, because 0 is one of all of them
That's not so. There are actually more than 2^53 representable numbers between 0..1 in IEEE 754. We have 52 bits of mantissa, then we have 11 bits of exponent, half of which is for negative exponents. Almost all negative exponents (rougly half of that 11 bit range) combined with mantissa gives us distinct value in 0..1 range.
Why we can't use full 0..1 range which IEEE allows us to generate random number? Because this range is not uniform (like the full double range is not uniform itself). For example there are more representable numbers in 0 .. 0.5 range than in 0.5 .. 1 range.
This is from a strictly academic persepective.
From Double Struct:
All floating-point numbers also have a limited number of significant
digits, which also determines how accurately a floating-point value
approximates a real number. A Double value has up to 15 decimal digits
of precision, although a maximum of 17 digits is maintained
internally. This means that some floating-point operations may lack
the precision to change a floating point value.
If only 15 decimal digits are significant, then your possible return values are:
0.000000000000000
To:
0.999999999999999
Said differently, you have 10^15 possible (comparably different, "distinct") values (see Permutations in the first answer):
10^15 = 1,000,000,000,000,000
Zero is just ONE of those possibilities:
1 / 1,000,000,000,000,000 = 0.000000000000001
Stated as a percentage:
0.0000000000001% chance of zero being randomly selected?
I think this is the closest "correct" answer you're going to get...
...whether it performs this way in practice is possibly a different story.
Just create a simple program, and let it run until you are satisfied by the number of tries done. (see: https://onlinegdb.com/ij1M50gRQ)
Random r = new Random();
Double d ;
int attempts=0;
int attempts0=0;
while (true) {
d = Math.Round(r.NextDouble(),3);
if(d==0) attempts0++;
attempts++;
if (attempts%1000000==0) Console.WriteLine($"Attempts: {attempts}, with {attempts0} times a 0 value, this is {Math.Round(100.0*attempts0/attempts,3)} %");
}
example output:
...
Attempts: 208000000, with 103831 times a 0 value, this is 0.05 %
Attempts: 209000000, with 104315 times a 0 value, this is 0.05 %
Attempts: 210000000, with 104787 times a 0 value, this is 0.05 %
Attempts: 211000000, with 105305 times a 0 value, this is 0.05 %
Attempts: 212000000, with 105853 times a 0 value, this is 0.05 %
Attempts: 213000000, with 106349 times a 0 value, this is 0.05 %
Attempts: 214000000, with 106839 times a 0 value, this is 0.05 %
...
Changing the value of d to be rounded to 2 decimals will return 0.5%
I'm in a computer systems course and have been struggling, in part, with two's complement. I want to understand it, but everything I've read hasn't brought the picture together for me. I've read the Wikipedia article and various other articles, including my text book.
What is two's complement, how can we use it and how can it affect numbers during operations like casts (from signed to unsigned and vice versa), bit-wise operations and bit-shift operations?
Two's complement is a clever way of storing integers so that common math problems are very simple to implement.
To understand, you have to think of the numbers in binary.
It basically says,
for zero, use all 0's.
for positive integers, start counting up, with a maximum of 2(number of bits - 1)-1.
for negative integers, do exactly the same thing, but switch the role of 0's and 1's and count down (so instead of starting with 0000, start with 1111 - that's the "complement" part).
Let's try it with a mini-byte of 4 bits (we'll call it a nibble - 1/2 a byte).
0000 - zero
0001 - one
0010 - two
0011 - three
0100 to 0111 - four to seven
That's as far as we can go in positives. 23-1 = 7.
For negatives:
1111 - negative one
1110 - negative two
1101 - negative three
1100 to 1000 - negative four to negative eight
Note that you get one extra value for negatives (1000 = -8) that you don't for positives. This is because 0000 is used for zero. This can be considered as Number Line of computers.
Distinguishing between positive and negative numbers
Doing this, the first bit gets the role of the "sign" bit, as it can be used to distinguish between nonnegative and negative decimal values. If the most significant bit is 1, then the binary can be said to be negative, where as if the most significant bit (the leftmost) is 0, you can say the decimal value is nonnegative.
"Sign-magnitude" negative numbers just have the sign bit flipped of their positive counterparts, but this approach has to deal with interpreting 1000 (one 1 followed by all 0s) as "negative zero" which is confusing.
"Ones' complement" negative numbers are just the bit-complement of their positive counterparts, which also leads to a confusing "negative zero" with 1111 (all ones).
You will likely not have to deal with Ones' Complement or Sign-Magnitude integer representations unless you are working very close to the hardware.
I wonder if it could be explained any better than the Wikipedia article.
The basic problem that you are trying to solve with two's complement representation is the problem of storing negative integers.
First, consider an unsigned integer stored in 4 bits. You can have the following
0000 = 0
0001 = 1
0010 = 2
...
1111 = 15
These are unsigned because there is no indication of whether they are negative or positive.
Sign Magnitude and Excess Notation
To store negative numbers you can try a number of things. First, you can use sign magnitude notation which assigns the first bit as a sign bit to represent +/- and the remaining bits to represent the magnitude. So using 4 bits again and assuming that 1 means - and 0 means + then you have
0000 = +0
0001 = +1
0010 = +2
...
1000 = -0
1001 = -1
1111 = -7
So, you see the problem there? We have positive and negative 0. The bigger problem is adding and subtracting binary numbers. The circuits to add and subtract using sign magnitude will be very complex.
What is
0010
1001 +
----
?
Another system is excess notation. You can store negative numbers, you get rid of the two zeros problem but addition and subtraction remains difficult.
So along comes two's complement. Now you can store positive and negative integers and perform arithmetic with relative ease. There are a number of methods to convert a number into two's complement. Here's one.
Convert Decimal to Two's Complement
Convert the number to binary (ignore the sign for now)
e.g. 5 is 0101 and -5 is 0101
If the number is a positive number then you are done.
e.g. 5 is 0101 in binary using two's complement notation.
If the number is negative then
3.1 find the complement (invert 0's and 1's)
e.g. -5 is 0101 so finding the complement is 1010
3.2 Add 1 to the complement 1010 + 1 = 1011.
Therefore, -5 in two's complement is 1011.
So, what if you wanted to do 2 + (-3) in binary? 2 + (-3) is -1.
What would you have to do if you were using sign magnitude to add these numbers? 0010 + 1101 = ?
Using two's complement consider how easy it would be.
2 = 0010
-3 = 1101 +
-------------
-1 = 1111
Converting Two's Complement to Decimal
Converting 1111 to decimal:
The number starts with 1, so it's negative, so we find the complement of 1111, which is 0000.
Add 1 to 0000, and we obtain 0001.
Convert 0001 to decimal, which is 1.
Apply the sign = -1.
Tada!
Like most explanations I've seen, the ones above are clear about how to work with 2's complement, but don't really explain what they are mathematically. I'll try to do that, for integers at least, and I'll cover some background that's probably familiar first.
Recall how it works for decimal: 2345 is a way of writing 2 × 103 + 3 × 102 + 4 × 101 + 5 × 100.
In the same way, binary is a way of writing numbers using just 0 and 1 following the same general idea, but replacing those 10s above with 2s. Then in binary, 1111is a way of writing 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20and if you work it out, that turns out to equal 15 (base 10). That's because it is 8+4+2+1 = 15.
This is all well and good for positive numbers. It even works for negative numbers if you're willing to just stick a minus sign in front of them, as humans do with decimal numbers. That can even be done in computers, sort of, but I haven't seen such a computer since the early 1970's. I'll leave the reasons for a different discussion.
For computers it turns out to be more efficient to use a complement representation for negative numbers. And here's something that is often overlooked. Complement notations involve some kind of reversal of the digits of the number, even the implied zeroes that come before a normal positive number. That's awkward, because the question arises: all of them? That could be an infinite number of digits to be considered.
Fortunately, computers don't represent infinities. Numbers are constrained to a particular length (or width, if you prefer). So let's return to positive binary numbers, but with a particular size. I'll use 8 digits ("bits") for these examples. So our binary number would really be 00001111or 0 × 27 + 0 × 26 + 0 × 25 + 0 × 24 + 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20
To form the 2's complement negative, we first complement all the (binary) digits to form 11110000and add 1 to form 11110001but how are we to understand that to mean -15?
The answer is that we change the meaning of the high-order bit (the leftmost one). This bit will be a 1 for all negative numbers. The change will be to change the sign of its contribution to the value of the number it appears in. So now our 11110001 is understood to represent -1 × 27 + 1 × 26 + 1 × 25 + 1 × 24 + 0 × 23 + 0 × 22 + 0 × 21 + 1 × 20Notice that "-" in front of that expression? It means that the sign bit carries the weight -27, that is -128 (base 10). All the other positions retain the same weight they had in unsigned binary numbers.
Working out our -15, it is -128 + 64 + 32 + 16 + 1 Try it on your calculator. it's -15.
Of the three main ways that I've seen negative numbers represented in computers, 2's complement wins hands down for convenience in general use. It has an oddity, though. Since it's binary, there have to be an even number of possible bit combinations. Each positive number can be paired with its negative, but there's only one zero. Negating a zero gets you zero. So there's one more combination, the number with 1 in the sign bit and 0 everywhere else. The corresponding positive number would not fit in the number of bits being used.
What's even more odd about this number is that if you try to form its positive by complementing and adding one, you get the same negative number back. It seems natural that zero would do this, but this is unexpected and not at all the behavior we're used to because computers aside, we generally think of an unlimited supply of digits, not this fixed-length arithmetic.
This is like the tip of an iceberg of oddities. There's more lying in wait below the surface, but that's enough for this discussion. You could probably find more if you research "overflow" for fixed-point arithmetic. If you really want to get into it, you might also research "modular arithmetic".
2's complement is very useful for finding the value of a binary, however I thought of a much more concise way of solving such a problem(never seen anyone else publish it):
take a binary, for example: 1101 which is [assuming that space "1" is the sign] equal to -3.
using 2's complement we would do this...flip 1101 to 0010...add 0001 + 0010 ===> gives us 0011. 0011 in positive binary = 3. therefore 1101 = -3!
What I realized:
instead of all the flipping and adding, you can just do the basic method for solving for a positive binary(lets say 0101) is (23 * 0) + (22 * 1) + (21 * 0) + (20 * 1) = 5.
Do exactly the same concept with a negative!(with a small twist)
take 1101, for example:
for the first number instead of 23 * 1 = 8 , do -(23 * 1) = -8.
then continue as usual, doing -8 + (22 * 1) + (21 * 0) + (20 * 1) = -3
Imagine that you have a finite number of bits/trits/digits/whatever. You define 0 as all digits being 0, and count upwards naturally:
00
01
02
..
Eventually you will overflow.
98
99
00
We have two digits and can represent all numbers from 0 to 100. All those numbers are positive! Suppose we want to represent negative numbers too?
What we really have is a cycle. The number before 2 is 1. The number before 1 is 0. The number before 0 is... 99.
So, for simplicity, let's say that any number over 50 is negative. "0" through "49" represent 0 through 49. "99" is -1, "98" is -2, ... "50" is -50.
This representation is ten's complement. Computers typically use two's complement, which is the same except using bits instead of digits.
The nice thing about ten's complement is that addition just works. You do not need to do anything special to add positive and negative numbers!
I read a fantastic explanation on Reddit by jng, using the odometer as an analogy.
It is a useful convention. The same circuits and logic operations that
add / subtract positive numbers in binary still work on both positive
and negative numbers if using the convention, that's why it's so
useful and omnipresent.
Imagine the odometer of a car, it rolls around at (say) 99999. If you
increment 00000 you get 00001. If you decrement 00000, you get 99999
(due to the roll-around). If you add one back to 99999 it goes back to
00000. So it's useful to decide that 99999 represents -1. Likewise, it is very useful to decide that 99998 represents -2, and so on. You have
to stop somewhere, and also by convention, the top half of the numbers
are deemed to be negative (50000-99999), and the bottom half positive
just stand for themselves (00000-49999). As a result, the top digit
being 5-9 means the represented number is negative, and it being 0-4
means the represented is positive - exactly the same as the top bit
representing sign in a two's complement binary number.
Understanding this was hard for me too. Once I got it and went back to
re-read the books articles and explanations (there was no internet
back then), it turned out a lot of those describing it didn't really
understand it. I did write a book teaching assembly language after
that (which did sell quite well for 10 years).
Two complement is found out by adding one to 1'st complement of the given number.
Lets say we have to find out twos complement of 10101 then find its ones complement, that is, 01010 add 1 to this result, that is, 01010+1=01011, which is the final answer.
Lets get the answer 10 – 12 in binary form using 8 bits:
What we will really do is 10 + (-12)
We need to get the compliment part of 12 to subtract it from 10.
12 in binary is 00001100.
10 in binary is 00001010.
To get the compliment part of 12 we just reverse all the bits then add 1.
12 in binary reversed is 11110011. This is also the Inverse code (one's complement).
Now we need to add one, which is now 11110100.
So 11110100 is the compliment of 12! Easy when you think of it this way.
Now you can solve the above question of 10 - 12 in binary form.
00001010
11110100
-----------------
11111110
Looking at the two's complement system from a math point of view it really makes sense. In ten's complement, the idea is to essentially 'isolate' the difference.
Example: 63 - 24 = x
We add the complement of 24 which is really just (100 - 24). So really, all we are doing is adding 100 on both sides of the equation.
Now the equation is: 100 + 63 - 24 = x + 100, that is why we remove the 100 (or 10 or 1000 or whatever).
Due to the inconvenient situation of having to subtract one number from a long chain of zeroes, we use a 'diminished radix complement' system, in the decimal system, nine's complement.
When we are presented with a number subtracted from a big chain of nines, we just need to reverse the numbers.
Example: 99999 - 03275 = 96724
That is the reason, after nine's complement, we add 1. As you probably know from childhood math, 9 becomes 10 by 'stealing' 1. So basically it's just ten's complement that takes 1 from the difference.
In Binary, two's complement is equatable to ten's complement, while one's complement to nine's complement. The primary difference is that instead of trying to isolate the difference with powers of ten (adding 10, 100, etc. into the equation) we are trying to isolate the difference with powers of two.
It is for this reason that we invert the bits. Just like how our minuend is a chain of nines in decimal, our minuend is a chain of ones in binary.
Example: 111111 - 101001 = 010110
Because chains of ones are 1 below a nice power of two, they 'steal' 1 from the difference like nine's do in decimal.
When we are using negative binary number's, we are really just saying:
0000 - 0101 = x
1111 - 0101 = 1010
1111 + 0000 - 0101 = x + 1111
In order to 'isolate' x, we need to add 1 because 1111 is one away from 10000 and we remove the leading 1 because we just added it to the original difference.
1111 + 1 + 0000 - 0101 = x + 1111 + 1
10000 + 0000 - 0101 = x + 10000
Just remove 10000 from both sides to get x, it's basic algebra.
The word complement derives from completeness. In the decimal world the numerals 0 through 9 provide a complement (complete set) of numerals or numeric symbols to express all decimal numbers. In the binary world the numerals 0 and 1 provide a complement of numerals to express all binary numbers. In fact The symbols 0 and 1 must be used to represent everything (text, images, etc) as well as positive (0) and negative (1).
In our world the blank space to the left of number is considered as zero:
35=035=000000035.
In a computer storage location there is no blank space. All bits (binary digits) must be either 0 or 1. To efficiently use memory numbers may be stored as 8 bit, 16 bit, 32 bit, 64 bit, 128 bit representations. When a number that is stored as an 8 bit number is transferred to a 16 bit location the sign and magnitude (absolute value) must remain the same. Both 1's complement and 2's complement representations facilitate this.
As a noun:
Both 1's complement and 2's complement are binary representations of signed quantities where the most significant bit (the one on the left) is the sign bit. 0 is for positive and 1 is for negative.
2s complement does not mean negative. It means a signed quantity. As in decimal the magnitude is represented as the positive quantity. The structure uses sign extension to preserve the quantity when promoting to a register [] with more bits:
[0101]=[00101]=[00000000000101]=5 (base 10)
[1011]=[11011]=[11111111111011]=-5(base 10)
As a verb:
2's complement means to negate. It does not mean make negative. It means if negative make positive; if positive make negative. The magnitude is the absolute value:
if a >= 0 then |a| = a
if a < 0 then |a| = -a = 2scomplement of a
This ability allows efficient binary subtraction using negate then add.
a - b = a + (-b)
The official way to take the 1's complement is for each digit subtract its value from 1.
1'scomp(0101) = 1010.
This is the same as flipping or inverting each bit individually. This results in a negative zero which is not well loved so adding one to te 1's complement gets rid of the problem.
To negate or take the 2s complement first take the 1s complement then add 1.
Example 1 Example 2
0101 --original number 1101
1's comp 1010 0010
add 1 0001 0001
2's comp 1011 --negated number 0011
In the examples the negation works as well with sign extended numbers.
Adding:
1110 Carry 111110 Carry
0110 is the same as 000110
1111 111111
sum 0101 sum 000101
SUbtracting:
1110 Carry 00000 Carry
0110 is the same as 00110
-0111 +11001
---------- ----------
sum 0101 sum 11111
Notice that when working with 2's complement, blank space to the left of the number is filled with zeros for positive numbers butis filled with ones for negative numbers. The carry is always added and must be either a 1 or 0.
Cheers
2's complement is essentially a way of coming up with the additive inverse of a binary number. Ask yourself this: Given a number in binary form (present at a fixed length memory location), what bit pattern, when added to the original number (at the fixed length memory location), would make the result all zeros ? (at the same fixed length memory location). If we could come up with this bit pattern then that bit pattern would be the -ve representation (additive inverse) of the original number; as by definition adding a number to its additive inverse always results in zero. Example: take 5 which is 101 present inside a single 8 bit byte. Now the task is to come up with a bit pattern which when added to the given bit pattern (00000101) would result in all zeros at the memory location which is used to hold this 5 i.e. all 8 bits of the byte should be zero. To do that, start from the right most bit of 101 and for each individual bit, again ask the same question: What bit should I add to the current bit to make the result zero ? continue doing that taking in account the usual carry over. After we are done with the 3 right most places (the digits that define the original number without regard to the leading zeros) the last carry goes in the bit pattern of the additive inverse. Furthermore, since we are holding in the original number in a single 8 bit byte, all other leading bits in the additive inverse should also be 1's so that (and this is important) when the computer adds "the number" (represented using the 8 bit pattern) and its additive inverse using "that" storage type (a byte) the result in that byte would be all zeros.
1 1 1
----------
1 0 1
1 0 1 1 ---> additive inverse
---------
0 0 0
Many of the answers so far nicely explain why two's complement is used to represent negative numbers, but do not tell us what two's complement number is, particularly not why a '1' is added, and in fact often added in a wrong way.
The confusion comes from a poor understanding of the definition of a complement number. A complement is the missing part that would make something complete.
The radix complement of an n digit number x in radix b is, by definition, b^n-x.
In binary 4 is represented by 100, which has 3 digits (n=3) and a radix of 2 (b=2). So its radix complement is b^n-x = 2^3-4=8-4=4 (or 100 in binary).
However, in binary obtaining a radix's complement is not as easy as getting its diminished radix complement, which is defined as (b^n-1)-y, just 1 less than that of radix complement. To get a diminished radix complement, you simply flip all the digits.
100 -> 011 (diminished (one's) radix complement)
to obtain the radix (two's) complement, we simply add 1, as the definition defined.
011 +1 ->100 (two's complement).
Now with this new understanding, let's take a look of the example given by Vincent Ramdhanie (see above second response):
Converting 1111 to decimal:
The number starts with 1, so it's negative, so we find the complement of 1111, which is 0000.
Add 1 to 0000, and we obtain 0001.
Convert 0001 to decimal, which is 1.
Apply the sign = -1.
Tada!
Should be understood as:
The number starts with 1, so it's negative. So we know it is a two's complement of some value x. To find the x represented by its two's complement, we first need find its 1's complement.
two's complement of x: 1111
one's complement of x: 1111-1 ->1110;
x = 0001, (flip all digits)
Apply the sign -, and the answer =-x =-1.
I liked lavinio's answer, but shifting bits adds some complexity. Often there's a choice of moving bits while respecting the sign bit or while not respecting the sign bit. This is the choice between treating the numbers as signed (-8 to 7 for a nibble, -128 to 127 for bytes) or full-range unsigned numbers (0 to 15 for nibbles, 0 to 255 for bytes).
It is a clever means of encoding negative integers in such a way that approximately half of the combination of bits of a data type are reserved for negative integers, and the addition of most of the negative integers with their corresponding positive integers results in a carry overflow that leaves the result to be binary zero.
So, in 2's complement if one is 0x0001 then -1 is 0x1111, because that will result in a combined sum of 0x0000 (with an overflow of 1).
2’s Complements: When we add an extra one with the 1’s complements of a number we will get the 2’s complements. For example: 100101 it’s 1’s complement is 011010 and 2’s complement is 011010+1 = 011011 (By adding one with 1's complement) For more information
this article explain it graphically.
Two's complement is mainly used for the following reasons:
To avoid multiple representations of 0
To avoid keeping track of carry bit (as in one's complement) in case of an overflow.
Carrying out simple operations like addition and subtraction becomes easy.
Two's complement is one of the ways of expressing a negative number and most of the controllers and processors store a negative number in two's complement form.
In simple terms, two's complement is a way to store negative numbers in computer memory. Whereas positive numbers are stored as a normal binary number.
Let's consider this example,
The computer uses the binary number system to represent any number.
x = 5;
This is represented as 0101.
x = -5;
When the computer encounters the - sign, it computes its two's complement and stores it.
That is, 5 = 0101 and its two's complement is 1011.
The important rules the computer uses to process numbers are,
If the first bit is 1 then it must be a negative number.
If all the bits except first bit are 0 then it is a positive number, because there is no -0 in number system (1000 is not -0 instead it is positive 8).
If all the bits are 0 then it is 0.
Else it is a positive number.
To bitwise complement a number is to flip all the bits in it. To two’s complement it, we flip all the bits and add one.
Using 2’s complement representation for signed integers, we apply the 2’s complement operation to convert a positive number to its negative equivalent and vice versa. So using nibbles for an example, 0001 (1) becomes 1111 (-1) and applying the op again, returns to 0001.
The behaviour of the operation at zero is advantageous in giving a single representation for zero without special handling of positive and negative zeroes. 0000 complements to 1111, which when 1 is added. overflows to 0000, giving us one zero, rather than a positive and a negative one.
A key advantage of this representation is that the standard addition circuits for unsigned integers produce correct results when applied to them. For example adding 1 and -1 in nibbles: 0001 + 1111, the bits overflow out of the register, leaving behind 0000.
For a gentle introduction, the wonderful Computerphile have produced a video on the subject.
The question is 'What is “two's complement”?'
The simple answer for those wanting to understand it theoretically (and me seeking to complement the other more practical answers): 2's complement is the representation for negative integers in the dual system that does not require additional characters, such as + and -.
Two's complement of a given number is the number got by adding 1 with the ones' complement of the number.
Suppose, we have a binary number: 10111001101
Its 1's complement is: 01000110010
And its two's complement will be: 01000110011
Reference: Two's Complement (Thomas Finley)
I invert all the bits and add 1. Programmatically:
// In C++11
int _powers[] = {
1,
2,
4,
8,
16,
32,
64,
128
};
int value = 3;
int n_bits = 4;
int twos_complement = (value ^ ( _powers[n_bits]-1)) + 1;
You can also use an online calculator to calculate the two's complement binary representation of a decimal number: http://www.convertforfree.com/twos-complement-calculator/
The simplest answer:
1111 + 1 = (1)0000. So 1111 must be -1. Then -1 + 1 = 0.
It's perfect to understand these all for me.
My apologies if this has been asked/answered before but I'm honestly not even sure how to word this as a question properly. I have the following bit pattern:
0110110110110110110110110110110110110110110110110110110110110110
I'm trying to perform a shift that'll preserve my underlying pattern; my first instinct was to use right rotation ((x >> count) | (x << (-count & 63))) but the asymmetry in my bit pattern results in:
0011011011011011011011011011011011011011011011011011011011011011 <--- wrong
The problem is that the most significant (far left) bit ends up being 0 instead of the desired 1:
1011011011011011011011011011011011011011011011011011011011011011 <--- right
Is there a colloquial name for this function I'm looking for? If not, how could I go about implementing this idea?
Additional Information:
While the question is language agnostic I'm currently trying to solve this using C#.
The bit patterns I'm using are entirely predictable and always have the same structure; the pattern starts with a single zero followed by n - 1 ones (where n is an odd number) and then repeats infinitely.
I'd like to accomplish this without conditional operations since they'd defeat the purpose of using bitwise manipulation in the first place but maybe I have no choice...
You've got a number structured like this:
B16 B15 B14 B13 B12 B11 B10 B09 B08 B07 B06 B05 B04 B03 B02 B01 B00
? 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0
The ? needs to appear in the MSB (B15, or B63, or whatever) after the shift. Where does it come from? Well, the closest copy is found n places to the right:
B13 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0
^--------------/
If your word has width w, this is 1 << (w-n)
*
So you can do:
var selector = 1 << (w-n);
var rotated = (val >> 1) | ((val & selector) << (n-1));
But you may want a multiple shift. Then we need to build a wider mask:
? 0 1 1 0 1 1 0 1 1 0 1 1 0 1 1 0
* * * * *
Here I've chosen to pretend n = 6, it just needs to be a multiple of the basic n, and larger than shift. Now:
var selector = ((1UL << shift) - 1) << (w - n);
var rotated = (val >> shift) | ((val & selector) << (n - shift));
Working demonstration using your pattern: http://rextester.com/UWYSW47054
It's easy to see that the output has period 3, as required:
1:B6DB6DB6DB6DB6DB
2:DB6DB6DB6DB6DB6D
3:6DB6DB6DB6DB6DB6
4:B6DB6DB6DB6DB6DB
5:DB6DB6DB6DB6DB6D
6:6DB6DB6DB6DB6DB6
7:B6DB6DB6DB6DB6DB
8:DB6DB6DB6DB6DB6D
9:6DB6DB6DB6DB6DB6
10:B6DB6DB6DB6DB6DB
11:DB6DB6DB6DB6DB6D
Instead of storing a lot of repetitions of a pattern, just store one recurrence and apply modulo operations on the indexes
byte[] pattern = new byte[] { 0, 1, 1 };
// Get a "bit" at index "i", shifted right by "shift"
byte bit = pattern[(i - shift + 1000000 * byte.Length) % byte.Length];
The + 1000000 * byte.Length must be greater than the greatest expected shift and ensures that we get a posistive sum.
This allows you to store patterns of virtually any length.
An optimization would be to store a mirrored version of the pattern. You could then shift left instead of right. This would simplify the index calculation
byte bit = pattern[(i + shift) % byte.Length];
Branchless Answer after a poke by #BenVoigt:
Get the last bit b by doing (n & 1);
Return n >> 1 | b << ((sizeof(n) - 1).
Original Answer:
Get the last bit b by doing (n & 1);
If b is 1, right shift the number by 1 bit and bitwise-OR it with 1 << (sizeof(n) - 1);
If b is 0, just right shift the number by 1 bit.
The problem was changed a bit through the comments.
For all reasonable n, the following problem can be solved efficiently after minimal pre-computation:
Given an offset k, get 64 bits starting at that position in the stream of bits that follows the pattern of (zero, n-1 ones) repeating.
Clearly the pattern repeats with a period of n, so only n different ulongs have to be produced for every given value of n. That could either be done explicitly, constructing all of them in pre-processing (they could be constructed in any obvious way, it doesn't really matter since that only happens once), or left more implicitly by storing only two ulongs per value for n (this works under the assumption that n < 64, see below) and then extracting a range from them with some shifting/ORing. Either way, use offset % n to compute which pattern to retrieve (since the offset is increasing in a predictable manner, no actual modulo operation is required[1]).
Even with the first method, memory consumption will be reasonable since this optimization is only an optimization for low n: in particular for n > 64 there will be fewer than 1 zero per word on average, so the "old fashioned way" of visiting every multiple of n and resetting that bit starts to skip work while the above trick would still visit every word and would not be able anymore to reset multiple bits at once.
[1]: if there are multiple n's in play at the same time, a possible strategy is keeping an array offsets where offsets[n] = offset % n, which could be updated according to: (not tested)
int next = offsets[n] + _64modn[n]; // 64 % n precomputed
offsets[n] = next - (((n - next - 1) >> 31) & n);
The idea being that n is subtracted whenever next >= n. Only one subtraction is needed since the offset and thing added to the offset are already reduced modulo n.
This offset-increment can be done with System.Numerics.Vectors, which is very feature-poor compared to actual hardware but is just about able to do this. It can't do the shift (yes, it's weird) but it can implement a comparison in a branchless way.
Doing one pass per value of n is easier, but touches lots of memory in a cache unfriendly manner. Doing lots of different n at the same time may not be great either. I guess you'd just have to bechmark that..
Also you could consider hard-coding it for some low numbers, something like offset % 3 is fairly efficient (unlike offset % variable). This does take manual loop-unrolling which is a bit annoying, but it's actually simpler, just big in terms of lines of code.
This question already has answers here:
What does the operator "<<" mean in C#?
(9 answers)
Closed 7 years ago.
i couldn't understand this code in c#
int i=4
int[] s =new int [1<<i];
Console.WriteLine(s.length);
the ouput is 16
i don't know why the output like that?
bit shift operator
From documentation
If first operand is an int or uint
(32-bit quantity), the shift count is
given by the low-order five bits of
second operand.
If first operand is a long or ulong
(64-bit quantity), the shift count is
given by the low-order six bits of
second operand.
Note that i<<1 and i<<33 give the same
result, because 1 and 33 have the same
low-order five bits.
This will be the same as 2^( the actual value of the lower 5 bits ).
So in your case it would be 2^4=16.
I'm assuming you mean i in place of r...
<<n means "shift left by n* bits". Since you start with 1=binary 00...00001, if you shift left 4 times you get binary 00...10000 = 16 (it helps if you are familiar with binary arithmetic - otherwise "calc.exe" has a binary converter).
Each bit moves left n places, filling (on the right) with 0s. *=note that n is actually "mod 32" for int, so (as a corner case) 1 << 33 = 2, not 0 which you might expect.
There is also >> (right shift), which moves for the right, filling with 0 for uints and +ve ints, and 1 for -ve ints.
<< is the left shift operator
x << y
means shift x to the left by y bits.
3 is 0011, 3<<1 is 0110 which 6.
It's usually used to multiply by 2 (shifting to the left is multiplying by 2)
As already mentioned, << is the left shift operator. In your particular example, the array size is being defined as a power of 2. The value 1 shifted left by some number is going to be 1, 2, 4, 8, 16, ...
I am working on a little Hardware interface project based on the Velleman k8055 board.
The example code comes in VB.Net and I'm rewriting this into C#, mostly to have a chance to step through the code and make sense of it all.
One thing has me baffled though:
At one stage they read all digital inputs and then set a checkbox based on the answer to the read digital inputs (which come back in an Integer) and then they AND this with a number:
i = ReadAllDigital
cbi(1).Checked = (i And 1)
cbi(2).Checked = (i And 2) \ 2
cbi(3).Checked = (i And 4) \ 4
cbi(4).Checked = (i And 8) \ 8
cbi(5).Checked = (i And 16) \ 16
I have not done Digital systems in a while and I understand what they are trying to do but what effect would it have to AND two numbers? Doesn't everything above 0 equate to true?
How would you translate this to C#?
This is doing a bitwise AND, not a logical AND.
Each of those basically determines whether a single bit in i is set, for instance:
5 AND 4 = 4
5 AND 2 = 0
5 AND 1 = 1
(Because 5 = binary 101, and 4, 2 and 1 are the decimal values of binary 100, 010 and 001 respectively.)
I think you 'll have to translate it to this:
i & 1 == 1
i & 2 == 2
i & 4 == 4
etc...
This is using the bitwise AND operator.
When you use the bitwise AND operator, this operator will compare the binary representation of the two given values, and return a binary value where only those bits are set, that are also set in the two operands.
For instance, when you do this:
2 & 2
It will do this:
0010 & 0010
And this will result in:
0010
0010
&----
0010
Then if you compare this result with 2 (0010), it will ofcourse return true.
Just to add:
It's called bitmasking
http://en.wikipedia.org/wiki/Mask_(computing)
A boolean only require 1 bit. In the implementation most programming language, a boolean takes more than a single bit. In PC this won't be a big waste, but embedded system usually have very limited memory space, so the waste is really significant. To save space, the booleans are packed together, this way a boolean variable only takes up 1 bit.
You can think of it as doing something like an array indexing operation, with a byte (= 8 bits) becoming like an array of 8 boolean variables, so maybe that's your answer: use an array of booleans.
Think of this in binary e.g.
10101010
AND
00000010
yields 00000010
i.e. not zero. Now if the first value was
10101000
you'd get
00000000
i.e. zero.
Note the further division to reduce everything to 1 or 0.
(i and 16) / 16 extracts the value (1 or 0) of the 5th bit.
1xxxx and 16 = 16 / 16 = 1
0xxxx and 16 = 0 / 16 = 0
And operator performs "...bitwise conjunction on two numeric expressions", which maps to '|' in C#. The '` is an integer division, and equivalent in C# is /, provided that both operands are integer types.
The constant numbers are masks (think of them in binary). So what the code does is apply the bitwise AND operator on the byte and the mask and divide by the number, in order to get the bit.
For example:
xxxxxxxx & 00000100 = 00000x000
if x == 1
00000x00 / 00000100 = 000000001
else if x == 0
00000x00 / 00000100 = 000000000
In C# use the BitArray class to directly index individual bits.
To set an individual bit i is straightforward:
b |= 1 << i;
To reset an individual bit i is a little more awkward:
b &= ~(1 << i);
Be aware that both the bitwise operators and the shift operators tend to promote everything to int which may unexpectedly require casting.
As said this is a bitwise AND, not a logical AND. I do see that this has been said quite a few times before me, but IMO the explanations are not so easy to understand.
I like to think of it like this:
Write up the binary numbers under each other (here I'm doing 5 and 1):
101
001
Now we need to turn this into a binary number, where all the 1's from the 1st number, that is also in the second one gets transfered, that is - in this case:
001
In this case we see it gives the same number as the 2nd number, in which this operation (in VB) returns true. Let's look at the other examples (using 5 as i):
(5 and 2)
101
010
----
000
(false)
(5 and 4)
101
100
---
100
(true)
(5 and 8)
0101
1000
----
0000
(false)
(5 and 16)
00101
10000
-----
00000
(false)
EDIT: and obviously I miss the entire point of the question - here's the translation to C#:
cbi[1].Checked = i & 1 == 1;
cbi[2].Checked = i & 2 == 2;
cbi[3].Checked = i & 4 == 4;
cbi[4].Checked = i & 8 == 8;
cbi[5].Checked = i & 16 == 16;
I prefer to use hexadecimal notation when bit twiddling (e.g. 0x10 instead of 16). It makes more sense as you increase your bit depths as 0x20000 is better than 131072.