I want to know the function of operator &. For example:
SELECT (8 & 16)
In the last code if I change the second value (16) to another one, like 10, the result changes.
I read about it, but I didn't found an accurate answer.
Thank's for your help.
For SQL Server you can find the answer here https://learn.microsoft.com/en-us/sql/t-sql/language-elements/bitwise-and-transact-sql?view=sql-server-ver15
The & bitwise operator performs a bitwise logical AND between the two expressions, taking each corresponding bit for both expressions. The bits in the result are set to 1 if and only if both bits (for the current bit being resolved) in the input expressions have a value of 1; otherwise, the bit in the result is set to 0.
If the left and right expressions have different integer data types (for example, the left expression is smallint and the right expression is int), the argument of the smaller data type is converted to the larger data type. In this case, the smallintexpression is converted to an int.
To come back to your Example:
If you compare 2 Integer numbers bitwise this will be calculated:
8 = 0 0 0 0 1 0 0 0
16 = 0 0 0 1 0 0 0 0
0 0 0 0 0 0 0 0 = 0
8 = 0 0 0 0 1 0 0 0
10 = 0 0 0 0 1 0 1 0
0 0 0 0 1 0 0 0 = 8
Hope that helps.
In MS SQL Server it is a bitwise AND operator.
If bits at any location are both 1, the result is 1.
1010 1010 = 170
0100 1011 = 75
0000 1010 = 10
Select 170 & 75
Returns 10
In most databases, & is the bitwise AND operator. What it does is translate its integer argument to their bit representation, and apply the AND operator to each individual bit (this is equivalent to multiplying each bit invidually):
For your operation 8 & 16:
8 is represented as 1000 in binary
16 corresponds to 10000
Performing the bitwise and:
01000
AND 10000
-----
00000 (0)
I got this query from some one and I queried this in SQL Server 2012
Select 20 | 25 as test
O/P : 29
select 20 & 25 as test
O/P : 16
select 10 | 25 as test
O/P : 27
select 10 & 25 as test
O/P : 8
I need step by step explanation for this output, can any one help me out.
Unlike binary calculators SQL Server does not add values if the
operator is OR, it uses pipe character (|) for bitwise OR
operator. It returns 1 if one or both bits are 1
AND returns 1 if both bits are 1. Either of the bit is 0 it returns 0
Step by Step Explanation as you were looking for
Select 20 | 25 as test <-- we will do it for this one first
First it will convert these values to Binary i.e
20 = 10100
25 = 11001
Now it returns 1 if one or both bits are 1 else it returns 0.
Considering this the result is 11101 which is 29
select 20 & 25 as test <-- Lets have a look into this now
Convert to Binary -
20 = 10100
25 = 11001
& returns 1 if both bits are 1 else returns 0
According to this the result is 10000 which is 16
Why does a double of 65555 converted to byte produce a result of 19 in Kotlin?
That's because of a numerical conversion from a wider type to a type of a smaller size. The Double (IEEE 754 double precision number) has its integral part factored to the powers of two as 65555 = 217 + 24 + 22 + 20 = 65536 + 16 + 2 + 1, which is stored in binary form as (higher bits to lower):
... 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 1 1
When this number is converted to Byte, only its lowest 8 bits are retained:
... _ _ _ _ _ _ _ _ _ _ 0 0 0 1 0 0 1 1
And that results into 24 + 22 + 20 = 16 + 2 + 1 = 19.
Because when you convert 65555 (or 65555.0) to a binary representation it takes more than one byte. So calling .toByte() takes the lowest one, which is 19.
65555 -> Binary == 1 00000000 00010011
^ ^^^^^^^^ ^^^^^^^^
1 0 19
Double 65555.0 gets converted to integer 65555 which is 0x10013. Conversion to Byte takes lower byte which is 0x13 (19 decimal).
I have to update the 32 bit register using these data
which includes bit shifting, I am confused about two things:
Which is the LSB and which is the MSB,
what is this operator |
Given the expression:
3 << 0 |
7 << 3 |
1 << 6 |
0 << 7 |
1 << 7 |
0 << 8 |
0 << 10 |
0 << 11 |
0 << 12 |
0 << 13 |
0 << 14
and the remaining 15 bits are 0.
How is the data shifted, assuming initial bits in register to be 0's?
011 111 1 0 1 0 0 0 0 0 0 X.......X
or
x .....X 0 0 0 0 0 0 1 0 1 111 011
The LSB (least-significant bit) is the bit whose value represents 1 (2^0) and the MSB is the bit whose value represents 2^(n-1), where n is the number of bits in the register. In general, when written out in binary, the MSB is the left-most bit and the LSB is the right-most bit. More often than not, the LSB is shown as bit 0 in hardware documentation, though I know one company that reverses the bit numbering so that the MSB is numbered 0.
<< is the C shift-left operator, shifting bit away from the LSB and toward the MSB. Therefore 7<<3 represents 111000 in binary.
| is the C bit-wise OR operator. This is used to combine values where resulting bits are one if either of the corresponding input bits is one.
Looking at your original value 3 << 0 | 7 << 3 | 1 << 6 | 0 << 7 | 1 << 7 | 0 << 8 | 0 << 10 | 0 << 11| 0 << 12 | 0 << 13 | 0 << 14
0000 0000 0000 0011 from 3<<0
0000 0000 0011 1000 from 7<<3
0000 0000 0100 0000 from 1<<6
0000 0000 0000 0000 from 0<<7
etc.
This type of construct is typically used to describe a value going into a register noting the individual fields of the register.
Closed. This question is off-topic. It is not currently accepting answers.
Want to improve this question? Update the question so it's on-topic for Stack Overflow.
Closed 11 years ago.
Improve this question
Write code to determine if a number is divisible by 3. The input to the function is a single bit, 0 or 1, and the output should be 1 if the number received so far is the binary representation of a number divisible by 3, otherwise zero.
Examples:
input "0": (0) output 1
inputs "1,0,0": (4) output 0
inputs "1,1,0,0": (6) output 1
This is based on an interview question. I ask for a drawing of logic gates but since this is stackoverflow I'll accept any coding language. Bonus points for a hardware implementation (verilog etc).
Part a (easy): First input is the MSB.
Part b (a little harder): First input is the LSB.
Part c (difficult): Which one is faster and smaller, (a) or (b)? (Not theoretically in the Big-O sense, but practically faster/smaller.) Now take the slower/bigger one and make it as fast/small as the faster/smaller one.
There's a fairly well-known trick for determining whether a number is a multiple of 11, by alternately adding and subtracting its decimal digits. If the number you get at the end is a multiple of 11, then the number you started out with is also a multiple of 11:
47278 4 - 7 + 2 - 7 + 8 = 0, multiple of 11 (47278 = 11 * 4298)
52214 5 - 2 + 2 - 1 + 4 = 8, not multiple of 11 (52214 = 11 * 4746 + 8)
We can apply the same trick to binary numbers. A binary number is a multiple of 3 if and only if the alternating sum of its bits is also a multiple of 3:
4 = 100 1 - 0 + 0 = 1, not multiple of 3
6 = 110 1 - 1 + 0 = 0, multiple of 3
78 = 1001110 1 - 0 + 0 - 1 + 1 - 1 + 0 = 0, multiple of 3
109 = 1101101 1 - 1 + 0 - 1 + 1 - 0 + 1 = 1, not multiple of 3
It makes no difference whether you start with the MSB or the LSB, so the following Python function works equally well in both cases. It takes an iterator that returns the bits one at a time. multiplier alternates between 1 and 2 instead of 1 and -1 to avoid taking the modulo of a negative number.
def divisibleBy3(iterator):
multiplier = 1
accumulator = 0
for bit in iterator:
accumulator = (accumulator + bit * multiplier) % 3
multiplier = 3 - multiplier
return accumulator == 0
Here... something new... how to check if a binary number of any length (even thousands of digits) is divisible by 3.
-->((0))<---1--->()<---0--->(1) ASCII representation of graph
From the picture.
You start in the double circle.
When you get a one or a zero, if the digit is inside the circle, then you stay in that circle. However if the digit is on a line, then you travel across the line.
Repeat step two until all digits are comsumed.
If you finally end up in the double circle then the binary number is divisible by 3.
You can also use this for generating numbers divisible by 3. And I wouldn't image it would be hard to convert this into a circuit.
1 example using the graph...
11000000000001011111111111101 is divisible by 3 (ends up in the double circle again)
Try it for yourself.
You can also do similar tricks for performing MOD 10, for when converting binary numbers into base 10 numbers. (10 circles, each doubled circled and represent the values 0 to 9 resulting from the modulo)
EDIT: This is for digits running left to right, it's not hard to modify the finite state machine to accept the reverse language though.
NOTE: In the ASCII representation of the graph () denotes a single circle and (()) denotes a double circle. In finite state machines these are called states, and the double circle is the accept state (the state that means its eventually divisible by 3)
Heh
State table for LSB:
S I S' O
0 0 0 1
0 1 1 0
1 0 2 0
1 1 0 1
2 0 1 0
2 1 2 0
Explanation: 0 is divisible by three. 0 << 1 + 0 = 0. Repeat using S = (S << 1 + I) % 3 and O = 1 if S == 0.
State table for MSB:
S I S' O
0 0 0 1
0 1 2 0
1 0 1 0
1 1 0 1
2 0 2 0
2 1 1 0
Explanation: 0 is divisible by three. 0 >> 1 + 0 = 0. Repeat using S = (S >> 1 + I) % 3 and O = 1 if S == 0.
S' is different from above, but O works the same, since S' is 0 for the same cases (00 and 11). Since O is the same in both cases, O_LSB = O_MSB, so to make MSB as short as LSB, or vice-versa, just use the shortest of both.
Here is an simple way to do it by hand.
Since 1 = 22 mod 3, we get 1 = 22n mod 3 for every positive integer.
Furthermore 2 = 22n+1 mod 3. Hence one can determine if an integer is divisible by 3 by counting the 1 bits at odd bit positions, multiply this number by 2, add the number of 1-bits at even bit posistions add them to the result and check if the result is divisible by 3.
Example: 5710=1110012.
There are 2 bits at odd positions, and 2 bits at even positions. 2*2 + 2 = 6 is divisible by 3. Hence 57 is divisible by 3.
Here is also a thought towards solving question c). If one inverts the bit order of a binary integer then all the bits remain at even/odd positions or all bits change. Hence inverting the order of the bits of an integer n results is an integer that is divisible by 3 if and only if n is divisible by 3. Hence any solution for question a) works without changes for question b) and vice versa. Hmm, maybe this could help to figure out which approach is faster...
You need to do all calculations using arithmetic modulo 3. This is the way
MSB:
number=0
while(!eof)
n=input()
number=(number *2 + n) mod 3
if(number == 0)
print divisible
LSB:
number = 0;
multiplier = 1;
while(!eof)
n=input()
number = (number + multiplier * n) mod 3
multiplier = (multiplier * 2) mod 3
if(number == 0)
print divisible
This is general idea...
Now, your part is to understand why this is correct.
And yes, do homework yourself ;)
The idea is that the number can grow arbitrarily long, which means you can't use mod 3 here, since your number will grow beyond the capacity of your integer class.
The idea is to notice what happens to the number. If you're adding bits to the right, what you're actually doing is shifting left one bit and adding the new bit.
Shift-left is the same as multiplying by 2, and adding the new bit is either adding 0 or 1. Assuming we started from 0, we can do this recursively based on the modulo-3 of the last number.
last | input || next | example
------------------------------------
0 | 0 || 0 | 0 * 2 + 0 = 0
0 | 1 || 1 | 0 * 2 + 1 = 1
1 | 0 || 2 | 1 * 2 + 0 = 2
1 | 1 || 0 | 1 * 2 + 1 = 0 (= 3 mod 3)
2 | 0 || 1 | 2 * 2 + 0 = 1 (= 4 mod 3)
2 | 1 || 2 | 2 * 2 + 1 = 2 (= 5 mod 3)
Now let's see what happens when you add a bit to the left. First, notice that:
22n mod 3 = 1
and
22n+1 mod 3 = 2
So now we have to either add 1 or 2 to the mod based on if the current iteration is odd or even.
last | n is even? | input || next | example
-------------------------------------------
d/c | don't care | 0 || last | last + 0*2^n = last
0 | yes | 1 || 0 | 0 + 1*2^n = 1 (= 2^n mod 3)
0 | no | 1 || 0 | 0 + 1*2^n = 2 (= 2^n mod 3)
1 | yes | 1 || 0 | 1 + 1*2^n = 2
1 | no | 1 || 0 | 1 + 1*2^n = 0 (= 3 mod 3)
1 | yes | 1 || 0 | 2 + 1*2^n = 0
1 | no | 1 || 0 | 2 + 1*2^n = 1
input "0": (0) output 1
inputs "1,0,0": (4) output 0
inputs "1,1,0,0": (6) output 1
shouldn't this last input be 12, or am i misunderstanding the question?
Actually the LSB method would actually make this easier. In C:
MSB method:
/*
Returns 1 if divisble by 3, otherwise 0
Note: It is assumed 'input' format is valid
*/
int is_divisible_by_3_msb(char *input) {
unsigned value = 0;
char *p = input;
if (*p == '1') {
value &= 1;
}
p++;
while (*p) {
if (*p != ',') {
value <<= 1;
if (*p == '1') {
ret &= 1;
}
}
p++;
}
return (value % 3 == 0) ? 1 : 0;
}
LSB method:
/*
Returns 1 if divisble by 3, otherwise 0
Note: It is assumed 'input' format is valid
*/
int is_divisible_by_3_lsb(char *input) {
unsigned value = 0;
unsigned mask = 1;
char *p = input;
while (*p) {
if (*p != ',') {
if (*p == '1') {
value &= mask;
}
mask <<= 1;
}
p++;
}
return (value % 3 == 0) ? 1 : 0;
}
Personally I have a hard time believing one of these will be significantly different to the other.
I think Nathan Fellman is on the right track for part a and b (except b needs an extra piece of state: you need to keep track of if your digit position is odd or even).
I think the trick for part C is negating the last value at each step. I.e. 0 goes to 0, 1 goes to 2 and 2 goes to 1.
A number is divisible by 3 if the sum of it's digits is divisible by 3.
So you can add the digits and get the sum:
if the sum is greater or equal to 10 use the same method
if it's 3, 6, 9 then it is divisible
if the sum is different than 3, 6, 9 then it's not divisible