Could you please advice what using "&" or "^" in SET command means (I haven't found any explanation by using Google).
For example, following Windows batch code block
SET V_COMMAND=3
SET /A V_FLAG="%V_COMMAND%&2"
echo VFlag is: %V_FLAG%
produces:
VFlag is: 2
But I haven't any opinion about what command above does.
Also there is another case with "^":
SET V_COMMAND=3
SET /A V_FLAG="%V_COMMAND%^3"
echo VFlag is: %V_FLAG%
For this case output is:
VFlag is: 0
Since you are using set /a, the indicated characters are bitwise operators:
& = bitwise AND = 1 if both bits are 1
^ = bitwise XOR = 1 if only one of the two bits is 1
So if a is 10 (1010 in binary) and b is 13 (1101 in binary)
1010 1010
1101 1101
---- ----
1000 = a & b 0111 = a ^ b
Or in your case with 3 dec = 11 bin and 2 dec = 10 bin
11 11
10 11
-- --
10 = 3&2 = 2 00 = 3^3 = 0
Those are bitwise operators - & is bitwise AND, and ^ is bitwise XOR. These bitwise operators are only available with SET /A.
Related
This question already has answers here:
Boolean operators vs Bitwise operators
(9 answers)
Closed 4 months ago.
What the different between logical operators and, or and bitwise analogs &, | in usage? Is there any difference in efficiency in various solutions?
Logical operators operate on logical values, while bitwise operators operate on integer bits. Stop thinking about performance, and use them for they're meant for.
if x and y: # logical operation
...
z = z & 0xFF # bitwise operation
Bitwise = Bit by bit checking
# Example
Bitwise AND: 1011 & 0101 = 0001
Bitwise OR: 1011 | 0101 = 1111
Logical = Logical checking or in other words, you can say True/False checking
# Example
# both are non-zero so the result is True
Logical AND: 1011 && 0101 = 1 (True)
# one number is zero so the result is False
Logical AND: 1011 && 0000 = 0 (False)
# one number is non-zero so the result is non-zero which is True
Logical OR: 1011 || 0000 = 1 (True)
# both numbers are zero so the result is zero which is False
Logical OR: 0000 || 0000 = 0 (False)
Logical operators are used for booleans, since true equals 1 and false equals 0. If you use (binary) numbers other than 1 and 0, then any number that's not zero becomes a one. Ex: int x = 5; (101 in binary) int y = 0; (0 in binary) In this case, printing x && y would print 0, because 101 was changed to 1, and 0 was kept at zero: this is the same as printing true && false, which returns false (0). On the other hand, bitwise operators perform an operation on every single bit of the two operands (hence the term "bitwise"). Ex: int x = 5; int y = 8; printing x | y (bitwise OR) would calculate this: 000101 (5)| 1000 (8) ----------- = 1101 (13) Meaning it would print 13.
In order to understand marching cube algorithm, I followed this page:
http://paulbourke.net/geometry/polygonise/
I have some questions:
what do bitwise & and | mean?, and how they will work with edge table to find the correct tringles?
if (grid.val[0] < isolevel) cubeindex |= 1;
if (edgeTable[cubeindex] & 1)
vertlist[0] =
VertexInterp(isolevel,grid.p[0],grid.p[1],grid.val[0],grid.val[1]);
Not going to read that text thats way to long. But here you can find how bit operators work https://en.wikipedia.org/wiki/Bitwise_operations_in_C. `
cubeindex |= 1 --> cubeindex = cubeindex | 1.
for example cubeindex = 26 (binary 11010) and 1 (binary 00001)
11010 | 00001 = 11011
Here you ad one 26->27.
For the followingedgeTable[cubeindex] & 1:
For example cubeindex = 17 (binary 10001) \
10001 & 00001 = 00001
This becomes 1. Used in the if statement, this just checks if the number edgeTable[cubeindex] contains the bit 00001 and returns true or false accordingly.
hope this helps :)
Cheers
Bitwise & (and) and | (or) operate on all the bits within an integer value. It operates on each bit independently, and is often used with you have a set of boolean truth values (aka flags) that indicate the state of various objects. It can also be used (as in your example) or test a specific flag while not modifying the others.
The bits in an integer represent powers of two values. If that bit is set to true, then that power of two is included in its value. If that bit is false, then it is no included. The least significant bit b0 represents 2^0, bit 1 represents 2^1, and so on.
For example, the value 5 = 101 in binary because 5 = 2^2 + 2^0 = 4 + 1 = 5
Bitwise OR works by setting the bits in the result to 1 if either of the operands contains a one in that position. Bitwise AND workds by setting each bit to 1 if and only if both operands have a 1 in that position.
For example:
Bitwise OR: 5 | 9 = 0101 | 1001 = 1101
Bitwise AND: 5 & 9 = 0101 & 1001 = 0001
Using these operators, if the set of bit values in cubeindex represent a set of states(true or false values), then you can use the same bitwise operands to test and if a specific state is true, while ignoring the other bits.
In your example:
cubeindex |= 1
sets the least significant bit in cubeindex to true (1) regardless of what it was before. This is because the bitwise OR of any bit value to 0, is the same bit value, while the bitwise OR of any bit value of 1, is always 1 and so is the same as setting that bit to 1 regardless of its previous stae. It is equivalent to:
cubeindex = cubeindex | 000000000000000001 = cubeindex with b0 set to 1
The logical AND is being used to test if the least significant bit of edgeTable[cubeindex] equals 1. The condition is only true if the b0 of this value is 1. Why? Because the other bits do not matter since they are being bitwise AND with zero, which is always zero.
edgeTable[cubeindex] & 1 = value of bit 0
Friends, I am trying to generate a random hex string on my web server. I am using Apache HTTPD on Windows 10. I am using the hex string for session ID's etc.
The code I have tried so far:
setlocal enabledelayedexpansion
set HESSTR=0123456789ABCDEF
set /a tempone="("%RANDOM%"*"16")"
set /a RANHEXS=tempone"/"32767
echo %RANHEXS%
set hexout=!HESSTR:~%RANHEXS%,1!
echo %hexout%
pause
endlocal
But this code Always returns 7... What am I doing wrong?
Thanks a lot!
Here is a method to create a random hexadecimal digit that relies on the undocumented built-in environment variable =ExitCode, which returns the exit code as an 8-digit hexadecimal number:
rem // Set ErrorLevel and exit code to a random number:
cmd /C exit %RANDOM%
rem // Return the last digit of the hexadecimal exit code:
echo %=ExitCode:~-1%
Note that the exit code can reach from 00000000 to FFFFFFFF. The value of RANDOM however covers the range from 0 to 32767, which is 0000 to 7FFF expressed in hexadecimal notation; so you can build a 3-digit hexadecimal number at most by using the last three digits.
#Set /a num=%random% %% 16 + 1
Echo %num% / 16
See set /?. % is defined in the C language as
% The result of the remainder operator is the remainder when the first operand is divided by the second
And we need to escape the % with another %, so %%.
So you divide a number by the range you want - so 165 %% 16 = 5 (the remainder). Then we add 1 to make it 1 to 16 else it would be 0 to 15.
PS CMD is unusual is being 0-32767, most random numbers are between 0 and 1 then you multiply. This is from VBScript's help - Int((upperbound - lowerbound + 1) * Rnd + lowerbound) so (16 - 1 + 1) * RandomNum + 1.
How do you find the largest power of 2 a number is divisible by using logic function
for example 144 is divisible by 16 which is 2^4.
How would one do this.
I know 144 in binary is 1001 0000 and I have to use a bitwise function.
But what would I use (and or andn orn ?) or perhaps something else and what can I use as my mask?
I know you have to look at the right most number to tell if it is divisible by 2.
Any help is appreciated
I would go with n & -n or with n & (~n + 1), in case you are worried about running across one's complement arithmetic, given the latter works with both arithmetics.
E.g.,
> 144 & (~144 + 1)
< 16
Now a short explanation.
The bitwise NOT (i.e., ~ operator) of a number n gives -(n + 1). It inverts all the bits of n. The number 2 is represented by 00000010 while its negation is 11111101 which equals to -3 (i.e., , see the two's complement representation of signed numbers).
Do not to confuse it with logical negation.
E.g., ~144 = -(144 + 1) = -145.
The bitwise AND (i.e., & operator) compares two bits of the inputs and generates a result of 1 if both are 1, otherwise it returns 0.
Now the main topic.
This is an old tricks that gives the highest power of 2 that n is divisible by. This means that it returns a number with a single one bit, specifically the bottom bit that was set in n.
For example the binary representation of 144 is 010010000. Its bottom 1 bit is the bit in fourth position (counting backward from right and starting at position 0). Thus the higher power of 2 that divides 144 is 16 (i.e., 00010000).
144 & (~144 + 1) = 144 & -144 = 16
16 & ( ~16 + 1) = 16 & - 16 = 16
10 & ( ~10 + 1) = 10 & - 10 = 2
12 & ( ~12 + 1) = 12 & - 12 = 4
11 & ( ~11 + 1) = 11 & - 11 = 1
3 & ( ~ 3 + 1) = 3 & - 3 = 1
Note that if n is not divisible by any power of 2 it returns 1.
Why it works?
The negative of n is produced by inverting its bits via ~, then adding 1 (again, see two's complement definition). This sum causes every 1 bit (starting from the bottom) to overflow until a 0 bit is encountered (let us call it the bit x). Here the overflow process stops, leaving remaining bits (those beyond the current x bit) unchanged. Thus performing & between n and its inverse will result in a binary string containing only the x bit.
An example follows.
010010000 | +144 ~
----------|-------
101101111 | -145 +
1 |
----------|-------
101110000 | -144
101110000 | -144 &
010010000 | +144
----------|-------
000010000 | 16
I would like to know what & does in the use case:
7 & 3
=> 3
8 & 3
=> 0
Or as seen in the general use case:
Integer & Integer
=> ??
I know that array & array2 gives the intersection between the two arrays, but I am unsure of exactly what is going on here when used with integers.
& is bitwise AND which examines the two operands bit-by-bit and sets each result bit to 1 if both the corresponding input bits are 1, and 0 otherwise. You can also think of it as bit-by-bit multiplication.
111 (7)
AND 011 (3)
------------
= 011 (3)
1000 (8)
AND 0011 (3)
------------
= 0000 (0)