I had this in part of the code. Could anyone explain how i & (i ^ (i - 1)) could be reduced to i & (-i)?
i ^ (i - 1) makes all bits after the last 1 bit of i becomes 1.
For example if i has a binary representation as abc...de10...0 then i - 1 will be abc...de01...1 in binary (See Why does (x-1) toggle all the bits from the rightmost set bit of x?). The part before the last 1 bit is not changed when subtracting 1 from i, so xoring with each other returns 0 in that part, while the remaining will be 1 because of the difference in i and i - 1. After that i & (i ^ (i - 1)) will get the last 1 bit of i
-i will inverse all bits up to the last 1 bit of i because in two's complement -i == ~i + 1, and i & (-i) results the same like the above
For example:
20 = 0001 0100
19 = 0001 0011
20 ^ 19 = 0000 0111 = 7
20 & 7 = 0000 0100
-20 = 1110 1100
20 & -20 = 0000 0100
See also
Why n bitwise and -n always return the right most bit (last bit)
What does (number & -number) mean in bit programming?
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I have a question, I would appreciate it if you helped me to understand it. Imagin I define the following number
c= 0x3FFFFFFF
and a = an arbitrary integer number=Q. My question is, why a &= c always is equal to "Q" and it does not change? for example, if I consider a=10 then the result of a &= c is 10 if a=256 the result of a &= c is 256. Could you please explain why? Thanks a lot.
Both a and c are integer types and are composed of 32 bits in a computer. The first digit of an integer in a computer is the sign bit.The first digit of a positive number is 0, and the first digit of a negative number is 1. 0x3FFFFFFF is a special value. The first two digits of this number are 0, and the other digits are all 1. 1 & 1 = 1, 1 & 0 = 0. So when the number a a is positive and less than c, a & 0x3FFFFFFF is still a itself
a &= c is the same as a = a & c, which calculates the binary and of a and b and then assign that value to a again - just in case you've mistaken what that operator does.
Now a contains almost only 1's. Then just think what each bit becomes: 1 & x will always be x. Since you try with such low numbers only, none of them will change.
Try with c=0xffffffff and you will get a different result.
You have not tested a &= c; with all possible values of a and are incorrect to assert it does not change the value of a in all cases.
a &= c; sets a to a value in which each bit is set if the two bits in the same position in a and in c are both set. If the two bits are not both set, 5he bit in the result is clear.
In 0x3FFFFFFF, the 30 least significant bits are set. When this is used in a &= c; with any number in which higher bits are set, such as 0xC0000000, the higher bits will be cleared.
If you know about bitwise & ("and") operation and how it works, then there should be no question about this. Say, you have two numbers a and b. Each of them are n-bits long. Look,
a => a_(n-1) a_(n-2) a_(n-3) ... a_i ... a_2 a_1 a_0
b => b_(n-1) b_(n-2) b_(n-3) ... b_i ... b_2 b_1 b_0
Where a_0 and b_0 are the least significant bits and a_(n-1) and b_(n-1) are the most significant bits of a and b respectively.
Now, take a look at the & operation on two single binary bits.
1 & 1 = 1
1 & 0 = 0
0 & 1 = 0
0 & 0 = 0
So, the result of the & operation is 1 only when all bits are 1. If at least one bit is 0, then the result is 0.
Now, for n-bits long number,
a & b = (a_i & b_i); where `i` is from 0 to `n-1`
For example, if a and b both are 4 bits long numbers and a = 5, b = 12, then
a = 5 => a = 0101
b = 12 => b = 1100
if c = (a & b), c_i = (a_i & b_i) for i=0..3, here all numbers are 4 bits(0..3)
now, c = c_3 c_2 c_1 c_0
so c_3 = a_3 & b_3
c_2 = a_2 & b_2
c_1 = a_1 & b_1
c_0 = a_0 & b_0
a 0 1 0 1
b 1 1 0 0
-------------
c 0 1 0 0 (that means c = 4)
therefore, c = a & b = 5 & 12 = 4
Now, what would happen, if all of the bits in one number are 1s?
Let's see.
0 & 1 = 0
1 & 1 = 1
so if any bit is fixed and it 1, then the result is the same as the other bit.
if a = 5 (0101) and b = 15 (1111), then
a 0 1 0 1 (5)
b 1 1 1 1 (15)
------------------
c 0 1 0 1 (5, which is equal to a=5)
So, if any of the numbers has all bits are 1s, then the & result is the same as the other number. Actually, for a=any value of 4-bits long number, you will get the result as a, since b is 4-bits long and all 4 bits are 1s.
Now another issue would happen, when a > 15 means a exceeds 4-bits
For the above example, expand the bit size to 1 and change the value of a is 25.
a = 25 (11001) and b = 15 (01111). Still, b is the same as before except the size. So the Most Significant Bit (MSB) is 0. Now,
a 1 1 0 0 1 (25)
b 0 1 1 1 1 (15)
----------------------
c 0 1 0 0 1 (9, not equal to a=25)
So, it is clear that we have to keep every single bit to 1 if we want to get the other number as the result of the & operation.
Now it is time to analyze the scenario you posted.
Here, a &= c is the same as a = a & c.
We assumed that you are using 32-bit integer variables.
You set c = 0x3FFFFFFF means c = (2^30) - 1 or c = 1073741823
a = 0000 0000 0000 0000 0000 0000 0000 1010 (10)
& c = 0011 1111 1111 1111 1111 1111 1111 1111 (1073741823)
----------------------------------------------------------------
a = 0000 0000 0000 0000 0000 0000 0000 1010 (10, which is equal to a=10)
and
a = 0000 0000 0000 0000 0000 0001 0000 0000 (256)
& c = 0011 1111 1111 1111 1111 1111 1111 1111 (1073741823)
----------------------------------------------------------------
a = 0000 0000 0000 0000 0000 0001 0000 0000 (256, which is equal to a=256)
but, if a > c, say a=0x40000000 (1073741824, c+1 in base 10), then
a = 0100 0000 0000 0000 0000 0001 0000 0000 (1073741824)
& c = 0011 1111 1111 1111 1111 1111 1111 1111 (1073741823)
----------------------------------------------------------------
a = 0000 0000 0000 0000 0000 0000 0000 0000 (0, which is not equal to a=1073741823)
So, your assumption ( the value of a after executing statement a &= c is the same as previous a) is true only if a <= c
PROGRAMMING LANGUAGE: C
I've a 8 bit data with only 3 bit used, for example:
0110 0001
Where 0 indicate unused bit that are always set to 0 and 1 indicate bits that change.
I want to convert this 0110 0001 8 bit to 3 bit that indicate this 3 used bits.
For example
0110 0001 --> 111
0010 0001 --> 011
0000 0000 --> 000
0100 0001 --> 101
How I can do that with minimal operations?
You can achieve this with a couple of bitwise operations:
((a >> 4) & 6) | (a & 1)
Assuming you start from xYYx xxxY, where x is a bit you don't care about and Y a bit to keep:
left shift by 4 of a will result in xYYx, then masking with 6 (binary 110) will make sure only the second and third bit are retained, resulting in YY0 and preventing flipped x bits from messing up.
a & 1 selects the LSB, resulting in Y.
the two parts, YY0 and Y are combined using a | bitwise or, resulting in YYY.
Now you have the 3 bits you asked. But keep in mind that you can't address single bits, so it will still be byte-aligned as 00000YYY
You can get the k'th bit of n: (where n is 011000001)
(n & ( 1 << k )) >> k
(More details about that at StackOverflow)
so you use that to get bit 1,6 and 7 and just add those:
r=bit1+bit6*16+bit7*32
Going through the book "Cracking the coding interview" by Gayle Laakmann McDowell, in bit manipulation chapter, it posts a question:
Find the value of (assuming numbers are represented by 4 bits):
1011 & (~0 << 2)
Now, ~0 = 1 and shifting it two times towards the left yields 100 ( = 0100 to complete the 4 bits). Anding 1011 with 0100 equals 0000.
However, the answer i have is 1000.
~0 is not 1 but 1111 (or 0xf). The ~ operator is a bitwise NOT operator, and not a logical one (which would be !).
So, when shifted by 2 places to the left, the last four bits are 1100. And 1100 & 1011 is exaclty 1000.
~0 does not equal 1. The 0 will default to being an integer, and the NOT operation will reverse ALL the bits, not just the first.
~ is the Bitwise Complement Operator.
The value of ~0 should be 1111 in 4 bits .
1011 & (~0 << 2)
= 1011 & ( 1111 << 2)
= 1011 & 1100
= 1000
1011 & (~0 << 2)
~0 is not 1 but rather 11112 or 0xF16.
Shifting 1111 to the left twice gives 1100 (the two leftmost bits have been dropped and filled in with 0s from the right).
Adding 1011 & 1100 gives 1 in each bit position for which the corresponding bit position is 1, otherwise 0. This follows that the result is 1000.
I tried to understand how if condition work with bitwise operators.
A way to check if a number is even or odd can be done by:
#include <iostream>
#include <string>
using namespace std;
string test()
{
int i = 8; //a number
if(i & 1)
return "odd";
else
return "even";
}
int main ()
{
cout << test();
return 0;
}
The Part I don't understand is how the if condition work. In this case if i = 8 then the in If statement it is doing 1000 & 1 which should gives back 1000 which equal 8.
If i = 7, then in if statement it should be doing 111 & 1 which gives back 111 which equal 7
Why is it the case that if(8) will return "even" and if(7) return "odd"? I guess I want to understand what the if statement is checking to be True and what to be False when dealing with bit-wise operators.
Just A thought when I wrote this question down is it because it's actually doing
for 8: 1000 & 0001 which gives 0
for 7: 0111 & 0001 which gives 1?
Yes, you are right in the last part. Binary & and | are performed bit by bit. Since
1 & 1 == 1
1 & 0 == 0
0 & 1 == 0
0 & 0 == 0
we can see that:
8 & 1 == 1000 & 0001 == 0000
and
7 & 1 == 0111 & 0001 == 0001
Your test function does correctly compute whether a number is even or odd though, because a & 1 tests whether there is a 1 in the 1s place, which there only is for odd numbers.
Actually, in C, C++ and other major programming languages the & operator do AND operations in each bit for integral types. The nth bit in a bitwise AND is equal to 1 if and only if the nth bit of both operands are equal to 1.
For example:
8 & 1 =
1000 - 8
0001 - 1
----
0000 - 0
7 & 1 =
0111 - 7
0001 - 1
----
0001 - 1
7 & 5 =
0111 - 7
0101 - 5
----
0101 - 5
For this reason doing a bitwise AND between an even number and 1 will always be equal 0 because only odd numbers have their least significant bit equal to 1.
if(x) in C++ converts x to boolean. An integer is considered true iff it is nonzero.
Thus, all if(i & 1) is doing is checking to see if the least-significant bit is set in i. If it is set, i&1 will be nonzero; if it is not set, i&1 will be zero.
The least significant bit is set in an integer iff that integer is odd, so thus i&1 is nonzero iff i is odd.
What you say the code is doing is actually how bit-wise operators are supposed to work. In your example of (8 & 1):
1000 & 0001 = 0000
because in the first value, the last bit is set to 0, while in the second value, the last bit is set to 1. 0 & 1 = 0.
0111 & 0001 = 0001
In both values, the last bit is set to 1, so the result is 1 since 1 & 1 = 1.
The expression i & 1, where i is an int, has type int. Its value is 1 or 0, depending on the value of the low bit of i. In the statement if(i & 1), the result of that expression is converted to bool, following the usual rule for integer types: 0 becomes false and non-zero becomes true.
I've got a sequence of bits, say
0110 [1011] 1111
Let's say I want to set that myddle nybble to 0111 as the new value.
Using a positional masking approach with AND or OR, I seem to have no choice but to first unset the original value to 0000, because if I trying ANDing or ORing against that original value of 1011, I'm not going to come out with the desired result of 0111.
Is there another logical operator I should be using to get the desired effect? Or am I locked into 2 operations every time?
The result after kindly assistance was:
inline void foo(Clazz* parent, const Uint8& material, const bool& x, const bool& y, const bool& z)
{
Uint8 offset = x | (y << 1) | (z << 2); //(0-7)
Uint64 positionMask = 255 << offset * 8; //255 = length of each entry (8 bits), 8 = number of bits per material entry
Uint64 value = material << offset * 8;
parent->childType &= ~positionMask; //flip bits to clear given range.
parent->childType |= value;
}
...I'm sure this will see further improvement, but this is the (semi-)readable version.
If you happen to already know the current values of the bits, you can XOR:
0110 1011 1111
^ 0000 1100 0000
= 0110 0111 1111
(where the 1100 needs to be computed first as the XOR between the current bits and the desired bits).
This is, of course, still 2 operations. The difference is that you could precompute the first XOR in certain circumstances.
Other than this special case, there is no other way. You fundamentally need to represent 3 states: set to 1, set to 0, don't change. You can't do this with a single binary operand.
You may want to use bit fields (and perhaps unions if you want to be able to access your structure as a set of bit fields and as an int at the same time) , something along the lines of:
struct foo
{
unsigned int o1:4;
unsigned int o2:4;
unsigned int o3:4;
};
foo bar;
bar.o2 = 0b0111;
Not sure if it translates into more efficient machine code than your clear/set...
Well, there's an assembly instruction in MMIX for this:
SETL $1, 0x06BF ; 0110 1011 1111
SETL $2, 0x0070 ; 0000 0111 0000
SETL rM, 0x00F0 ; set mask register
MUX $1,$2,$1 ; result is 0110 0111 1111
But in C++ here's what you're probably thinking of as 'unsetting the previous value'.
int S = 0x6BF; // starting value: 0110 1011 1111
int M = 0x0F0; // value mask: 0000 1111 0000
int V = 0x070; // value: 0000 0111 0000
int N = (S&~M) | V; // new value: 0110 0111 1111
But since the intermediate result 0110 0000 1111 from (S&~M) is never stored in a variable anywhere I wouldn't really call it 'unsetting' anything. It's just a bitwise boolean expression. Any boolean expression with the same truth table will work. Here's another one:
N = ((S^V) & M) ^ A; // corresponds to Oli Charlesworth's answer
The related truth tables:
S M V (S& ~M) | V ((S^V) & M) ^ S
0 0 0 0 1 0 0 0 0
* 0 0 1 0 1 1 1 0 0
0 1 0 0 0 0 0 0 0
0 1 1 0 0 1 1 1 1
1 0 0 1 1 1 1 0 1
* 1 0 1 1 1 1 0 0 1
1 1 0 0 0 0 1 1 0
1 1 1 0 0 1 0 0 1
^ ^
|____________________|
The rows marked with '*' don't matter because they won't occur (a bit in V will never be set when the corresponding mask bit is not set). Except for those rows, the truth tables for the expressions are the same.