I am having trouble with the last problem of my bit twiddling homework exercise. The function is supposed to return 1 if any odd bit is set to 1.
Here is what I have so far:
int anyOddBit(int x) {
return (x & 0xaaaaaaaa) != 0;
}
That works perfectly, but I am not allowed to use a constant that large (only allowed 0 through 255, 0xFF). I am also not allowed to use !=
Specifically, this is what I am limited to using:
Each "Expr" is an expression using ONLY the following:
1. Integer constants 0 through 255 (0xFF), inclusive. You are
not allowed to use big constants such as 0xffffffff.
2. Function arguments and local variables (no global variables).
3. Unary integer operations ! ~
4. Binary integer operations & ^ | + << >>
I can't figure out how to do this within in those restrictions and I'd really appreciate it if someone could point me in the right direction. Thanks in advance!
You could do your ORs ahead of ANDs:
((x>>0) | (x>>8) | (x>>16) | (x>>24)) & 0xaa
The initial shift (x >> 0) will be optimized out - it's there for consistent look.
You can use:
!!(( ( x & 0xff)
| ((x >> 8) & 0xff)
| ((x >> 16) & 0xff)
| ((x >> 24) & 0xff)
) & 0xaa)
The "inner" bit, which ORs together each source octet, will give you an octet where each bit is set if the equivalent bit is set in any of the source octets. So, if one of the odd bits is set in the source octets, it will also be set in the target one.
Then, by simply ANDing that with 0xaa, you get a zero value if no odd bits are set, or a non-zero value if any of the odd bits are set.
Then, since you need 0 or 1, and you can't use !=, you can acheive a similar effect with !!, two logical not operators. It works because !(any-non-zero-value) gives 0 and !0 gives 1.
In order to do it with 12 operators only (rather than 13 as per my original solution above), you can remove the & 0xff for the >> 24 value since it's not actually necessary (zero-bits are shifted in from the left):
!!(( ( x & 0xff)
| ((x >> 8) & 0xff)
| ((x >> 16) & 0xff)
| ((x >> 24) )
) & 0xaa)
In fact, you can do even better than that. The final & 0xaa will clear out all the upper 24 bits anyway so no & 0xff sections are needed (it also fits on one line as well):
!!((x | (x >> 8) | (x >> 16) | (x >> 24)) & 0xaa)
That gets it down to nine operators.
0xaaaaaaaa is basically (0xaa << 24) | (0xaa << 16) | (0xaa << 8) | (0xaa), and that is allowed, isn't it?
Related
I've seen the answer here: http://clc-wiki.net/wiki/K%26R2_solutions:Chapter_2:Exercise_6
and i've tested the first, but in this part:
x = 29638;
y = 999;
p = 10;
n = 8;
return (x & ((~0 << (p + 1)) | (~(~0 << (p + 1 - n)))))
in a paper it give to me a 6, but in the program it return 28678...
in this part:
111001111000110
&000100000000111
in the result, the left-most three bits has to be 1's like in x but the bitwise operator & says:
The output of bitwise AND is 1 if the corresponding bits of all operands is 1. If either bit of an operand is 0, the result of corresponding bit is evaluated to 0.
so why it returns the number with thats 3 bits in 1?
Here we go, one step at a time (using 16-bit numbers). We start with:
(x & ((~0 << (p + 1)) | (~(~0 << (p + 1 - n)))))
Substituting in numbers (in decimal):
(29638 & ((~0 << (10 + 1)) | (~(~0 << (10 + 1 - 8)))))
Totalling up the bit shift amounts gives:
(29638 & ((~0 << 11) | (~(~0 << 3))))
Rewriting numbers as binary and applying the ~0s...
(0111001111000110 & ((1111111111111111 << 1011) | (~(1111111111111111 << 0011))))
After performing the shifts we get:
(0111001111000110 & (1111100000000000 | (~ 1111111111111000)))
Applying the other bitwise-NOT (~):
(0111001111000110 & (1111100000000000 | 0000000000000111))
And the bitwise-OR (|):
0111001111000110 & 1111100000000111
And finally the bitwise-AND (&):
0111000000000110
So we then have binary 0111000000000110, which is 2 + 4 + 4096 + 8192 + 16384, which is 28678.
What does the & mean in this code:
(number >> 9) & 0b111
I know about & in terms of pointers. But not sure how it works in the code above
Lets break it down:
(number >> 9) & 0b111
| | | | |
| | | | Binary '7'*
| | | Binary AND
| | Number to shift by
| Binary shift operator
Variable
We'll start with the expression in the parenthesis:
(number >> 9)
This performs a binary right-shift by 9 places. For example:
1101101010010011 will be shifted to become:
0000000001101101
The & symbol is Binary AND. Where the bits are both 1 in both of the source variables, the returned value will have those bits set:
01101
& 11010
= 01000
So your code shifts your number by 9 places and performs AND on the result against b111. As the three least significant bits are all set in the second input, the result of this operation will be the bits that are set in the bottom three bits of the shifted input.
Example:
number = 1101101010010011
number >> 9 = 0000000001101101
(number >> 9) & '111' = 0000000000000101
An alternate way of thinking about it is as follows: The line extracts bits 10-12 and returns them as the result.
XXXXbbbXXXXXXXXX -> bbb
A common use for this is to apply a mask to a value to extract the bits. E.g. some libraries allow you to pass parameters with enumerable types like this:
set_params(option_a | option_b);
which sets both option_a and option_b.
Whether a parameter is set can be read by:
set_params(unsigned int params)
{
if (params & option_a)
{ /* do option_a stuff */}
}
*assuming your compiler has a binary extension to the C spec. otherwise you could use 0x7 (hex 7) or just 7
It is the bitwise AND operator.
More info here:
Wikipedia link
& is Bitwise AND
The C operators are here:
https://en.wikipedia.org/wiki/Operators_in_C_and_C%2B%2B
Okay i know this is a pretty mean task from which i got nightmares but maybe ..i'll crack that code thanks to someone of you.
I want to compare if number is between 0 and 10 with bitwise operators. Thats the thing.. it is between 0 and 10 and not for example between 0 and 2, 0 and 4, 0 and 8 and so on..
Reference for number/binary representation with 0-4 bits. (little endian)
0 0
1 1
2 10
3 11
4 100
5 101
6 110
7 111
8 1000
9 1001
10 1010
11 1011
12 1100
13 1101
14 1110
15 1111
Trying to figure out something like
if(((var & 4) >> var) + (var & 10))
I attempt to solve it with bitwise operators only (no addition).
The expression below will evaulate to nonzero if the number (v) is out of the 0 - 10 inclusive range:
(v & (~0xFU)) |
( ((v >> 3) & 1U) & ((v >> 2) & 1U) ) |
( ((v >> 3) & 1U) & ((v >> 1) & 1U) & (v & 1U) )
The first line is nonzero if the number is above 15 (any higher bit than the first four is set). The second line is nonzero if in the low 4 bits it is between 12 and 15 inclusive. The third line is nonzero if in the low 4 bits the number is either 11 or 15.
It was not clear in the question, but if the number to test is limited between the 0 - 15 inclusive range (only low 4 bits), then something nicer is possible here:
((~(v >> 3)) & 1U) |
( ((~(v >> 2)) & 1U) & (( ~v ) & 1U) ) |
( ((~(v >> 2)) & 1U) & ((~(v >> 1)) & 1U) )
First line is 1 if the number is between 0 and 7 inclusive. Second line is 1 if the number is one of 0, 2, 8 or 10. Third line is 1 if the number is one of 0, 1, 8 or 9. So OR combined the expression is 1 if the number is between 0 and 10 inclusive. Relating this solution, you may also check out the Karnaugh map, which can assist in generating these (and can also be used to prove there is no simpler solution here).
I don't think I could get any closer stricly using only bitwise operators in a reasonable manner. However if you can use addition it becomes a lot simpler as Pat's solution shows it.
Assuming that addition is allowed, then:
(v & ~0xf) | ((v+5) & ~0xf)
is non-zero if v is out-of-range. The first term tests if v is outside the range 0..15, and the second shifts the unwanted 11, 12, 13, 14, 15 outside the 0..15 range.
When addition is allowed and the range is 0..15, a simple solution is
(v - 11) & ~7
which is nonzero when v is in the range 0..10. Using shifts instead, you can use
(1<<10) >> v
which is also nonzero if the input is in the range 0..10. If the input range is unrestricted and the shift count is modulo 32, like on most CPUs, you can use
((1<<11) << ~v) | (v & ~15)
which is nonzero if the input is not in the range (the opposite is difficult since already v == 0 is difficult with only bitops). If other arithmetic operations are allowed, then
v / 11
can be used, which is also nonzero if the input is not in the range.
bool b1 = CheckCycleStateWithinRange(cycleState, 0b0, 0b1010); // Note *: 0b0 = 0 and 1010 = 10
bool CheckCycleStateWithinRange(int cycleState, int minRange, int maxRange) const
{
return ((IsGreaterThanEqual(cycleState, minRange) && IsLessThanEqual(cycleState, maxRange)) ? true : false );
}
int IsGreaterThanEqual(int cycleState, int limit) const
{
return ((limit + (~cycleState + 1)) >> 31 & 1) | (!(cycleState ^ limit));
}
int IsLessThanEqual(int cycleState, int limit) const
{
return !((limit + (~cycleState + 1)) >> 31 & 1) | (!(cycleState ^ limit));
}
All of the bit shuffling algorithms I've found deal with 16-bit or 32-bit, which means that even if I use only the first 25-bits of an int, the shuffle will leave bits outside. This function is in an inner loop of a CPU-intensive process so I'd prefer it to be as fast as possible. I've tried modifying the code of the Hacker's Delight 32-bit shuffle algorithm
x = (x & 0x0000FF00) << 8 | (x >> 8) & 0x0000FF00 | x & 0xFF0000FF;
x = (x & 0x00F000F0) << 4 | (x >> 4) & 0x00F000F0 | x & 0xF00FF00F;
x = (x & 0x0C0C0C0C) << 2 | (x >> 2) & 0x0C0C0C0C | x & 0xC3C3C3C3;
x = (x & 0x22222222) << 1 | (x >> 1) & 0x22222222 | x & 0x99999999;
but am having difficulty in doing some partly because I'm not sure where the masks come from. I tried shifting the number and re-shuffling but so far the results are all for naught. Any help would be GREATLY appreciated!
(I am using C but I can convert an algorithm from another language)
First, for the sake of evenness, we can extend the problem to a 26-bit shuffle by remembering that bit 25 will appear at the end of the interleaved list, so we can trim it off after the interleaving operation without affecting the positions of the other bits.
Now we want to interleave the first and second sets of 13 bits; but we only have an algorithm to interleave the first and second sets of 16 bits.
A straightfoward approach might be to just move the high and low parts of x into more workable positions before applying the standard algorithm:
x = (x & 0x1ffe000) << 3 | x & 0x00001fff;
x = (x & 0x0000FF00) << 8 | (x >> 8) & 0x0000FF00 | x & 0xFF0000FF;
x = (x & 0x00F000F0) << 4 | (x >> 4) & 0x00F000F0 | x & 0xF00FF00F;
x = (x & 0x0C0C0C0C) << 2 | (x >> 2) & 0x0C0C0C0C | x & 0xC3C3C3C3;
x = (x & 0x22222222) << 1 | (x >> 1) & 0x22222222 | x & 0x99999999;
The zeroes at the top of each half will be interleaved and appear at the top of the result.
~ & ^ | + << >> are the only operations I can use
Before I continue, this is a homework question, I've been stuck on this for a really long time.
My original approach: I thought that !x could be done with two's complement and doing something with it's additive inverse. I know that an xor is probably in here but I'm really at a loss how to approach this.
For the record: I also cannot use conditionals, loops, ==, etc, only the functions (bitwise) I mentioned above.
For example:
!0 = 1
!1 = 0
!anything besides 0 = 0
Assuming a 32 bit unsigned int:
(((x>>1) | (x&1)) + ~0U) >> 31
should do the trick
Assuming x is signed, need to return 0 for any number not zero, and 1 for zero.
A right shift on a signed integer usually is an arithmetical shift in most implementations (e.g. the sign bit is copied over). Therefore right shift x by 31 and its negation by 31. One of those two will be a negative number and so right shifted by 31 will be 0xFFFFFFFF (of course if x = 0 then the right shift will produce 0x0 which is what you want). You don't know if x or its negation is the negative number so just 'or' them together and you will get what you want. Next add 1 and your good.
implementation:
int bang(int x) {
return ((x >> 31) | ((~x + 1) >> 31)) + 1;
}
The following code copies any 1 bit to all positions. This maps all non-zeroes to 0xFFFFFFFF == -1, while leaving 0 at 0. Then it adds 1, mapping -1 to 0 and 0 to 1.
x = x | x << 1 | x >> 1
x = x | x << 2 | x >> 2
x = x | x << 4 | x >> 4
x = x | x << 8 | x >> 8
x = x | x << 16 | x >> 16
x = x + 1
For 32 bit signed integer x
// Set the bottom bit if any bit set.
x |= x >> 1;
x |= x >> 2;
x |= x >> 4;
x |= x >> 8;
x |= x >> 16;
x ^= 1; // Toggle the bottom bit - now 0 if any bit set.
x &= 1; // Clear the unwanted bits to leave 0 or 1.
Assuming e.g. an 8-bit unsigned type:
~(((x >> 0) & 1)
| ((x >> 1) & 1)
| ((x >> 2) & 1)
...
| ((x >> 7) & 1)) & 1
You can just do ~x & 1 because it yields 1 for 0 and 0 for everything else