In every case (including divisible by 2): if (number % n == 0) do(); Anding with a mask of low order bits is just obfuscation, and with a modern compiler will not be any faster than writing the code in a readable fashion. If you have to test all of the cases, you might improve … Read more
If you know the size of type, you could do something like: uint i = 17; uint j = i << 1 | i >> 31; … which would perform a circular shift of a 32 bit value. As a generalization to circular shift left n bits, on a b bit variable: /*some unsigned numeric … Read more
Setting a bit Use the bitwise OR operator (|) to set a bit. number |= 1UL << n; That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit. Use 1ULL if number … Read more
After looking into the two functions from the question (“shr9” and “shr11”) and merging/tweaking the good parts, I finally found the solution. All tests passed (I even added more in the demo), and it also works for shifts by a negative number. [Live Demo] function unsignedRightShift($a, $b) { if ($b >= 32 || $b < … Read more
XSLT is Turing-complete, see for example here or here, hence it can be done. But I have used XSLT only one or two times and can give no solution. UPDATE I just read a tutorial again and found a solution using the following fact. bitset(x, n) returns true, if the n-th bit of x is … Read more
Fastest way? I’d use something like this: template <class T> constexpr T make_mask(std::size_t pos, std::size_t len) { return ((static_cast<T>(1) << len)-1) << pos; }
The second part of the question, about the special case of a 1-bit deposit, requires two steps. In the first step, we need to determine the bit index r of the single 1-bit in val, with a suitable response in case val is zero. This can easily be accomplished via the POSIX function ffs, or … Read more
The code is treating the first bit as the low bit of the word, so you end up with each word reversed. As a quick-and-dirty fix, try this: bytes[byteIndex] |= (byte)(1 << (7-bitIndex)); That puts the first bit in the array at the highest position in the first byte, etc.
You have to shift and compare elementwise. for(i = 0; i < len; ++i) array[i] >>= 3; for example. If you want to move the bits shifted out of one element to the next, it’s more complicated, say you’re shifting right, then unsigned char bits1 = 0, bits2 = 0; for(i = len-1; i >= … Read more
In Python 3.10+, there is int.bit_count(): >>> 123 .bit_count() 6 Python 2.6 or 3.0: def bitsoncount(x): return bin(x).count(‘1’) Example: >>> x = 123 >>> bin(x) ‘0b1111011′ >>> bitsoncount(x) 6 Or Matt Howells’s answer in Python: def bitsoncount(i): assert 0 <= i < 0x100000000 i = i – ((i >> 1) & 0x55555555) i = (i … Read more