a |= b; is the same as a = (a | b); It calculates the bitwise OR of the two operands, and assigns the result to the left operand. To explain your example code: for (String search : textSearch.getValue()) matches |= field.contains(search); I presume matches is a boolean; this means that the bitwise operators behave … Read more
Here’s some Octave (MATLAB) code that computes offsets for positive n assuming the conjectures below. Seems to work for 2 and 3, but I suspect that one of the assumptions breaks down when n is too large. No time to investigate right now. % finds the best offset for positive n % assuming that the … Read more
y is getting sign-extended in the first snippet, which would overwrite x with -1 whenever y < 0. In the second snippet, the cast to int is done before the shift, so x actually gets the value of y. long l = (((long)x) << 32) | (y & 0xffffffffL); int x = (int)(l >> 32); … Read more
It works best if you provide a mask. That way you specify how far to sign extend. >>> bin(-27 & 0b1111111111111111) ‘0b1111111111100101’ Or perhaps more generally: def bindigits(n, bits): s = bin(n & int(“1″*bits, 2))[2:] return (“{0:0>%s}” % (bits)).format(s) >>> print bindigits(-31337, 24) 111111111000010110010111 In basic theory, the actual width of the number is a … Read more
XOR operator (also called “logical addition”) is defined like this: a b a^b ———– 0 0 0 0 1 1 1 0 1 1 1 0 So a^0 leaves a intact while a^1 toggles it. For multiple-bit values, the operation is performed bitwise, i.e. between corresponding bits of the operands.
See also Set all bits in CPU register to 1 efficiently which covers AVX, AVX2, and AVX512 zmm and k (mask) registers. You obviously didn’t even look at the asm output, which is trivial to do: #include <immintrin.h> __m256i all_ones(void) { return _mm256_set1_epi64x(-1); } compiles to with GCC and clang with any -march that includes … Read more
Branchless (at the language level) code that maps negative to -1, zero to 0 and positive to +1 looks as follows int c = (n > 0) – (n < 0); if you need a different mapping you can simply use an explicit map to remap it const int MAP[] = { 1, 0, 2 … Read more
Here’s an elegant solution: void getBin(int num, char *str) { *(str+5) = ‘\0’; int mask = 0x10 << 1; while(mask >>= 1) *str++ = !!(mask & num) + ‘0’; } Here, we start by making sure the string ends in a null character. Then, we create a mask with a single one in it (its … Read more
Use a bitmask. It is sort of like: new_value = 0, 1, 2 or 3 // (this is the value you will set in) bit_mask = (3<<5) // (mask of the bits you want to set) reg_data = (reg_data & (~bit_mask)) | (new_value<<5) This preserves the old bits and OR’s in the new ones.
The logarithmic version of the adamk function: int isNotZero(unsigned int n){ n |= n >> 16; n |= n >> 8; n |= n >> 4; n |= n >> 2; n |= n >> 1; return n & 1; }; And the fastest one, but in assembly: xor eax, eax sub eax, n // … Read more