c++,bit-manipulation,bit-shift,bitwise-or,bitboard

You need to use 1LL as 64 bit value before you use shift operator << to get 64 bit result: #include <stdint.h> uint64_t kings = 0ULL; kings |= 1ULL << i; ...

c,overflow,frama-c,nitrogen,bitwise-or

which language is that? This is Why either in its version 2 or in its version 3 (the transition was about at the time of the version you use). the following condition cannot be proved […]: 0 <= bw_or(integer_of_uint32(result7), integer_of_uint32(result8)) It is always difficult to decide how to axiomatize...

You can apply the^ operator on two ints, not on an int and an int array. Based on the error message, Kc[n] is an array of int. You can apply the operator on two ints : r[temp1][n] = r[temp1][n]^Kc[temp1][n]; I have no idea if the indices make sense (since I...

sql-server,ssis,bitwise-operators,bitwise-or

I don't have a whole lot of experience with that flavor of SQL, but a bitwise OR is not the same thing as an OR clause in the WHERE statement. The bitwise OR will OR each bit of the integer together to produce a new integer For example, the numbers...

A halfway decent compiler will tell you: warning: format ‘%d’ expects argument of type ‘int *’, but argument 2 has type ‘short int *’ scanf("%d", &curVal); You need to use %hd to scan a short, or (in this case better) change your variable to int. If you're using GCC, add...

That's easy to explain: 0x6B as you write it gets interpreted as a default integer (probably 32 bit). So (ch|0x6B)== 0x000000FF==0b00000000000000000000000011111111. Hence, ~(ch|0x6B) == 0b11111111111111111111111100000000, which is not 0, hence is true. If you, however, put that result into a char, only the lower 8 bits are saved, and hence...

One vectorized approach - [m,n] = size(X) %// Get size of input array bd = dec2bin(X)-'0' %// Get binary digits %// Get cumulative "OR-ed" version with ANY(..,1) cum_or = reshape(any(permute(reshape(bd,m,n,[]),[2 3 1]),1),8,[]) %// Finally convert to decimals U = 2.^(7: -1:0)*cum_or ...