Here is an applet where you can exercise some bit-operations, including shifting. You have a collection of bits, and you move some of them beyond their bounds: 1111 1110 << 2 1111 1000 It is filled from the right with fresh zeros. 🙂 0001 1111 >> 3 0000 0011 Filled from the left. A special … Read more
The standard way to do division is by implementing binary long-division. This involves subtraction, so as long as you don’t discount this as not a bit-wise operation, then this is what you should do. (Note that you can of course implement subtraction, very tediously, using bitwise logical operations.) In essence, if you’re doing Q = … Read more
& is binary and. If you have a binary value, and you and with another binary value, then the result will be the bitwise and of the two. An example: 01101010 & 01011001 = 01001000 The rightmost bit is either a 1 (and in that case the number is an odd number) or it is … Read more
The ~ unary operator is bitwise negation. If you need fewer bits than what fits in an int then you’ll need to mask it with & after the fact.
Almost any environment worth its salt will optimize this away for you. And if it doesn’t, you’ve got bigger fish to fry. Seriously, do not waste one more second thinking about this. You will know when you have performance problems. And after you run a profiler, you will know what is causing it, and it … Read more
long may be a 64-bit type, but 1 is still an int. You need to make 1 a long int using the L suffix: unsigned long x = 1UL << 32; (You should also make it unsigned using the U suffix as I’ve shown, to avoid the issues of left shifting a signed integer. There’s … Read more
These are Bitwise Operators (reference). x & 1 produces a value that is either 1 or 0, depending on the least significant bit of x: if the last bit is 1, the result of x & 1 is 1; otherwise, it is 0. This is a bitwise AND operation. x >>= 1 means “set x … Read more
Right shift of a negative signed number has implementation-defined behaviour. If your 8 bits are meant to represent a signed 8 bit value (as you’re talking about a “signed 32 bit integer” before switching to 8 bit examples) then you have a negative number. Shifting it right may fill “empty” bits with the original MSB … Read more
Using bitwise operations on strings (characters) Convert letter to lowercase: OR by space => (x | ‘ ‘) Result is always lowercase even if letter is already lowercase eg. (‘a’ | ‘ ‘) => ‘a’ ; (‘A’ | ‘ ‘) => ‘a’ Convert letter to uppercase: AND by underline => (x & ‘_’) Result is always … Read more
sounds a bit like homework, but: bool IsBitSet(byte b, int pos) { return (b & (1 << pos)) != 0; } pos 0 is least significant bit, pos 7 is most.