I know the best way to divide a number by 2 is to move one bit to the left. What do I do if I am dividing by a multiple of 2 (for example 8), do I move over by 3 bits, here are my questions:
I know these operations can be done at the assembly level because we are dealing with registers, I just don't know if we have access to such things in C++.
Accessing the higher/lower bytes of an integer, and swapping them can be done in at least two ways. Either a combination of >> and |, or with a union.
For example something like:
short swapped = (original<<8)|(original>>8); will give you the two bytes of a 2-byte integer swapped. If you have a larger integer (e.g. 4 bytes), some more masking and shifting will be required, if all bytes are needed in some particularly shuffled order.
Optimizing divisions by multiples of 2 with right shift (>>) is a no-optimization. You should write readable code that gives a clear idea of what is intended. The compiler will trivially perform such micro-optimizations.
256 and 8 are literals. The compiler will simply replace that with int x = 32; If you wrote something like x = y / 8; the compiler would turn it into x = y >> 3;. It is not necessary to write obfuscated code like that, the compiler knows to do these things.(65 -1)/2 is also just an expression that contains nothing but literals. This is evaluated at compile time and replaced with a single integer value. No calculation happening at runtime.
<<and>>and|and&.x / 2. See this question or this one or this other one.onebit shift divides by 2(2^1), thenthreebit shifts divides by ?