当通过移位计算零时，零的按位“非”等于零

Question

当size = 64时，为什么size在1（含）和64（含~((~0ULL) >> size << size)之间的~((~0ULL) >> size << size)取值为0？

Answer 1

Shifting by a value greater than or equal to the size of the value is undefined behaviour.

This is probably in the standard to keep the shift operation fast enough and just mapping underlying hardware operations.

For reasons that are not entirely clear to me for example even in x86 assembly shifting is done modulo the size of the register. I'm no hardware engineer but probably that was the most efficient thing to do.

Another unfortunate limitation is that shifting by a negative count is not the same as shifting by a positive count in the opposite direction and is instead just UB. This sometimes requires adding extra conditionals in computations.

Answer 2

Because what you are doing is undefined behavior. Don't shift a 64 bit integer by 64 places.

Edited to add: Why does left shift operation invoke Undefined Behaviour when the left side operand has negative value?