int左移32位的行为未定义
Is Shifting more than 32 bits of a uint64_t integer on an x86 machine Undefined Behavior?
uint64_t s = 1 << 32;
uint64_t s = 1ULL << 32;
Note, however, that if you write a literal that fits into 32 bits, e.g. uint64_t s = 1 << 32 as surmised by @drhirsch,
you don't actually shift a 64-bit value but a 32-bit one. That is undefined behaviour. The most common results are a
shift by shift_distance % 32 or 0, depending on what the hardware does.