问题
The following code triggers a gcc warning (gcc 4.2.1):
#include <boost/cstdint.hpp>
boost::uint64_t x = 1 << 32; // warning: left shift count >= width of type
Shouldn't it be fine since the type has 64 bits?
回答1:
How to shift >= 32 bits in
uint64_t?
If your compiler supports long long:
boost::uint64_t x = 1LL << 32;
Otherwise:
boost::uint64_t x = boost::uint64_t(1) << 32;
Shouldn't it be fine since the type has 64 bits?
No. Even though x is 64 bits, 1 isn't. 1 is 32 bits. How you use a result has no effect on how that result is generated.
来源：https://stackoverflow.com/questions/14846566/how-to-shift-32-bits-in-uint64-t