How to offset> = 32 bits in uint64_t?

Question

The following code triggers a warning gcc

(gcc 4.2.1):

#include <boost/cstdint.hpp>
boost::uint64_t x = 1 << 32; // warning: left shift count >= width of type

Shouldn't this be accurate since the type is 64 bits?

+3

Frank 13 Feb At 4:48 am

1 answer

Robᵩ · Accepted Answer · 2013-02-13T04:50:08+0000

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 this be accurate since the type is 64 bits?

Not. Even if it x

's 64 bits, 1

no. 1

- 32 bits. How you use the result does not affect how the result is created.