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?
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.
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