warning: left shift count >= width of type

unsigned long x = 1UL << 31;

Not show the error message. Because before you specify the 32, is not true because only limited to 0-31.

The accepted solution is fine for [constant]ULL<<32 but no good for existing variables - e.g. [variable]<<32. The complete solution for variables is: ((unsigned long long)[variable]<<32). Aside: My personal opinion of this warning is that it is totally unnecessary in the first place. The compiler can see what the receiving data type is and knows the width of the parameters from the definitions in headers or constant values. I believe Apple could make the clang compiler a little more intelligent than it is regarding this warning.

long may be a 64-bit type, but 1 is still an int. You need to make 1 a long int using the L suffix:

unsigned long x = 1UL << 32;

(You should also make it unsigned using the U suffix as I've shown, to avoid the issues of left shifting a signed integer. There's no problem when a long is 64 bits wide and you shift by 32 bits, but it would be a problem if you shifted 63 bits)

unsigned long is 32 bit or 64 bit which depends on your system. unsigned long long is always 64 bit. You should do it as follows:

unsigned long long x = 1ULL << 32