It seems like I should be able to perform bit shift in C/C++ by more than 32 bits provided the left operand of the shift is a long. But this doesn't seem to work, at least with the g++ compiler.
Example:
unsigned long A = (1L << 37)
gives
A = 0
which isn't what I want. Am I missing something or is this just not possible?
-J
Well, it depends on the actual size of the type long (more precisely, its width in bits). Most likely on your platform long has width of 32 bits, so you get 0 as the result (also, see P.S. below). Use a bigger type. long long maybe?
P.S. As an additional note, shifting a type by more bits than its width (or equal number of bits) produces undefined behavior in C and C++ (C++ uses the term length instead of width). So, you are not guaranteed to get 0 from neither 1L << 37 nor 1L << 32 on a platform where long has width 32.
A is equal to 0 because A only has 32-bits, so of course you are shifting all the bits off to the left leaving only 0 bits left. You need to make A 64-bit:
unsigned long long A = (1ULL << 37);
Or if you intend to use Visual C++:
unsigned __int64 A = (1ULL << 37);
Re-try this using a variable of type uint64_t (from stdint.h) instead of long. uint64_t is guaranteed to be 64 bits long and should behave as you expect.
Are you sure that a long is 64 bits with your particular OS and compiler ? Use stdint.h and try it like this:
#include <stdint.h>
uint64_t x = (1ULL << 37);
New C++ Standard introduces LL and ULL suffixes for integer literals. You could try to use them since all latest compilers supports them. But you should know that it is not part of current C++ Standard.
long long A = (1LL << 37)