How do I bit shift a long by more than 32 bits? [duplicate]

Answer 1

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.

Answer 2

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);

Answer 3

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)

Answer 4

You have the arguments reversed.

unsigned long A1 = (1L << 37); // is 2 to the 37th 
unsigned long A = (37UL<<1UL); // has just multiplied 37 by 2

Answer 5

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.

Answer 6

In your case you're restricted to the language base types. A generic solution for [quasi] arbitrary size integers is using Integer class under Crypto++.