Olivier Langlois's blog

I was expecting:

int main( int argc, char *argv[] ) 
{ 
   unsigned char m = 32; 
   register unsigned mask = (1<<m); 
   std::cout << std::hex << mask << '\n'; 
   return 0; 
} 

to print 0 but instead this program compiled with g++ (and VC++.NET2003 too) prints 1!

If I change (1<<m) by (1<<32) or if change the program for:

int main( int argc, char *argv[] ) 
{ 
   unsigned char m = 31; 
   register unsigned mask = (1<<m)<<1; 
   std::cout << std::hex << mask << '\n'; 
   return 0; 
} 

it gives me the expected 0.

In the C++ standard document, section 5.8. It is written

"The behavior is undefined if the right operand is negative, or greater than or equal to the length in bits of the promoted left operand."

The root for this behavior probably originates from C and the safe way to perform a bit shift when the number of bits to shift may exceed the length of the left operand is to implement the shift operation in a function:

Example, almost but not universally portable:

#include <climits> 

unsigned int safe_uint_shift(unsigned int value,
                             unsigned int bits) 
{ 
    if( bits > (CHAR_BIT*std::sizeof(unsigned int) )
    {
      return 0;
    }
    else
    {
        return value << bits; 
    }
}  

Put it in a header and make it inline if you like.

This solution has been proposed by Jack Klein.

The other way that I have used is to promote the left operand to an integer type having enough bits:

register unsigned mask =
  (static_cast<unsigned long long>(1)<<m)<<1;