Hello,
I want to toggle bits in an unsigned long variable. But I get an overflow with bits higher then 31.
Any ideas how to solve this? Thank you.
Code:unsigned long n = 63; unsigned long c = 0x8000000000000000; c ^= (1 << n); Output: 01111111 | 11111111 | 11111111 | 11111111 | 10000000 | 00000000 | 00000000 | 00000000
Of course you do. Typically, a long is 32 bits long.
Originally Posted by Adak
If I change my code to the following:
shouldn't it work then? I still have the same problem.Code:unsigned long long n = 63; unsigned long long c = 0x8000000000000000; c ^= (1 << n);
The value "1" is 32 bits. You shift that left 63 and you get zero. Write "1LL" instead.
Code://try //{ if (a) do { f( b); } while(1); else do { f(!b); } while(1); //}
This works as expected:
Also, IIRC, shifting a signed number is implementation defined.Code:int main() { unsigned long long n = 63; unsigned long long c = 0x8000000000000000; unsigned long long mask = 1ull << n; c ^= mask; }
EDIT: More exactly, shifting a negative signed number is either implementation defined or undefined, so avoid it.
Last edited by Elysia; 02-19-2011 at 10:15 AM.
Thank you guys for your help
Try:
Code:c ^= (1ull << n);
Better yet:
Code:#include <limits> template < typename Integer > Integer msb( void ) { static Integer msk = Integer( 1 ) << sizeof( Integer ) * std::numeric_limits< unsigned char >::digits - 1; return msk; } // Example: #include <iostream> int main( void ) { using namespace std; cout << int( msb< unsigned char >( ) ) << endl; cout << msb< short >( ) << endl; cout << msb< unsigned long >( ) << endl; }
Code:#include <cmath> #include <complex> bool euler_flip(bool value) { return std::pow ( std::complex<float>(std::exp(1.0)), std::complex<float>(0, 1) * std::complex<float>(std::atan(1.0) *(1 << (value + 2))) ).real() < 0; }
