Bit manipulation for __int128

Question

I tried to add bits to an __int128 (using clang compiler and 64-bit system), but failed.

__int128 x = 0;                           //DECLARING AND INITIALIZING X
x |= ((static_cast<__int128>(1)) << 95);  //ADDING A '1' TO 95th BIT (from right)
std::cerr<< std::bitset<100>(x) << std::endl;  //PRINTING BITSET

prints:

0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

I know there are short forms I can use for unsigned int or long long int, such as
(1u << 15) or (1LL << 15), but I haven't found any 128-bit equivalent.

What do I need to do make this work?

Answer 1

If we use braces std::bitset<100>{x} instead of parentheses std::bitset<100>(x), the program fails to compile because of a narrowing conversion:

error: non-constant-expression cannot be narrowed from type '__int128' to 'unsigned long long' in initializer list [-Wc++11-narrowing]

What's happening here is the __int128 is working correctly, but the conversion to std::bitset is incorrect. You'll have to split the __int128 into 64-bit chunks:

std::bitset<100> hi{static_cast<unsigned long long>(x >> 64)},
                 lo{static_cast<unsigned long long>(x)},
                 bits{(hi << 64) | lo};
std::cout << bits << '\n';