How to print __int128 in g++?

Question

I am using the GCC built-in type __int128 for a few things in my C++ program, nothing really significant, at least not enough to justify to use BigInt library o

Answer 1

If you don't need any of the fancy formatting options, writing your own << operator is trivial. Formally, I suspect that writing one for __int128_t would be considered undefined behavior, but practically, I think it would work, up until the library starts providing actual support for it (at which point, you'd retire your conversion operator).

Anyway, something like the following should work:

std::ostream&
operator<<( std::ostream& dest, __int128_t value )
{
    std::ostream::sentry s( dest );
    if ( s ) {
        __uint128_t tmp = value < 0 ? -value : value;
        char buffer[ 128 ];
        char* d = std::end( buffer );
        do
        {
            -- d;
            *d = "0123456789"[ tmp % 10 ];
            tmp /= 10;
        } while ( tmp != 0 );
        if ( value < 0 ) {
            -- d;
            *d = '-';
        }
        int len = std::end( buffer ) - d;
        if ( dest.rdbuf()->sputn( d, len ) != len ) {
            dest.setstate( std::ios_base::badbit );
        }
    }
    return dest;
}

Note that this is just a quicky, temporary fix, until the time the g++ library supports the type. It counts on 2's complement, wrap around on overflow, for __int128_t, but I'd be very surprised if that wasn't the case (formally, it's undefined behavior). If not, you'll need to fix up the initialization of tmp. And of course, it doesn't handle any of the formatting options; you can add as desired. (Handling padding and the adjustfield correctly can be non-trivial.)