Is there a 128 bit integer in C++?

Question

I need to store a 128 bits long UUID in a variable. Is there a 128-bit datatype in C++? I do not need arithmetic operations, I just want to easily store and read the value v

Answer 1

There is no 128-bit integer in Visual-C++ because the Microsoft calling convention only allows returning of 2 32-bit values in the RAX:EAX pair. The presents a constant headache because when you multiply two integers together with the result is a two-word integer. Most load-and-store machines support working with two CPU word-sized integers but working with 4 requires software hack, so a 32-bit CPU cannot process 128-bit integers and 8-bit and 16-bit CPUs can't do 64-bit integers without a rather costly software hack. 64-bit CPUs can and regularly do work with 128-bit because if you multiply two 64-bit integers you get a 128-bit integer so GCC version 4.6 does support 128-bit integers. This presents a problem with writing portable code because you have to do an ugly hack where you return one 64-bit word in the return register and you pass the other in using a reference. For example, in order to print a floating-point number fast with Grisu we use 128-bit unsigned multiplication as follows:

#include <cstdint>
#if defined(_MSC_VER) && defined(_M_AMD64)
#define USING_VISUAL_CPP_X64 1
#include <intrin.h>
#include <intrin0.h>
#pragma intrinsic(_umul128)
#elif (__GNUC__ > 4 || (__GNUC__ == 4 && __GNUC_MINOR__ >= 6))
#define USING_GCC 1
#if defined(__x86_64__)
#define COMPILER_SUPPORTS_128_BIT_INTEGERS 1
#endif
#endif

#if USING_VISUAL_CPP_X64
    UI8 h;
    UI8 l = _umul128(f, rhs_f, &h);
    if (l & (UI8(1) << 63))  // rounding
      h++;
    return TBinary(h, e + rhs_e + 64);
#elif USING_GCC
    UIH p = static_cast<UIH>(f) * static_cast<UIH>(rhs_f);
    UI8 h = p >> 64;
    UI8 l = static_cast<UI8>(p);
    if (l & (UI8(1) << 63))  // rounding
      h++;
    return TBinary(h, e + rhs_e + 64);
#else
    const UI8 M32 = 0xFFFFFFFF;
    const UI8 a = f >> 32;
    const UI8 b = f & M32;
    const UI8 c = rhs_f >> 32;
    const UI8 d = rhs_f & M32;
    const UI8 ac = a * c;
    const UI8 bc = b * c;
    const UI8 ad = a * d;
    const UI8 bd = b * d;
    UI8 tmp = (bd >> 32) + (ad & M32) + (bc & M32);
    tmp += 1U << 31;  /// mult_round
    return TBinary(ac + (ad >> 32) + (bc >> 32) + (tmp >> 32), e + rhs_e + 64);
#endif
  }

Answer 2

I would recommend using std::bitset<128> (you can always do something like using UUID = std::bitset<128>;). It will probably have a similar memory layout to the custom struct proposed in the other answers, but you won't need to define your own comparison operators, hash etc.

Answer 3

use the TBigInteger template and set any bit range in the template array like this TBigInt<128,true> for being a signed 128 bit integer or TBigInt<128,false> for being an unsigned 128 bit integer. Hope that helps maybe a late reply and someone else found this method already.

Answer 4

GCC and Clang support __int128

Answer 5

Although GCC does provide __int128, it is supported only for targets (processors) which have an integer mode wide enough to hold 128 bits. On a given system, sizeof() intmax_t and uintmax_t determine the maximum value that the compiler and the platform support.

Answer 6

Checkout boost's implementation:

#include <boost/multiprecision/cpp_int.hpp>

using namespace boost::multiprecision;

int128_t v = 1;

This is better than strings and arrays, especially if you need to do arithmetic operations with it.