128 bit integer on cuda?

Question

I just managed to install my cuda SDK under Linux Ubuntu 10.04. My graphic card is an NVIDIA geForce GT 425M, and I'd like to use it for some heavy computational problem. What I wonder is: is there any way to use some unsigned 128 bit int var? When using gcc to run my program on the CPU, I was using the __uint128_t type, but using it with cuda doesn't seem to work. Is there anything I can do to have 128 bit integers on cuda?

Answer 1

For best performance, one would want to map the 128-bit type on top of a suitable CUDA vector type, such as uint4, and implement the functionality using PTX inline assembly. The addition would look something like this:

typedef uint4 my_uint128_t;
__device__ my_uint128_t add_uint128 (my_uint128_t addend, my_uint128_t augend)
{
    my_uint128_t res;
    asm ("add.cc.u32      %0, %4, %8;\n\t"
         "addc.cc.u32     %1, %5, %9;\n\t"
         "addc.cc.u32     %2, %6, %10;\n\t"
         "addc.u32        %3, %7, %11;\n\t"
         : "=r"(res.x), "=r"(res.y), "=r"(res.z), "=r"(res.w)
         : "r"(addend.x), "r"(addend.y), "r"(addend.z), "r"(addend.w),
           "r"(augend.x), "r"(augend.y), "r"(augend.z), "r"(augend.w));
    return res;
}

The multiplication can similarly be constructed using PTX inline assembly by breaking the 128-bit numbers into 32-bit chunks, computing the 64-bit partial products and adding them appropriately. Obviously this takes a bit of work. One might get reasonable performance at the C level by breaking the number into 64-bit chunks and using __umul64hi() in conjuction with regular 64-bit multiplication and some additions. This would result in the following:

__device__ my_uint128_t mul_uint128 (my_uint128_t multiplicand, 
                                     my_uint128_t multiplier)
{
    my_uint128_t res;
    unsigned long long ahi, alo, bhi, blo, phi, plo;
    alo = ((unsigned long long)multiplicand.y << 32) | multiplicand.x;
    ahi = ((unsigned long long)multiplicand.w << 32) | multiplicand.z;
    blo = ((unsigned long long)multiplier.y << 32) | multiplier.x;
    bhi = ((unsigned long long)multiplier.w << 32) | multiplier.z;
    plo = alo * blo;
    phi = __umul64hi (alo, blo) + alo * bhi + ahi * blo;
    res.x = (unsigned int)(plo & 0xffffffff);
    res.y = (unsigned int)(plo >> 32);
    res.z = (unsigned int)(phi & 0xffffffff);
    res.w = (unsigned int)(phi >> 32);
    return res;
}

Below is a version of the 128-bit multiplication that uses PTX inline assembly. It requires PTX 3.0, which shipped with CUDA 4.2, and the code requires a GPU with at least compute capability 2.0, i.e. a Fermi or Kepler class device. The code uses the minimal number of instructions, as sixteen 32-bit multiplies are needed to implement a 128-bit multiplication. By comparison, the variant above using CUDA intrinsics compiles to 23 instructions for an sm_20 target.

__device__ my_uint128_t mul_uint128 (my_uint128_t a, my_uint128_t b)
{
    my_uint128_t res;
    asm ("{\n\t"
         "mul.lo.u32      %0, %4, %8;    \n\t"
         "mul.hi.u32      %1, %4, %8;    \n\t"
         "mad.lo.cc.u32   %1, %4, %9, %1;\n\t"
         "madc.hi.u32     %2, %4, %9,  0;\n\t"
         "mad.lo.cc.u32   %1, %5, %8, %1;\n\t"
         "madc.hi.cc.u32  %2, %5, %8, %2;\n\t"
         "madc.hi.u32     %3, %4,%10,  0;\n\t"
         "mad.lo.cc.u32   %2, %4,%10, %2;\n\t"
         "madc.hi.u32     %3, %5, %9, %3;\n\t"
         "mad.lo.cc.u32   %2, %5, %9, %2;\n\t"
         "madc.hi.u32     %3, %6, %8, %3;\n\t"
         "mad.lo.cc.u32   %2, %6, %8, %2;\n\t"
         "madc.lo.u32     %3, %4,%11, %3;\n\t"
         "mad.lo.u32      %3, %5,%10, %3;\n\t"
         "mad.lo.u32      %3, %6, %9, %3;\n\t"
         "mad.lo.u32      %3, %7, %8, %3;\n\t"
         "}"
         : "=r"(res.x), "=r"(res.y), "=r"(res.z), "=r"(res.w)
         : "r"(a.x), "r"(a.y), "r"(a.z), "r"(a.w),
           "r"(b.x), "r"(b.y), "r"(b.z), "r"(b.w));
    return res;
}

Answer 2

A much-belated answer, but you could consider using this library:

https://github.com/curtisseizert/CUDA-uint128

which defines a 128-bit-sized structure, with methods and freestanding utility functions to get it to function as expected, which allow it to be used like a regular integer. Mostly.

Answer 3

CUDA doesn't support 128 bit integers natively. You can fake the operations yourself using two 64 bit integers.

Look at this post:

typedef struct {
  unsigned long long int lo;
  unsigned long long int hi;
} my_uint128;

my_uint128 add_uint128 (my_uint128 a, my_uint128 b)
{
  my_uint128 res;
  res.lo = a.lo + b.lo;
  res.hi = a.hi + b.hi + (res.lo < a.lo);
  return res;
} 

Answer 4

For posterity, note that as of 11.5, CUDA and nvcc support __int128_t in device code when the host compiler supports it (e.g., clang/gcc, but not MSVC). 11.6 added support for debug tools with __int128_t.

See:

• https://developer.nvidia.com/blog/cuda-11-6-toolkit-new-release-revealed/
• https://developer.nvidia.com/blog/implementing-high-precision-decimal-arithmetic-with-cuda-int128/