128 bit integer on cuda?
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;
}
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;
}
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.