-1

我正在尝试在 CUDA 中实现多精度乘法。为此,我实现了一个内核,它应该计算uint32_t类型操作数与 256 位操作数的乘法,并将结果放入 288 位数组中。到目前为止,我已经想出了这个代码:

__device__ __constant__ UN_256fe B_const;

 __global__ void multiply32x256Kernel(uint32_t A, UN_288bite* result){

uint8_t tid = blockIdx.x * blockDim.x + threadIdx.x;
//for managing warps
//uint8_t laineid = tid % 32; 
//allocate partial products into array of uint64_t 
__shared__ uint64_t partialMuls[8];
uint32_t carry, r;
if((tid < 8) && (tid != 0)){
    //compute partial products
    partialMuls[tid] = A * B_const.uint32[tid];

    //add partial products and propagate carry
    result->uint32[8] = (uint32_t)partialMuls[7];
    r = (partialMuls[tid] >> 32) + ((uint32_t)partialMuls[tid - 1]);
    carry = r < (partialMuls[tid] >> 32);
    result->uint32[0] = (partialMuls[0] >> 32);
    while(__any(carry)){

        r = r + carry;
        //new carry?        
        carry = r < carry;  
    } 
result->uint32[tid] = r;

}

我的数据类型是:

typedef struct UN_256fe{

uint32_t uint32[8];

}UN_256fe;

typedef struct UN_288bite{

uint32_t uint32[9];

}UN_288bite;

我的内核工作,但它给了我错误的结果。我无法在内核中调试,所以如果有人让我知道问题出在哪里或者我如何在内核中调试我的代码,我将不胜tegra-ubuntu感激cuda-6.0。谢谢

4

1 回答 1

2

这个答案与CUDA本身无关,而是一个通用的C实现。

我不能完全理解你在做什么(尤其是使用carry),但你可以根据我自己的大 num 函数尝试这个片段。我定义dtype为更容易使用较小的字段进行测试。请注意,我没有专门使用 a carry,而是继承了部分产品。

// little-endian
#include <stdio.h>
#include <stdint.h>
#include <limits.h>

#define dtype uint8_t           // for testing
//#define dtype uint32_t        // for proper ver

#define SHIFTS (sizeof(dtype)*CHAR_BIT)
#define NIBBLES (SHIFTS/4)
#define ARRLEN 8

typedef struct UN_256fe {
    dtype uint[ARRLEN];
} UN_256fe;

typedef struct UN_288bite {
    dtype uint[ARRLEN+1];
} UN_288bite;

void multiply(UN_288bite *product, UN_256fe *operand, dtype multiplier)
{
    int i;
    uint64_t partial = 0;
    for (i=0; i<ARRLEN; i++) {
        partial = partial + (uint64_t)multiplier * operand->uint[i];
        product->uint[i] = (dtype)partial;
        partial >>= SHIFTS;                     // carry
    }
    product->uint[i] = (dtype)partial;
}

int main(void)
{
    int i;
    dtype multiplier = 0xAA;
    UN_256fe operand = { 1, 2, 3, 4, 5, 6, 7, 8};
    UN_288bite product;

    multiply(&product, &operand, multiplier);

    for(i=ARRLEN-1; i>=0; i--)
        printf("%0*X", NIBBLES, operand.uint[i]);
    printf("\n * %0*X = \n", NIBBLES, multiplier);
    for(i=ARRLEN; i>=0; i--)
        printf("%0*X", NIBBLES, product.uint[i]);
    printf("\n");

    return 0;
}

程序输出为uint8_t

0807060504030201
 * AA =
0554A9FF54A9FF54AA
于 2015-12-08T20:14:10.237 回答