CUDA中运行次获取结果的时间差

Question

我正在尝试使用 CUDA 在 GPU 上实现我的算法。这个程序运行良好，但有一个问题。当我尝试打印结果时，它们显示得太晚了。这是我的一些代码。假设真实结果无关紧要。

__device__ unsigned char dev_state[128];

__device__ unsigned char GMul(unsigned char a, unsigned char b) { // Galois Field (256) Multiplication of two Bytes
    unsigned char p = 0;
    int counter;
    unsigned char hi_bit_set;
    for (counter = 0; counter < 8; counter++) {
        if ((b & 1) != 0) {
            p ^= a;
        }
        hi_bit_set = a & 0x80;
        a <<= 1;
        if (hi_bit_set != 0) {
            a ^= 0x1b; /* x^8 + x^4 + x^3 + x + 1 */
        }
        b >>= 1;
    }
    return p;
}


__global__ void AESROUND()
{
    __shared__ unsigned char dev_rkey;
    __shared__ unsigned char dev_sh_state;
    int state_idx = blockIdx.x;
    int offset = ((state_idx / 4)) *4;

    for (int i = 0; i < 512; i++)
    {
        dev_rkey = dev_state[state_idx];

        dev_sh_state= GMul(dev_state[state_idx], 0x02) ^ GMul(dev_state[(state_idx + 5) % 16], 0x03) ^ dev_state[(offset + 5) % 16] ^ dev_state[(offset + 5) % 16];


        dev_state[state_idx] = dev_sh_state ^ dev_rkey;
    }


}

调用 AESROUND

int main()
{

    unsigned char p[] = { 0x19, 0x3d, 0xe3, 0xbe, 0xa0, 0xf4, 0xe2, 0x2b, 0x9a, 0xc6, 0x8d, 0x2a, 0xe9, 0xf8, 0x48, 0x08 };

unsigned char h_state[128];
for (long long i = 0; i < 128; i++)
    h_state[i] = p[i%16];

cudaMemcpyToSymbolAsync(dev_state, h_state, 128, 0, cudaMemcpyHostToDevice);

clock_t start, finish;

start = clock();
for (long long i = 0; i < 1024; i++)

     AESROUND << <128, 128 >> >();
finish = clock();

float Time = finish - start;

printf("\n\nprocessing time: %2.15f (ms)\n", Time);

cudaMemcpyFromSymbolAsync(h_state, dev_state, 128, 0, cudaMemcpyDeviceToHost);
printf("\n\state After Encryption:\n ");
for (int i = 0; i < 16; i++)
    printf("%x ", h_state[i]);

getchar();
return 0;
}

这是结果：

processing time: 1.0000000000000 (ms)

-经过很长一段时间（~5秒），将显示下一行-

state after encryption:
88 91 23 09 78 65 11 87 65 43 56 71 20 93 18 70

如您所见，处理时间太快，但 128Byte 将显示为时已晚。为什么会这样？这与GPU有关吗？我该如何解决？

Answer 1

这里的混乱似乎是由于使用基于主机的计时方法来计时（主要是）设备 activity。

内核启动是异步的。主机代码启动内核，然后不等待内核完成就继续运行。因此这种时机：

start = clock();
for (long long i = 0; i < 1024; i++)

     AESROUND << <128, 128 >> >();
finish = clock();

仅测量内核启动时间。（即使在循环中重复启动内核的情况下也是如此。如果没有超出某些设备队列，则每次内核启动将是异步的，从而允许主机线程（即 for 循环）继续。）

为了测量完整的设备执行时间，您可以这样做：

start = clock();
for (long long i = 0; i < 1024; i++)

     AESROUND << <128, 128 >> >();
cudaDeviceSynchronize();   //wait for device to finish
finish = clock();

CUDA中运行次获取结果的时间差

the Difference between running time and time of obtaining results in CUDA

cuda

gpu

nvidia