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当我测试以下示例时，我发现通过增加 blocksPerGrid 和 threadsPerBlock 内核延迟会增加

这样如果

int threadsPerBlock = 1;    
int blocksPerGrid = 1;

blocksPerGrid 和 threadsPerBlock 等于 1 内核的延迟 = .0072 ms

但是当我进行以下操作时，延迟变得更高= .049 ms

int threadsPerBlock = 1024;
int blocksPerGrid = (N+threadsPerBlock-1) / threadsPerBlock;

在哪里

N = 50000; //the no. of array elements

下面是完整的 VecAdd 示例。你可以测试一下

// Includes
#include <stdio.h>
#include <cutil_inline.h>
#include <shrQATest.h>

// Variables
float* h_A;
float* h_B;
float* h_C;
float* d_A;
float* d_B;
float* d_C;
bool noprompt = false;

// Functions
void CleanupResources(void);
void RandomInit(float*, int);
void ParseArguments(int, char**);

 // Device code
 __global__ void VecAdd(const float* A, const float* B, float* C, int N)
{
  int i = blockDim.x * blockIdx.x + threadIdx.x;
  if (i < N)
      C[i] = A[i] + B[i];
 } 

 // Host code
 int main(int argc, char** argv)
 {
     shrQAStart(argc, argv);

cudaEvent_t event1, event2;
cudaEventCreate(&event1);
cudaEventCreate(&event2);


 printf("Vector Addition\n");
 int N = 50000;
 size_t size = N * sizeof(float);
 ParseArguments(argc, argv);

// Allocate input vectors h_A and h_B in host memory
h_A = (float*)malloc(size);
if (h_A == 0) CleanupResources();
h_B = (float*)malloc(size);
if (h_B == 0) CleanupResources();
h_C = (float*)malloc(size);
if (h_C == 0) CleanupResources();

// Initialize input vectors
RandomInit(h_A, N);
RandomInit(h_B, N);

// Allocate vectors in device memory
cutilSafeCall( cudaMalloc((void**)&d_A, size) );
cutilSafeCall( cudaMalloc((void**)&d_B, size) );
cutilSafeCall( cudaMalloc((void**)&d_C, size) );

// Copy vectors from host memory to device memory
cutilSafeCall( cudaMemcpy(d_A, h_A, size, cudaMemcpyHostToDevice) );
cutilSafeCall( cudaMemcpy(d_B, h_B, size, cudaMemcpyHostToDevice) );

 // Invoke kernel
 int threadsPerBlock = 1024;
 int blocksPerGrid = (N+threadsPerBlock-1) / threadsPerBlock;

 cudaEventRecord(event1, 0); 
 VecAdd<<<blocksPerGrid, threadsPerBlock>>>(d_A, d_B, d_C, N);
 cudaEventRecord(event2, 0);

cudaEventSynchronize(event1); //optional
cudaEventSynchronize(event2);

float dt_ms;
cudaEventElapsedTime(&dt_ms, event1, event2);

printf("delay_time = %f\n", dt_ms);

   cutilCheckMsg("kernel launch failure");
 #ifdef _DEBUG
  cutilSafeCall( cutilDeviceSynchronize() );
 #endif

  // Copy result from device memory to host memory
  // h_C contains the result in host memory
  cutilSafeCall( cudaMemcpy(h_C, d_C, size, cudaMemcpyDeviceToHost) );

  // Verify result
  int i;
  for (i = 0; i < N; ++i) {
      float sum = h_A[i] + h_B[i];
      if (fabs(h_C[i] - sum) > 1e-5)
          break;
  }

     CleanupResources();
     shrQAFinishExit(argc, (const char **)argv, (i==N) ? QA_PASSED : QA_FAILED);

 }

  void CleanupResources(void)
 {
  // Free device memory
  if (d_A)
     cudaFree(d_A);
  if (d_B)
     cudaFree(d_B);
  if (d_C)
     cudaFree(d_C);

  // Free host memory
  if (h_A)
     free(h_A);
  if (h_B)
     free(h_B);
  if (h_C)
     free(h_C);

  cutilDeviceReset();
 }

  // Allocates an array with random float entries.
  void RandomInit(float* data, int n)
 {
    for (int i = 0; i < n; ++i)
    data[i] = rand() / (float)RAND_MAX;
 }

 // Parse program arguments
 void ParseArguments(int argc, char** argv)
 {
   for (int i = 0; i < argc; ++i) {
      if (strcmp(argv[i], "--noprompt") == 0 ||
          strcmp(argv[i], "-noprompt") == 0) 
      {
        noprompt = true;
        break;
        }
    }
  }

谁能为我解释一下这是什么意思？

4

1 回答 1

2

在情况 1 中，启动大小为 1 的内核线程并执行 2 次读取和 1 次写入操作。在案例 2 中，启动大小为 50176 线程的内核并执行 100,000 次读取和 50,000 次写入操作。将工作负载增加 50,000 会使执行时间增加约 7 倍。两次发射所做的工作明显不同。

于 2013-01-07T23:25:43.123 回答