Memcpy time consumption

...

...

...

cudaMalloc((void**)&d_total, n*(n-1)/2*sizeof(unsigned int));

for(float i=0; i<100; i+=0.0005)

{

  ...

  ...

  ...

  RunKernel<<<blockNum, threadNum>>>(n, d_total, d_a, d_b, d_c, d_d, d_e, d_f);

  cutStartTimer(hTimer);

  cudaMemcpy(_total, d_total, n*(n-1)/2*sizeof(unsigned int), cudaMemcpyDeviceToHost);

  cutStopTimer(hTimer);

}

cudaFree(d_total)

cudaFree...

...

...
Time consumption

    .

    .

    .

25.351915

25.511745

25.542639

25.459280

25.367662

25.597145

25.694475

589.564331

589.044983

588.816833

592.139099

591.546204

591.162537

    .

    .

When “i=0, 0.0005, 0.001, 0.0015… to i=20”, the time consumption for each value that is about 25 milliseconds, but when “i” is large than 20 or the other number, I will get about 589 milliseconds.

I have a qustion:

The size of d_total that is fixed. -> n*(n-1)/2*sizeof(unsigned int)

Why will the time consumption of Memcpy be changed so largely?

Please forgive my poor English.

Many thanks.

Thank you.

:blink:

Call cudaThreadSynchronize() before cudaStartTimer. As you have it now, you may or may not be timing the kernel execution along with the memcpy.

I already called cudaThreadSynchronize() before cudaStartTimer, it’s ok.

But I don’t understand yet, could you explain the reason to me?

Thank you.

:huh: