thrust doesn't provided the expected result using thrust::minimum

Question

consider the following code, when p is a pointer allocated GPU-side.

thrust::device_ptr<float> pWrapper(p); thrust::device_ptr<float> fDevPos = thrust::min_element(pWrapper, pWrapper + MAXX * MAXY, thrust::minimum<float>()); fRes = *fDevPos; *fDicVal = fRes;

after applying the same thing on cpu side.

float *hVec = new float[MAXX * MAXY]; cudaMemcpy(hVec, p, MAXX*MAXY*sizeof(float), cudaMemcpyDeviceToHost); float min = 999; int index = -1; for(int i = 0 ; i < MAXX* MAXY; i++) { if(min > hVec[i]) { min = hVec[i]; index = i; } } printf("index :%d a wrapper : %f, as vectorDevice : %f\n",index, fRes, min); delete hVec;

i get that min != fRes. what am i doing wrong here?

Jared Hoberock · Accepted Answer · 2012-03-15 21:51:04Z

thrust::minimum_element requires the user to supply a comparison predicate. That is, a function which answers the yes-or-no question "is x smaller than y?"

thrust::minimum is not a predicate; it answers the question "which of x or y is smaller?".

To find the smallest element using minimum_element, pass the thrust::less predicate:

ptr_to_smallest_value = thrust::min_element(first, last, thrust::less<T>());

Alternatively, don't pass anything. thrust::less is the default:

ptr_to_smallest_value = thrust::min_element(first, last);

If all you're interested in is the value of the smallest element (not an iterator pointing to the smallest element), you can combine thrust::minimum with thrust::reduce:

smallest_value = thrust::reduce(first, last, std::numeric_limits<T>::max(), thrust::minimum<T>());

and which way is faster? i used the way you wrote me last night and it returned a pointer (device_ptr). on which i had to use the * operator to extract the value.
They should be nearly the same speed as they're both limited by bandwidth of reading the array.

Collectives™ on Stack Overflow

thrust doesn't provided the expected result using thrust::minimum

1 Answer 1

2 Comments

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

2 Comments

Related