Why is my call of the CUDA math library sqrt() function failing?

Question

I am new to Cuda, I have the following function:

__global__ void square(float *myArrayGPU)
{
   myArrayGPU[threadIdx.x] = sqrt(threadIdx.x);
}

I want to use the cuda math library, I tried to #include "math.h" but I still get the error

error: calling a __host__ function("__sqrt") from a __global__ function("square") is not allowed

Any idea what library should I include to use the sqrt?

score 29 · Accepted Answer · edited Dec 01 '13 at 23:16

threadIdx.x is of type int. CUDA math library is overloaded only for single precision (float) and double precision (double). You need to supply either a 'float' or 'double' type parameter to sqrt() for the CUDA version of sqrt() to be called.

Change

myArrayGPU[threadIdx.x] = sqrt(threadIdx.x);

into

myArrayGPU[threadIdx.x] = sqrt( (float) threadIdx.x);

For more detailed information, take a look at the CUDA sqrt() prototype documentation.

score 9 · Answer 2 · answered Dec 01 '13 at 12:51

9

sqrt expects a floating type variable. Try sqrt((float)(threadIdx.x))

answered Dec 01 '13 at 12:51

Avi Ginsburg

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1

Good answer, but unfortunately Harshil Sharma beat you to the 'textbook version'... :-) – einpoklum Dec 01 '13 at 23:17

Why is my call of the CUDA math library sqrt() function failing?

2 Answers2