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

c++ c math cuda square-root
Avraam Mavridis picture Avraam Mavridis · Dec 1, 2013 · Viewed 15.6k times · Source

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?

Answer

Harshil Sharma picture Harshil Sharma · Dec 1, 2013

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.

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