Can an array of shorts be passed into a CUDA kernel - cuda

Question

I have written a CUDA kernel and when I copy of an array of shorts to device memory and then pass it to the kernel it doesn't work. The simplified code below expresses my issue.
KernelCaller()
{
const int size = 1;
short hostArray[size]{41};
short* devPointer;
cudaMalloc((void**)&devicePointer, size * sizeof(short));
cudaMemcpy(devPointer, hostArray, size * sizeof(short), cudaMemcpyHostToDevice);
cudaKernel<<<1,1>>>(devPointer);
}
__global__
void cudaKernel(short* arr)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
short val = arr[idx];
}
At this point the value of val is 1063714857 and what I want it to be is 41.
I assume the issue is 41 in hex is 0x29 and the value I have is 0x3F670029 so it looks like it read too many bytes cause the 0x29 is at the beginning. When I switch to an array of floats it works perfectly, but I was trying to save memory. Does CUDA not allow an array of shorts?

Answer 1

I have implemented your code and getting the output as expected.
Here's the code
#include<stdio.h>
__global__ void cudaKernel(short* arr)
{
int idx = blockIdx.x * blockDim.x + threadIdx.x;
short val = arr[idx];
# if __CUDA_ARCH__>=200
printf("Inside kernel %d\n",val);
#endif
arr[idx] = val;
}
int main()
{
const int size = 1;
short hostArray[size]{41};
printf("Before kernel call %d\n",hostArray[0]);
short *devPointer;
cudaMalloc((void**)&devPointer, size * sizeof(short));
cudaMemcpy(devPointer, hostArray, size * sizeof(short), cudaMemcpyHostToDevice);
cudaKernel<<<1,1>>>(devPointer);
cudaMemcpy(hostArray, devPointer, size * sizeof(short), cudaMemcpyDeviceToHost);
printf("After kernel call %d\n",hostArray[0]);
cudaFree(devPointer);
return 0;
}
And the output is
Before kernel call 41
Inside kernel 41
After kernel call 41
So, yes we can pass array of shorts into a CUDA kernel.