Newb question: Different results for dbg, emu and release

Accelerated Computing CUDA CUDA Programming and Performance
#1

OK I don’t understand the difference in results from the code at the bottom of this post. The results from,

nvcc -I /opt/NVIDIA_CUDA_SDK/common/inc/ Main.cu -deviceemu

are,

host: (2,2)

host: (2,2)

host: (2,2)

host: (2,2)

host: (2,2)

host: (2,2)

---------------

host: (4,4)

host: (4,4)

host: (4,4)

host: (4,4)

host: (4,4)

as I would expect. The results from

nvcc -I /opt/NVIDIA_CUDA_SDK/common/inc/ Main.cu -g -G

are,

host: (2,2)

host: (2,2)

host: (2,2)

host: (2,2)

host: (2,2)

---------------

host: (2,2)

host: (2,2)

host: (2,2)

host: (2,2)

host: (2,2)

Not as I’d expect. The results from just,

nvcc -I /opt/NVIDIA_CUDA_SDK/common/inc/ Main.cu

are,

host: (2,2)

host: (2,2)

host: (2,2)

host: (2,2)

host: (2,2)

---------------

host: (512,2)

host: (512,2)

host: (512,2)

host: (512,2)

host: (512,2)

So what have I done wrong?

Thanks very much.

#include <stdio.h>

#include <assert.h>

#include <cuda.h>

#include <cutil.h>

__global__ void testDouble2(double2* vaid, int N) {

	

	int tx = threadIdx.x;

	int ty = threadIdx.y;

	int bx = blockIdx.x;

	int by = blockIdx.y;

	//printf("(%i,%i)-(%i,%i)\n",bx,by,tx,ty);

	vaid[tx].x = vaid[tx].x + vaid[tx].x;

	vaid[tx].y = vaid[tx].y + vaid[tx].y;

}

int main(void) {

	double2 host[] = {

		make_double2(2, 2),

		make_double2(2, 2),

		make_double2(2, 2),

		make_double2(2, 2),

		make_double2(2, 2),

	};

	for (int i = 0; i < 5;i++){

	  printf("host: (%g,%g)\n", host[i].x, host[i].y);

	}

	printf("---------------\n");

	double2* device;

	cudaMalloc((void **) &device, sizeof(double2)*5);

	cudaMemcpy(device, &host,sizeof(double2)*5, cudaMemcpyHostToDevice);

	dim3 threads(5, 1);

	dim3 grid(1,1);

	testDouble2<<< grid,threads >>> (device, 5);

	

	cudaMemcpy(&host, device, sizeof(double2)*5, cudaMemcpyDeviceToHost);

	for (int i = 0; i < 5;i++){

	 printf("host: (%g,%g)\n", host[i].x, host[i].y);

	}

}
#2

compile with -arch sm_13

#3

Thanks for taking the time to reply. I tried with that and got the same result. I realise now that my card (9500 GT) supports only compute capability 1.1 and double precision was implemented in 1.3.

I guess I’ll have to stick to single precision for now.

Is there no way to make the GPU treat doubles as floats without need to change the code? I also tried sm_11, sm_10, compute_10 and compute_11.

Thanks.

#4

no, making the GPU treat doubles as floats is essentially meaningless.

#5

I guess in some circumstances it might be as bad as treating them as ints.

Thanks for your help clearing this up.