Take Garbage Value wrong output how to use shared memory in a program

Accelerated Computing CUDA CUDA Programming and Performance
#1

Hi all,

I read a program to find the Multiplication of SquareMatrix Using shared Memory.But It gives the wrong result(outputsnapshot)and my expected result is other(shapshot expect).

Tell me?

How Can I get a exacted output?

Thanks in Advance

#include<stdio.h>

#include<cuda.h>

__global__ void Shar(float *a,float *b,float *c,int n)

{

	__shared__ float aTile[4][4],bTile[4][4];

	int row=blockIdx.y*blockDim.y+threadIdx.y;

	int col=blockIdx.x*blockDim.x+threadIdx.x;

	float sum=0.0;

	aTile[threadIdx.y][threadIdx.x]=a[row*n+threadIdx.x];

	bTile[threadIdx.y][threadIdx.x]=b[threadIdx.y*n+col];

	__syncthreads();

	for(int i=0;i<n;i++)

	{

		sum+=aTile[threadIdx.y][i]*bTile[i][threadIdx.x];

	}

	c[row*n+col]=sum;

}

int main()

{

	float *a_h,*b_h,*c_h,*a_d,*b_d,*c_d;

	int i,n;

	n=4;

	size_t size=sizeof(float)*(n*n);

	a_h=(float*)malloc(size);

	b_h=(float*)malloc(size);

	c_h=(float*)malloc(size);

	cudaMalloc((void**)&a_d,size);

	cudaMalloc((void**)&b_d,size);

	cudaMalloc((void**)&c_d,size);

	

	for(i=0;i<(n*n);i++)

	{

		a_h[i]=1;

	}

	cudaMemcpy(a_d,a_h,size,cudaMemcpyHostToDevice);

	 

		for(i=0;i<(n*n);i++)

		{

			 b_h[i]=1;

		}

		cudaMemcpy(b_d,b_h,size,cudaMemcpyHostToDevice);

	int blocksize=4;

	int nblock=n/blocksize+(n%blocksize==0?0:1);

	int TILE_DIM=4;

	Shar<<<nblock,blocksize,TILE_DIM>>>(a_d,b_d,c_d,n);

	cudaMemcpy(c_h,c_d,size,cudaMemcpyDeviceToHost);

	printf("\nMultiplication Of Matrix");

	for(i=0;i<(n*n);i++)

	{

		printf("\n%f",c_h[i]);

	}

	free(a_h);

	free(b_h);

	free(c_h);

	cudaFree(a_d);

	cudaFree(b_d);

	cudaFree(c_d);

	return 0;

}

expect.png
output.png

#2

You forgot to add:

cudaThreadSynchronize();

after kernels execution. Therefore you might copy arrays before kernels stop operation on them.

Additionally If someone use shared memory, extra care should be taken to prevent collisions. I have many time received unstable results because one of the threads was reading from shared memory befor other have put something there.

#3

No, CUDA implicitly synchronizes between operations on the same stream. (If you don’t specify a stream, you are on stream 0.) You do not need cudaThreadSynchronize() for anything but benchmarking kernel runtime (and perhaps one obscure use of zero-copy memory).

You should check the return codes on your CUDA calls to see if your kernel is being run at all.