在CUDA中的块之间分配线程 [英] Distribute the threads between blocks in CUDA
问题描述
我正在使用CUDA中的项目。第一次我使用 Dim 8 * 8 作为我的矩阵只有一个块。然后我计算索引如下：
I'm working on a project in CUDA. The first time I used only one block with Dim 8*8 as my matrix. And then I calculated the index as follows:
int idx = blockIdx.x * blockDim.x + threadIdx.x;
int idy = blockIdx.y * blockDim.y + threadIdx.y;
它给了我一个正确的答案。之后，我想在块之间分配线程来衡量性能。我使网格变暗为（2,1），块变暗为（4,8）。
And it gave me a correct answer. After that I want to distribute the threads between blocks to measure the performance. I make the grid dim to be (2,1) and the block dim to be (4,8).
当我手动调试代码时，给我正确的索引，而不改变上面提到的公式。但是当我运行程序时，屏幕挂起，结果都是零。
When I debug the code by hand, it seems to give me the correct index without changing the formula mentioned above. But when I run the program, the screen hangs and the results are all zero.
我做错了什么，如何解决这个问题？
What did I do wrong, and how can I fix this?
这是内核函数
__global__ void cover_fault(int *a,int *b, int *c, int *d, int *mulFV1, int *mulFV2, int *checkDalU1, int *checkDalU2, int N)
{
//Fig.2
__shared__ int f[9][9];
__shared__ int compV1[9],compV2[9];
int dalU1[9] , dalU2[9];
int Ra=2 , Ca=2;
for (int i = 0 ; i < N ; i++)
for (int j = 0 ; j < N ; j++)
f[i][j]=0;
f[3][0] = 1;
f[0][2] = 1;
f[0][6] = 1;
f[3][7] = 1;
f[2][4] = 1;
f[6][4] = 1;
f[7][1] = 1;
int t =0 ,A = 1,B = 1 , UTP = 5 , LTP = -5 , U_max = 40 , U_min = -160;
bool flag = true;
int sumV1, sumV2;
int checkZero1 , checkZero2;
int idx = blockIdx.x * blockDim.x + threadIdx.x;
int idy = blockIdx.y * blockDim.y + threadIdx.y;
while ( flag == true)
{
if ( c[idy] == 0 )
compV1[idy] = 1;
else if ( c[idy]==1)
compV1[idy] = 0 ;
if ( d[idy] == 0 )
compV2[idy] = 1;
else if ( d[idy]==1 )
compV2[idy] = 0 ;
sumV1 = reduce ( c, N );
sumV2 = reduce ( d, N );
if (idx<N && idy <N)
{
if(idx==0)
mulFV1[idy]=0;
if(idy==0)
mulFV2[idx]=0;
__syncthreads();
atomicAdd(&(mulFV1[idy]),f[idy][idx]*compV2[idx]);
atomicAdd(&(mulFV2[idx]),f[idy][idx]*compV1[idy]);
}
dalU1[idy] = ( -1*A*( sumV1 - Ra )) + (B * mulFV1[idy] * compV1[idy]) ;
dalU2[idy] = ( -1*A*( sumV2 - Ca )) + (B * mulFV2[idy] * compV2[idy]) ;
a[idy] = a[idy] + dalU1[idy];
b[idy] = b[idy] + dalU2[idy];
if ( a[idy] > U_max )
a[idy] = U_max;
else
if (a[idy] < U_min )
a[idy] = U_min;
if ( b[idy] > U_max )
b[idy] = U_max;
else
if (b[idy] < U_min )
b[idy] = U_min;
if (dalU1[idy]==0)
checkDalU1[idy]=0;
else
checkDalU1[idy]=1;
if (dalU2[idy]==0)
checkDalU2[idy]=0;
else
checkDalU2[idy]=1;
__syncthreads();
checkZero1 = reduce(checkDalU1,N);
checkZero2 = reduce(checkDalU2,N);
if ( checkZero1==0 && checkZero2==0)
flag = false;
else
{
if ( a[idy] > UTP )
c[idy] = 1;
else
if ( a[idy] < LTP )
c[idy] = 0 ;
if ( b[idy] > UTP )
d[idy] = 1;
else
if ( b[idy] < LTP )
d[idy] = 0 ;
t++;
}//end else
sumV1=0;
sumV2=0;
mulFV1[idy]=0;
mulFV2[idy]=0;
} //end while
}//end function
推荐答案
在你的索引计算中， idx 会给你列索引和 idy 行索引。是否正在访问您的矩阵 M [idy] [idx] ？
In your index computation, idx will give you the column index and idy the row index. Are you accessing your matrix as M[idy][idx]?
cuda线程根据正交系统：X是水平的，Y是垂直的。因此，如果你在实际矩阵中说点M [0] [1]，它是M [1] [0]。
The cuda threads are organized according to the orthogonal system: X is horizontal and Y is vertical. So if you say the point M[0][1] in the actual matrix it's M[1][0].
这篇关于在CUDA中的块之间分配线程的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
