使用CUDA创建2D数组

Question

在cuda c编程指南文档中有一个显示2d数组的示例:

// Kernel definition
__global__ void MatAdd(float A[N][N], float B[N][N], float C[N][N])
{
int i = blockIdx.x * blockDim.x + threadIdx.x;
int j = blockIdx.y * blockDim.y + threadIdx.y;
if (i < N && j < N)
C[i][j] = A[i][j] + B[i][j];
}

int main()
{
...
// Kernel invocation
dim3 threadsPerBlock(16, 16);
dim3 numBlocks(N / threadsPerBlock.x, N / threadsPerBlock.y);
MatAdd<<<numBlocks, threadsPerBlock>>>(A, B, C);
...
}

我使用下面的2d数组形式,并正常工作:

dim3 grid[COLUMNS][ROWS];
kernel_Matrix<<<grid,1>>>(dev_strA, dev_strB, dev_Matrix);

__global__ void add(int *a, int *b, int *c)
{
int x = blockIdx.x;
int y = blockIdx.y;
int i = (COLUMNS*y) + x;
c[i] = a[i] + b[i];
}

有一种方法用[] []定义实现2d数组？我以这种方式测试但不起作用.

Answer 1

dim3不是数组,而是在CUDA头文件(vector_types.h)中定义的结构.此结构用于在全局函数的执行配置中指定GRID的维度,即in <<< >>>.它不会保留"真正的"块,它只是配置了许多将要执行的块.

初步化这种结构的两种方式(据我所知)是:
1.2 dim3 grid(x, y, z);
.dim3 grid = {x, y, z};

编辑: 主机代码dim3初始化并将数组传递给内核函数,您将能够通过[][]以下方式访问其元素:

float A[N][N];
float B[N][N];
float C[N][N];

float (*d_A)[N]; //pointers to arrays of dimension N
float (*d_B)[N];
float (*d_C)[N];

for(int i = 0; i < N; i++) {
    for(int j = 0; j < N; j++) {
        A[i][j] = i;
        B[i][j] = j;
    }
}       

//allocation
cudaMalloc((void**)&d_A, (N*N)*sizeof(float));
cudaMalloc((void**)&d_B, (N*N)*sizeof(float));
cudaMalloc((void**)&d_C, (N*N)*sizeof(float));

//copying from host to device
cudaMemcpy(d_A, A, (N*N)*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(d_B, B, (N*N)*sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(d_C, C, (N*N)*sizeof(float), cudaMemcpyHostToDevice);

// Kernel invocation
dim3 threadsPerBlock(16, 16);
dim3 numBlocks(N / threadsPerBlock.x, N / threadsPerBlock.y);
MatAdd<<<numBlocks, threadsPerBlock>>>(d_A, d_B, d_C);

//copying from device to host
cudaMemcpy(A, (d_A), (N*N)*sizeof(float), cudaMemcpyDeviceToHost);
cudaMemcpy(B, (d_B), (N*N)*sizeof(float), cudaMemcpyDeviceToHost);
cudaMemcpy(C, (d_C), (N*N)*sizeof(float), cudaMemcpyDeviceToHost);