我正在尝试计算行数和列数不相等的2D矩阵的 Summed Area Table 。我遇到了一个小问题，我的代码似乎在行和列相等的情况下运行正常，但是当行和列不相等时，它无法计算最终输出的最后一行。问题是我无法弄清楚为什么会这样。
基本上，在积分和中，每个像素或索引元素计算其上方和后方的所有矩阵元素的总和。例如，对于具有以下元素的3x2输入数组：
[5, 2|
|5, 2|
|5, 2]
输出数组中的积分和将为：
[5, 7|
|10, 14|
|15, 21]
基本上以下是我在CUDA C中尝试做的事情：
for(int matrixElement_y_index=0; matrixElement_y_index<=total_rows-1; matrixElement_y_index++)
{
//matrixElement_x_index and matrixElement_y_index represent (x,y) indices of each matrix element
for(int matrixElement_x_index=0; matrixElement_x_index<=total_columns-1; matrixElement_x_index++)
{
int temp=0;
for(int r=0;r<=(matrixElement_y_index);r++)
{
for(int c=0; c<=matrixElement_x_index;c++)
{
temp=temp+input[c][r];
}
}
output[matrixElement_y_index][matrixElement_x_index]=temp;
}
}
我到目前为止提出的CUDA C代码如下：
#include <iostream>
#include <cuda_runtime.h>
using namespace std;
__global__ void image_integral(int *a, int*b, int width_x,int width_y)
{
// Thread Ids equal to block Ids because the each blocks contains one thread only.
int gidx = blockIdx.x;
int gidy = blockIdx.y;
int temp=0;
if(gidx>=width_x || gidy>=width_y)
{
//Return the threads which exceed the input array's X or Y dimension.
return;
}
else
//Compute the Integral Image or Summed Area Table
{
// The first loop iterates from zero to the Y index of the thread which represents the corresponding element of the output/input array.
for(int counter=0;counter<=gidy;counter++)
{
// The first loop iterates from zero to the X index of the thread which represents the corresponding element of the output/input array
for(int counter_two=0; counter_two<=gidx; counter_two++)
{
temp = temp+a[counter*width_x+counter_two];
}
}
}
//Transfer the final result to the output array
b[gidy*width_x+gidx]=temp;
}
void main()
{
//M is number of rows
//N is number of columns
int M=3,N=2, m_e=0;
int total_e=M*N;
int widthstep=total_e*sizeof(int);
int * matrix_a= (int *)malloc(widthstep);
int * matrix_b= (int *)malloc(widthstep);
cout<<"Enter elements for "<< M<<"x"<<N<<" matrix";
for(int r=0;r<=M-1;r++)
{
for(int c=0; c<=N-1;c++)
{
cout<<"Enter Matrix element [ "<<c<<","<<r<<"]";
cin>>m_e;
matrix_a[r*M+c]=m_e;
matrix_b[r*M+c]=0;
}
}
int * d_matrix_a, * d_matrix_b;
cout<<"Input:"<<endl;
for(int kk=0;kk<=M-1;kk++)
{
for(int jj=0;jj<=N-1;jj++){
cout<<matrix_a[kk*M+jj]<<" ";}
cout<<endl;
}
cout<<endl;
cudaMalloc(&d_matrix_a,widthstep);
cudaMalloc(&d_matrix_b,widthstep);
cudaMemcpy(d_matrix_a,matrix_a,widthstep,cudaMemcpyHostToDevice);
cudaMemcpy(d_matrix_b,matrix_b,widthstep,cudaMemcpyHostToDevice);
//Creating a grid where the number of blocks are equal to the number of pixels or input matrix elements.
//Each block contains only one thread.
dim3 grid(M,N);
image_integral<<<grid,1>>>(d_matrix_a, d_matrix_b,M,N);
cudaThreadSynchronize();
cudaMemcpy(matrix_b,d_matrix_b,widthstep,cudaMemcpyDeviceToHost);
cout<<"The Summed Area table is: "<<endl;
for(int kk=0;kk<=M-1;kk++)
{
for(int jj=0;jj<=N-1;jj++)
cout<<matrix_b[kk*M+jj]<<" ";
cout<<endl;
}
system("pause");
cudaFree(d_matrix_a);
cudaFree(d_matrix_b);
free(matrix_a);
free(matrix_b);
}
非常感谢!!
答案 0 :(得分：5)
您的主要问题是内存使用和存储错误。使用你的代码你也破坏了堆！ 我通过使用行主要排序来改变你的代码，因为它通常用在c / c ++中。
将输入写入主机内存matrix_a[r*M+c]时，会出现第一个错误。由于r范围来自0..M(3)且c范围来自0..N(2)，因此最大索引为2*3+1=7。但你的矩阵只有6个元素 - 最大指数是5！因此，我改变了所有矩阵访问。
通过这些更改，我也必须适合您的网格设置。现在是dim3 grid(N,M);。
如果您不确定变量代表什么或如何使用它，请使用良好的代表名称，就像在c参考代码中所做的那样！
随着这一点的改变你的代码对我有用。请注意，矩阵的输入方式也已改变了！
更改完整代码之上： 内核函数：
__global__ void image_integral(int *a, int*b, int rowsTotal,int colsTotal)
{
// Thread Ids equal to block Ids because the each blocks contains one thread only.
int col = blockIdx.x;
int row = blockIdx.y;
int temp=0;
if(col < colsTotal && row < rowsTotal)
{
// The first loop iterates from zero to the Y index of the thread which represents the corresponding element of the output/input array.
for(int r=0;r<=row;r++)
{
// The second loop iterates from zero to the X index of the thread which represents the corresponding element of the output/input array
for(int c=0; c<=col; c++)
{
temp = temp+a[r*colsTotal+c];
}
}
}
//Transfer the final result to the output array
b[row*colsTotal+col]=temp;
}
主持人实施：
void main()
{
//M is number of rows
//N is number of columns
int M=3,N=2, m_e=0;
int total_e=M*N;
int widthstep=total_e*sizeof(int);
int * matrix_a= (int *)malloc(widthstep);
int * matrix_b= (int *)malloc(widthstep);
cout<<"Enter elements for "<< M<<"x"<<N<<" matrix";
for(int r=0;r<M;r++)
{
for(int c=0; c<N;c++)
{
cout<<"Enter Matrix element [ "<<r<<","<<c<<"]";
cin>>m_e;
matrix_a[r*N+c]=m_e;
matrix_b[r*N+c]=0;
}
}
int * d_matrix_a, * d_matrix_b;
cout<<"Input:"<<endl;
for(int r=0;r<M;r++)
{
for(int c=0; c<N;c++)
{
cout << matrix_a[r*N+c]<<" ";
}
cout << endl;
}
cout<<endl;
cudaMalloc(&d_matrix_a,widthstep);
cudaMalloc(&d_matrix_b,widthstep);
cudaMemcpy(d_matrix_a,matrix_a,widthstep,cudaMemcpyHostToDevice);
cudaMemcpy(d_matrix_b,matrix_b,widthstep,cudaMemcpyHostToDevice);
//Creating a grid where the number of blocks are equal to the number of pixels or input matrix elements.
//Each block contains only one thread.
dim3 grid(N,M);
image_integral<<<grid,1>>>(d_matrix_a, d_matrix_b,M,N);
cudaThreadSynchronize();
cudaMemcpy(matrix_b,d_matrix_b,widthstep,cudaMemcpyDeviceToHost);
cout<<"The Summed Area table is: "<<endl;
for(int r=0;r<M;r++)
{
for(int c=0; c<N;c++)
{
cout << matrix_b[r*N+c]<<" ";
}
cout << endl;
}
system("pause");
cudaFree(d_matrix_a);
cudaFree(d_matrix_b);
free(matrix_a);
free(matrix_b);
}