Question

我想将网格和nextgrid数组转换为1d数组。因此，我可以在CUDA中更容易地实现每个单元格的线程化。目前，当从命令行传递特定维度时，会导致分段错误。例如。如果gridWidth = 6且gridHeight = 8。但是会在某些价值观上按预期工作。例如。如果他们都被传入6。

所以，我想知道如何使用pair值正确初始化1d数组，然后如何在嵌套for循环中实现它。

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <cstdlib>
int main (int argc, char *argv[]) 
{

using namespace std;
//delcare variables
int numIterate = atoi (argv[1]);

int gridWidth = atoi (argv[2]);

int gridHeight = atoi (argv[3]);
cout << "iterate: " << numIterate;

cout << " w: " << gridWidth;

cout << " h: " << gridHeight;

cout << endl;

  //make grid

  int grid[gridWidth][gridHeight];
  int nextgrid[gridWidth][gridHeight];
  int genCount=0;

//set all values in grid as 0
    for(int i=0;i<gridHeight;i++){
        for(int j=0;j<gridWidth;j++){
            grid[i][j]=0;

        }
    }
    //insert glider pattern
    grid[2][3]=1;
    grid[3][2]=1;
    grid[4][2]=1;
    grid[4][3]=1;
    grid[4][4]=1;

    for(int k=0; k< numIterate+1; k++){

      cout << endl;
        cout << endl;

        cout << "GENERATION " << genCount <<endl;
    for(int i=0;i<gridHeight;i++){
          cout << endl;
        for(int j=0;j<gridWidth;j++){

            int count=0;
         count+= grid[i][(j-1+gridWidth)%gridWidth] //left
         +grid[i][(j+1+gridWidth)%gridWidth]    //right
         +grid[(i-1+gridHeight)%gridHeight][j]    //up
         +grid[(i+1+gridHeight)%gridHeight][j]    //down
         +grid[(i-1+gridHeight)%gridHeight][ (j-1+gridWidth)%gridWidth]  //up-left
         +grid[(i+1+gridHeight)%gridHeight][ (j+1+gridWidth)%gridWidth]  //down-right
         +grid[(i+1+gridHeight)%gridHeight][ (j-1+gridWidth)%gridWidth]  //down-left
         +grid[(i-1+gridHeight)%gridHeight][ (j+1+gridWidth)%gridWidth] ;//up-right
      //return count;
      //cout << count;
      printf("%d", grid[i][j]);

                // Cell is lonely and dies
                if ((grid[i][j] == 1) && (count < 2)){
                    nextgrid[i][j] = 0;
                }
                // Cell dies due to over population
                else if((grid[i][j] == 1) && (count > 3)){
                    nextgrid[i][j] = 0;
                }
                // A new cell is born
                else if ((grid[i][j] == 0) && (count == 3)){
                    nextgrid[i][j] = 1;
                }
                // Remains the same
                else{
                    nextgrid[i][j] = grid[i][j];
            }
        }
    }
//assign current grid to be the next grid
  for(int i=0;i<gridHeight;i++){
        for(int j=0;j<gridWidth;j++){
      grid[i][j] = nextgrid[i][j];
    }
    }
    genCount++;
    }
}

Answer 1

说grid1d[z]是grid[x][y]的一维表示，其中x，y是二维矩阵的大小，z是大小的z = x * y 1d数组（也是gird[x1][x2]）。假设您要使用与0 <=x1 < x,0 <= y1 < y（grid1d[z1]）对应的元素，那么您z1 = x1 * y + y1位于OLDJAVA=$(ssh -q $server "java -version 2>&1 | cut -s -d '\"' -f2")。

将2d数组循环转换为1d矩阵数组C ++

1 个答案: