为什么我的下降数组值消失了？答案

【问题标题】：Why is my falling array value disappearing?为什么我的下降数组值消失了？
【发布时间】：2022-08-17 13:27:26
【问题描述】：

我在 C 中有一个二维数组，我试图使正值或1 值下降直到它到达数组的底部，但是由于某种原因，一旦我运行代码，1 就会向左移动并且然后下来。它的工作方式是将当前值变为0，底部值变为1，重复此操作并产生直线下降的效果，此代码在fallDown()功能。

代码：

#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
#ifdef __unix__
# include <unistd.h>
#elif defined _WIN32
# include <windows.h>
#define sleep(x) Sleep(1000 * (x))
#endif

int sizeX = 20;
int sizeY = 20;

int grid[20][20] = {{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}};

int fallDown(int grid[sizeX][sizeY], int x, int y){
    grid[x][y]=0;
 grid[x][y+1]=1;
}

int main(void){
 int neighbour_count[sizeX][sizeY];
    int x,y,iterations;

 for(iterations=0;iterations<500;iterations++){
  system(\"cls\"); //Clear screen
  for(x=0;x<sizeX;x++){
    printf(\"\\n\");
    for(y=0;y<sizeY;y++){
     if(grid[x][y]==1){
      printf(\"@\");
             }
    else{
      printf(\" \");
    }
         }
  }
  for(y=0;y<sizeY;y++){
    for(x=0;x<sizeX;x++){
        if(grid[x][y] == 1){
         fallDown(grid, x, y);
    }
   }
        }
  printf(\"\\n\");
  sleep(1);
 }
}

fallDown 没有任何边界检查。使用您的循环，可能会使用y == sizeY - 1 调用它。那么y + 1 就会越界。
@Some程序员老兄仍然，我只是想了解为什么会发生这种向左走的奇怪行为以及如何解决它。界限问题是我打算稍后解决的问题。

标签： arrays c matrix

【解决方案1】：

x 和 y 需要在 fallDown 中交换。
在main 的第二组循环中，当找到1 时，循环必须停止，否则x 循环的下一次执行将一次又一次地fallDown...直到循环结束。
我正在使用 Linux，必须使用 clear 来清除屏幕。

#include<stdio.h>
#include<stdlib.h>
#include <unistd.h>

int sizeX = 20;
int sizeY = 20;

int grid[20][20] = {{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}};

void fallDown(int grid[sizeX][sizeY], int x, int y){
    grid[x][y]=0;
    ++x;
    if ( x == 20) {
        x = 0; // back to top
    }
    grid[x][y]=1;
}

int main(void){
    int x,y,iterations;

    for(iterations=0;iterations<500;iterations++){
        system("clear"); //Clear screen
        for(x=0;x<sizeX;x++){
            printf("\n");
            for(y=0;y<sizeY;y++){
                if(grid[x][y]==1){
                    printf("@");
                }
                else{
                    printf(" ");
                }
            }
        }
        for(y=0;y<sizeY;y++){
            for(x=0;x<sizeX;x++){
                if(grid[x][y] == 1){
                    fallDown(grid, x, y);
                    y = sizeY;
                    break;
                }
            }
        }
        printf("\n");
        sleep(1);
    }
}

【讨论】：

【解决方案2】：

另一种方法是使用终端转义码。
（只要终端支持它们。）
这会包裹运动，因此如果@ 超出边缘，它会出现在相对边缘。不确定这是否是我们想要的？

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <time.h>

#define MAXROW 20
#define MAXCOL 20

typedef struct at_s {
    int row;
    int col;
} at_t;

void moveat ( at_t *at, int ud, int rl) {
    // moves up down right left diagonal
    at->row += ud;
    if ( at->row > MAXROW) {
        at->row = 1;
    }
    if ( at->row < 1) {
        at->row = MAXROW;
    }
    at->col += rl;
    if ( at->col > MAXCOL) {
        at->col = 1;
    }
    if ( at->col < 1) {
        at->col = MAXCOL;
    }
}

void showat ( at_t at, int show) {
    printf ( "\033[%d;%dH%c", at.row, at.col, show);
    fflush ( stdout);
}

int main ( void) {
    int chgrow = 0;
    int chgcol = 0;
    at_t at = { 4, 12};

    srand ( time ( NULL));
    printf ( "\033[2J");//clear screen

    for ( int moving = 0; moving < 500; ++moving) {
        showat ( at, ' ');
        chgrow = rand ( ) % 3; // range of 0 1 or 2
        --chgrow; // range is now -1 0 or 1

        chgcol = rand ( ) % 3;
        --chgcol;
        moveat ( &at, chgrow, chgcol);
        showat ( at, '@');
        printf ( "\033[22;1H\n");
        sleep ( 1);
    }

    printf ( "\033[22;1H\n");

    return 0;
}

【讨论】：

【解决方案3】：

int fallDown(int grid[sizeX][sizeY], int x, int y){
    grid[x][y] = 0;
    grid[x][y+1] = 1;
}

sizex 是行数， sizeY 是列数

所以你应该像这样改变fallDown函数

int fallDown(int grid[sizeX][sizeY], int x, int y){
      if(x == 19 || grid[x][y] == 0) return;
      grid[x+1][y] = 1;
      grid[x][y] = 0;
}

【讨论】：

它仍然不起作用，它只是保持静止。
现在它所做的只是传送到底部而忽略了旨在减慢它的 sleep(1) 函数
只需交换 X 和 Y 即可解决问题。
根据二维数组的结构，X 表示行，Y 表示列。我们必须增加 X 才能向下移动。
我需要更改哪些行，fallDown 函数中的行，因为当我这样做时 @ 不会移动，很抱歉给您带来麻烦。

【解决方案4】：

我找到了一个更简单的答案，您只需在每次移动@ 时更新屏幕即可。

#include<stdio.h>
#include<stdlib.h>
#include<conio.h>
#ifdef __unix__
# include <unistd.h>
#elif defined _WIN32
# include <windows.h>
#define sleep(x) Sleep(1000 * (x))
#endif

int sizeX = 20;
int sizeY = 20;

int grid[20][20] = {{0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0},
                    {0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0}};
                    
int drawScreen(int x, int y){
    system("cls"); //Clear screen
  for(x=0;x<sizeX;x++){
    printf("\n");
    for(y=0;y<sizeY;y++){
     if(grid[x][y]==1){
      printf("@");
     }
    else{
     printf(" ");
    }
   }
  }
}

int main(void){
 int x,y,iterations;

 for(iterations=0;iterations<10;iterations++){
    for(y=0;y<sizeY;y++){
        for(x=0;x<sizeX;x++){
            if(grid[y][x] >= 1){
                grid[y][x] = 0;
                grid[y+1][x] = 1;
                drawScreen(x,y);
                break;
      }
        }
    }
    printf("\n");
 }
}

【讨论】：