我如何为这段代码编写函数

Question

我正在尝试为代码编写函数或方法。我真的不知道如何在下面的输出中打印出出生数字和死亡数字的循环。

这是我的代码：

#include <stdio.h>

char grid[10][10];

// Moved this function here
int occ(int x, int y) {
    int i, j, count;
    char gridCopy[10][10];


// You probably are going to do this often in the future
// Make a function for it instead of having the same code often
void printGrid() {
    int i, j;
    for (i = 0; i < 10; i++) {
        for (j = 0; j < 10; j++)
            printf("%c", grid[i][j]);
        printf("\n");
    }
}

Answer 1

occ 函数中不需要 gridCopy。 请记住，数组是转置的。当您访问元素 (x,y) 时，您需要使用grid[y][x]

下面是函数generateNext() 的工作代码，它为整个grid 数组生成下一个状态。这是您需要girdCopy 的地方。结果与您的示例完全相同。

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

char grid[10][10];
int  born,died,generation;

// Moved this function here
int occ(int x, int y) {
    int  count;
    int  xm1,xp1,ym1,yp1;

    xm1=x-1;xp1=x+1;ym1=y-1;yp1=y+1;
    if (xm1<0) // handle boundary cases - wrap around
        xm1=9;
    if (ym1<0)
        ym1=9;
    if (xp1>9)
        xp1=0;
    if (yp1>9)
        yp1=0;

    // Checking on the value of the neighboring cells
    count = 0;
    if (grid[ym1][x] == '*')
        count++;
    if (grid[ym1][xp1] == '*')
        count++;
    if (grid[y][xp1] == '*')
        count++;
    if (grid[yp1][xp1] == '*')
        count++;
    if (grid[yp1][x] == '*')
        count++;
    if (grid[yp1][xm1] == '*')
        count++;
    if (grid[y][xm1] == '*')
        count++;
    if (grid[ym1][xm1] == '*')
        count++;

    return count;
}

void generateNext()
{
    int x,y;
    char gridCopy[10][10];

    born=0;died=0;
    generation++;

    for (y=0; y<10; y++) {
        for (x=0; x<10; x++) {
            gridCopy[y][x]=grid[y][x];
        }
    }

    for (y=0; y<10; y++) {
        for (x=0; x<10; x++) {
            if (grid[y][x]=='*' && occ(x,y)<2)
                {gridCopy[y][x]='-';died++;}
            if (grid[y][x]=='*' && occ(x,y)>3)
                {gridCopy[y][x]='-';died++;}
            if (grid[y][x]=='-' && occ(x,y)==3)
                {gridCopy[y][x]='*';born++;}
        }
    }

    for (y=0; y<10; y++) {
        for (x=0; x<10; x++) {
            grid[y][x]=gridCopy[y][x];
        }
    }
}

// You probably are going to do this often in the future
// Make a function for it instead of having the same code often
void printGrid() {
    int x, y;
    for (y = 0; y < 10; y++) {
        for (x = 0; x < 10; x++)
            printf("%c", grid[y][x]);
        printf("\n");
    }
}


/*
 * 
 */
int main(int argc, char** argv) {

    int x, y, answ;

    // Setting entire grid to '-'
    for (y = 0; y < 10; y++)
        for (x = 0; x < 10; x++)
            grid[y][x] = '-'; // No need for a variable, just use '-' directly

    // Printing out grid
    printf("This is the original array\n");
    printGrid();

    // Setting initial state
    grid[5][4] = '*'; // No need for a variable, just use '*' directly
    grid[5][5] = '*'; 
    grid[5][6] = '*'; 
    grid[4][4] = '*'; grid[3][4] = '*';
    grid[4][6] = '*'; grid[3][6] = '*';

    // Printing out grid
    printf("This is the new array\n");
    printGrid();

    //printf("The number of neighbors is: %d\n", occ(3, 3));

    generation=0;
    answ=1;

    while(answ)
    {
        generateNext();
        printf("\n\nGeneration number %d",generation);
        printf("\nBorn=%d,Died=%d\n",born,died);
        printGrid();
        printf("\nPrint next generation-1,Exit-0:");
        scanf("%d",&answ);
    }

    return (EXIT_SUCCESS);
}

Answer 2

计数类似于打印或复制整个网格所必须执行的操作：使用相同的循环，但在每个单元格中，根据其中的内容添加到适当的计数器。

或者，如果保留死细胞和活细胞的数量，您可以在更新网格时更新它们：如果细胞发生变化，则在细胞变得活跃或死亡时适当地更新计数器。如果您需要报告每轮出生人数和死亡人数，也可以使用此选项。