有段代码，调试后的结果怎么都理解不了！求解惑

[ImAmelie]

这是自己写8皇后问题代码
调试的时候有一段操作的返回值怎么都理解不了
问题我用注释的形式说明了！

#include <stdio.h>
#define block(a,b,c) change(a,b,2,c)
#define unblock(xy) rechange(xy)
typedef struct
{
	int i;
	int j;
}XY;
void queen(int, int);
void change(int m, int n, int num, XY *p);
void rechange(XY *p);
int matrix[8][8] = { 0 };
int num = 0;
int main()
{
	queen(8, 0);
	return 0;
}
void queen(int n,int i)
{
	if (n == 0)
	{
		num++;
		printf("第%d种解法\n", num);
		for (int i = 0; i < 7; i++)
		{
			for (int j = 0; j < 7; j++)
			{
				if (matrix[i][j] == 0 || matrix[i][j] == 2)
					putchar(' ');
				else
					putchar('*');
			}
			putchar('\n');
		}
		return;
	}
	for (int j = 0; j != 8; j++)
	{
		if (matrix[i][j] == 0)
		{
			XY xy[13] = { 0 };
			matrix[i][j] = 1;//在这里设置断点1
			printf("%d\n%d\n", n, i);//n=8 i=0
			block(i, j, xy);//执行完此语句n和i的值会变化？？？？
			//宏展开change(i,j,2,xy);
			printf("%d\n%d\n", n, i);//n=6 i=7   问题出在这里，change函数是怎么改变n和i的值的？
			queen(n - 1, i + 1);//在这里设置断点2
			matrix[i][j] = 0;
			unblock(xy);
		}
	}
}
void change(int m, int n, int num,XY *p)
{
	int i, j;
	//上
	for (i = m - 1, j = n; i != -1; i--)
	{
		if (matrix[i][j] == 2)
			break;
		matrix[i][j] = num;
		p->i = i;
		p->j = j;
		p++;
	}
	//下
	for (i = m + 1, j = n; i != 8; i++)
	{
		if (matrix[i][j] == 2)
			break;
		matrix[i][j] = num;
		p->i = i;
		p->j = j;
		p++;
	}
	//左
	for (i = m, j = n - 1; j != -1; j--)
	{
		if (matrix[i][j] == 2)
			break;
		matrix[i][j] = num;
		p->i = i;
		p->j = j;
		p++;
	}
	//右
	for (i = m, j = n + 1; j != 8; j++)
	{
		if (matrix[i][j] == 2)
			break;
		matrix[i][j] = num;
		p->i = i;
		p->j = j;
		p++;
	}
	//左上
	for (i = m - 1, j = n - 1; i != -1 && j != -1; i--, j--)
	{
		if (matrix[i][j] == 2)
			break;
		matrix[i][j] = num;
		p->i = i;
		p->j = j;
		p++;
	}
	//左下
	for (i = m + 1, j = n - 1; i != 8 && j != -1; i++, j--)
	{
		if (matrix[i][j] == 2)
			break;
		matrix[i][j] = num;
		p->i = i;
		p->j = j;
		p++;
	}
	//右上
	for (i = m - 1, j = n + 1; i != -1 && j != 8; i--, j++)
	{
		if (matrix[i][j] == 2)
			break;
		matrix[i][j] = num;
		p->i = i;
		p->j = j;
		p++;
	}
	//右下
	for (i = m + 1, j = n + 1; i != 8 && j != 8; i++, j++)
	{
		if (matrix[i][j] == 2)
			break;
		matrix[i][j] = num;
		p->i = i;
		p->j = j;
		p++;
	}
	p->i = -1;
	p->j = -1;
}
void rechange(XY *p)
{
	for (; p->i != -1; p++)
		matrix[p->i][p->j] = 0;
}

[ImAmelie]

给你发私信了，请查收！

引用 2 楼 zhao4zhong1 的回复:

[赵4老师]

“给定一个小点的输入，完整单步跟踪(同时按Alt+7键查看Call Stack里面从上到下列出的对应从里层到外层的函数调用历史)一遍。”是理解递归函数工作原理的不二法门！ 递归函数关注以下几个因素 ·退出条件 ·参数有哪些 ·返回值是什么 ·局部变量有哪些 ·全局变量有哪些 ·何时输出 ·会不会导致堆栈溢出

[赵4老师]

仅供参考

#include <stdio.h>
int d[8][8];
int n;
void queen8(int L) {
    int x,y;

    if (8==L) {
        n++;
        printf("%d\n",n);
        for (y=0;y<8;y++) {
            for (x=0;x<8;x++) {
//              printf("%d",d[y][x]);
                if (d[y][x]) printf("■");
                else         printf("□");
            }
            printf("\n");
        }
        return;
    }
    for (x=0;x<8;x++) {
        for (y=0;y<L;y++) {
            if  (              d[y][x      ]
             || (x-(L-y)>=0 && d[y][x-(L-y)])
             || (x+(L-y)< 8 && d[y][x+(L-y)]))
                break;
        }
        if (y>=L) {
            d[L][x]=1;
            queen8(L+1);
            d[L][x]=0;
        }
    }
}
int main() {
    int y,x;
    for (y=0;y<8;y++) {
        for (x=0;x<8;x++) {
            d[y][x]=0;
        }
    }
    queen8(0);
    return 0;
}

[baidu_18258203]

p溢出鸟..