我试图编写一个完美的迷宫生成器,但由于递归导致迷宫太大时出现segfult,所以代码中几乎没有问题。以下是代码的主要部分:
t_maze *init_maze(int w, int h)
{
t_maze *maze;
int j;
int i;
if ((maze = malloc(sizeof(t_maze))) == NULL)
return (NULL);
maze->w = w;
maze->h = h;
if ((maze->cells = malloc(sizeof(char *) * maze->h)) == NULL)
return (NULL);
j = -1;
while (++j < maze->h)
{
if ((maze->cells[j] = malloc(sizeof(char) * maze->w)) == NULL)
return (NULL);
i = -1;
while (++i < maze->w)
maze->cells[j][i] = (j % 2 == 1 || i % 2 == 1) ? (1) : (0);
}
return (maze);
}
void detect_neighbours(t_maze *maze, char *neighbours, int x,
int y)
{
int i;
// I fill the array with 1 (means there is no neighbours)
// If there is a neighours, I set the cell to 0
// In this order: Top, right, bottom, left
i = -1;
while (++i < 4)
neighbours[i] = 1;
if (y - 2 >= 0 && x >= 0 && y - 2 < maze->h
&& x < maze->w && maze->cells[y - 2][x] == 0)
neighbours[0] = 0;
if (x + 2 >= 0 && x + 2 < maze->w && y >= 0 && y < maze->h
&& maze->cells[y][x + 2] == 0)
neighbours[1] = 0;
if (y + 2 < maze->h && y + 2 >= 0 && x >= 0
&& x < maze->w && maze->cells[y + 2][x] == 0)
neighbours[2] = 0;
if (x - 2 >= 0 && x - 2 < maze->w && y >= 0 && y < maze->h
&& maze->cells[y][x - 2] == 0)
neighbours[3] = 0;
}
int there_is_no_neighbours(char *neighbours)
{
int i;
// this function returns 0 if there is at least 1 neigbours
i = -1;
while (++i < 4)
if (neighbours[i] == 0)
i = 41;
if (i == 42)
return (0);
return (1);
}
void set_maze_protected(t_maze *maze, int y, int x, int val)
{
// To prevent segfault when I put values in the maze,
// I check the x and y keys
if (x >= 0 && y >= 0 && x < maze->w && y < maze->h)
maze->cells[y][x] = val;
}
int build_maze(t_maze *maze, int x, int y)
{
char neighbours[4];
int i;
int ret;
ret = 0;
detect_neighbours(maze, neighbours, x, y);
if (there_is_no_neighbours(neighbours) == 1)
return (0);
i = rand() % 4;
while (neighbours[i] == 1)
i = rand() % 4;
if (i == 0)
{
set_maze_protected(maze, y - 1, x, 2);
set_maze_protected(maze, y - 2, x, 2);
ret = build_maze(maze, x, y - 2);
}
if (i == 1)
{
set_maze_protected(maze, y, x + 1, 2);
set_maze_protected(maze, y, x + 2, 2);
ret = build_maze(maze, x + 2, y);
}
if (i == 2)
{
set_maze_protected(maze, y + 1, x, 2);
set_maze_protected(maze, y + 2, x, 2);
ret = build_maze(maze, x, y + 2);
}
if (i == 3)
{
set_maze_protected(maze, y, x - 1, 2);
set_maze_protected(maze, y, x - 2, 2);
ret = build_maze(maze, x - 2, y);
}
while (ret != 0)
ret = build_maze(maze, x, y);
return (1);
}
int main()
{
t_maze *maze;
int w;
int h;
w = 50;
h = 50;
srand(time(NULL) * getpid());
if ((maze = init_maze(w, h)) == NULL)
return (1);
maze->cells[0][0] = 2;
build_maze(maze, 0, 0);
// display_maze shows values in the 2D array (maze->cells)
display_maze(maze);
return (0);
}
我用这个调用在main中调用这个函数:
build_maze(maze, 0, 0);
该函数检测的是该单元格有邻居,如果有邻居,则该函数随机调用其中一个邻居,并打开两个邻居之间的门。
例如,如果x和y参数大于2500,则它将断开。(如果小于2500,效果会很好)
怎么解决这个问题?
我学会了尾声,但我忽略了在这种情况下如何实现尾声,
谢谢您,
致意
最佳答案
您可以增加堆栈大小。
在POSIX系统上,可以使用以下代码。
#include<stdio.h>
#include <sys/resource.h>
#define required_stack_size 0x8000000 // change this to the stack size you need
int main (int argc, char **argv)
{
struct rlimit rl;
int result;
if((result = getrlimit(RLIMIT_STACK, &rl)) < 0)
{
fprintf(stderr, "getrlimit returned result %d\n", result);
return -1;
}
if(rl.rlim_cur<required_stack_size)
{
rl.rlim_cur = required_stack_size;
if((result = setrlimit(RLIMIT_STACK, &rl)) < 0)
{
fprintf(stderr, "setrlimit returned result = %d\n", result);
return -1;
}
}
//the rest code
return 0;
}