【问题标题】:C : Get a segmentation fault when generating random mazeC:生成随机迷宫时出现分段错误
【发布时间】:2017-11-25 21:04:34
【问题描述】:

我尝试随机生成一个迷宫,但是在编译下面的程序时出现分段错误

这是代码

void spread(int v, int x, int y, int *t,int w, int *count){
    //table of directions(Right,Down,Left,Up)
    int d[4][2] = {{1,0},{0,1},{-1,0},{0,-1}}; 
    int i;
    t[y * w + x] = v;

    if(v == 0 && (x & 1) && (y & 1))
        (*count)++; /*increments at every box which has value as zero with x odd and y also*/

    //spread the value to all directions
    for(i = 0; i < 4; i++){
        if(v < t[(y+d[i][1]) * w + x +d[i][0]]){
            spread(v,x + d[i][0],y+d[i][0],t,w,count);
        }
    }
}

int *init(int m, int n){
        //Initializing the maze
        int *t = NULL, mp = 2 * m +1, np = 2 * n + 1;
        int x,y,k,d;
        int count = 1;
        t = malloc(mp * np * sizeof *t);
        assert(t);
        for(y = k = 0; y < np ;++y){
            for(x = 0; x < mp; ++x){
                if((x & 1) && (y & 1))
                    t[y * mp + x] = k++;
                else
                    t[y * mp + x] = -1; 
            }

        }
        //Make a labyrinth randomly
        while(count < (m * n)){
            srand(time(NULL));
            if(myRand(2)){ // Up/Down separator 
                do{
                    x = myRand(m) * 2 + 1;
                    y = (myRand(n - 1) + 1) * 2;

                }while(t[(y - 1) * mp + x] == t[(y + 1) * mp + x]); /*Don't select the ones which are equal*/

                d = t[(y - 1) * mp + x] - t[(y + 1) * mp + x];
                //d selects the lowest one
                if(d > 0){
                    t[y * mp +x] = t[(y + 1) * mp + x];
                    spread(t[(y + 1) * mp +x],x,y-1,t,mp,&count);
                }

                else if(d < 0){
                    t[y * mp +x] = t[(y - 1) * mp + x];
                    spread(t[(y - 1) * mp +x],x,y+1,t,mp,&count);
                }   
            }
            else{   //Right/Left separator
                do{         
                    x = (myRand(m - 1) + 1) * 2;
                    y = myRand(n) * 2 + 1;

                }while(t[y * mp + x - 1] == t[y * mp + x + 1]);

                d = t[y * mp + x - 1] - t[y * mp + x + 1];

                if(d > 0){
                    t[y * mp +x] = t[y * mp + x + 1];
                    spread(t[y * mp + x + 1],x-1,y,t,mp,&count);
                }

                else if(d < 0){
                    t[y * mp +x] = t[y * mp +x - 1];
                    spread(t[y * mp + x - 1],x+1,y,t,mp,&count);
                }   
            }
        }   
        return t;
}

迷宫首先被初始化(这里是 t),其值为:-1 表示墙壁,v > 0 表示节点

然后为了在节点之间建立连接(这是为了有一个迷宫),随机选择一个框,但是行或列必须是奇数才能选择代表分隔符的墙(上/下或右/左)

所以“d”取墙壁周围盒子之间的最小值(表示墙壁上方盒子的值和墙壁下方盒子的值之间的最小值,与右/左相同)

函数spread:将一个值传播到所有方向(右、下、左、上)

这是一个例子:

+  +  +  +  +  +  +  +  +  
+  0  +  1  +  2  +  3  +  
+  +  +  +  +  +  +  +  +  
+  4  +  5  +  6  +  7  +  
+  +  +  +  +  +  +  +  +  
+  8  +  9  +  10 +  11 +  
+  +  +  +  +  +  +  +  + 

将某些价值观传播给其他人时

+  +  +  +  +  +  +  +  +  
+  0  +  1  1  1  +  3  +  
+  +  +  +  +  +  +  +  +  
+  4  +  5  +  6  +  7  +  
+  +  +  +  +  +  +  +  +  
+  8  +  9  9  9  +  11 +  
+  +  +  +  +  +  +  +  + 

我试图通过调试程序来解决问题,它工作了一会儿我的意思是显示上面的内容,但后来我得到了这个

Program received signal SIGSEGV, Segmentation fault.
0x00000000004008e4 in spread (
   v=<error reading variable: Cannot access memory at address 0x7fffff7fefec>, x=<error reading variable: Cannot access memory at address 0x7fffff7fefe8>, 
    y=<error reading variable: Cannot access memory at address 0x7fffff7fefe4>, t=<error reading variable: Cannot access memory at address 0x7fffff7fefd8>, 
    w=<error reading variable: Cannot access memory at address 0x7fffff7fefe0>, count=<error reading variable: Cannot access memory at address 0x7fffff7fefd0>)
    at Lab.c:34
#1  0x0000000000400a1a in spread (v=8, x=4, y=6, t=0x603010, w=9, 
    count=0x7fffffffddd4) at Lab.c:48
#2  0x0000000000400a1a in spread (v=8, x=4, y=6, t=0x603010, w=9, 
    count=0x7fffffffddd4) at Lab.c:48
#3  0x0000000000400a1a in spread (v=8, x=4, y=6, t=0x603010, w=9, 
    count=0x7fffffffddd4) at Lab.c:48
#4  0x0000000000400a1a in spread (v=8, x=4, y=6, t=0x603010, w=9, 
    count=0x7fffffffddd4) at Lab.c:48

【问题讨论】:

  • 怀疑无限递归
  • 什么保证y * w + xt 的范围内?
  • x 不能超过 mp(y 也不能超过 np)所以我认为 y * w + x 也不能超过 mp * np 的值
  • 为什么你认为它超出了t的限制
  • 如果我理解得很好,在spread函数中,如果v != 0,你跳过第一个if语句,进入for循环。然后,如果当前块旁边的任何块(与值 v 相关联)与小于或等于 v 的值相关联,则使用新的当前块(与前一个块相邻)输入 spread 函数. BUT 由于您不评估 xy*count 的值,因此您可能会:或用完数组的限制,或者再次转轮并再次直到出现Stack Overflow。 (我完全不确定我在说什么)。有人能告诉我我是对还是错吗?

标签: c segmentation-fault maze


【解决方案1】:
void spread(int v, int x, int y, int *t,int w, int *count){
    //table of directions(Right,Down,Left,Up)
    int d[4][2] = {{1,0},{0,1},{-1,0},{0,-1}}; 
    int i;
    t[y * w + x] = v;

    if(v == 0 && (x & 1) && (y & 1))
        (*count)++; /*increments at every box which has value as zero with x odd and y also*/

    //spread the value to all directions
    for(i = 0; i < 4; i++){
        if(v < t[(y+d[i][1]) * w + x +d[i][0]]){
            spread(v,x + d[i][0],y+d[i][1],t,w,count);
        }
    }
}

int *init(int m, int n){
        //Initializing the maze
        int *t = NULL, mp = 2 * m +1, np = 2 * n + 1;
        int x,y,k,d;
        int count = 1;
        t = malloc(mp * np * sizeof *t);
        assert(t);
        for(y = k = 0; y < np ;++y){
            for(x = 0; x < mp; ++x){
                if((x & 1) && (y & 1))
                    t[y * mp + x] = k++;
                else
                    t[y * mp + x] = -1; 
            }

        }
        //Make a labyrinth randomly
        srand(time(NULL));
        while(count < (m * n)){

            if(myRand(2)){ // Up/Down separator 
                do{
                    x = myRand(m) * 2 + 1;
                    y = (myRand(n - 1) + 1) * 2;

                }while(t[(y - 1) * mp + x] == t[(y + 1) * mp + x]); /*Don't select the ones which are equal*/

                d = t[(y - 1) * mp + x] - t[(y + 1) * mp + x];
                //d selects the lowest one
                if(d > 0){
                    t[y * mp +x] = t[(y + 1) * mp + x];
                    spread(t[(y + 1) * mp +x],x,y-1,t,mp,&count);
                }

                else if(d < 0){
                    t[y * mp +x] = t[(y - 1) * mp + x];
                    spread(t[(y - 1) * mp +x],x,y+1,t,mp,&count);
                }   
            }
            else{   //Right/Left separator
                do{         
                    x = (myRand(m - 1) + 1) * 2;
                    y = myRand(n) * 2 + 1;

                }while(t[y * mp + x - 1] == t[y * mp + x + 1]);

                d = t[y * mp + x - 1] - t[y * mp + x + 1];

                if(d > 0){
                    t[y * mp +x] = t[y * mp + x + 1];
                    spread(t[y * mp + x + 1],x-1,y,t,mp,&count);
                }

                else if(d < 0){
                    t[y * mp +x] = t[y * mp +x - 1];
                    spread(t[y * mp + x - 1],x+1,y,t,mp,&count);
                }   
            }
        }   
        return t;
}

【讨论】:

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