Hello. I'm having troubles with getting a program to work. It has to return how many steps is takes to get out of the labyrinth or a sentence "You won't escape". The user types in: sizes of the labyrinth x * y, the positions of start and exit and the labyrinth itself ( where -5 is start/exit, -1 is a wall and 0 is a corridor). For example: 3 5 0 0 2 4 -5 0 0 0 0 0 -1 -1 -1 0 0 0 0 0 -5 The program should return: 6 I have written a program. It uses a FIFO queue. It seems to work but when i typed: 6 20 5 2 2 17 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 -1 0 0 0 0 0 0 -1 -1 -5 0 0 0 -1 0 0 -1 0 0 0 0 -1 0 0 -1 0 0 0 0 -1 0 0 0 0 0 0 0 0 -1 0 0 -1 0 0 0 -1 0 0 0 0 0 0 0 0 -5 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 The program has crushed. Compiling shows no errors. Typed in: 5 5 2 4 4 2 -1 0 0 0 0 -1 -1 -1 -1 0 0 -1 -1 -1 -5 -1 0 0 0 0 -1 -1 -5 0 -1 Program returned 6, while it should 4. Something's wrong. Any help would be appreciated. Code: #include <stdio.h> #include <cstdlib> struct FIFOQueue { int** buffer; int begin, end; FIFOQueue(int n) { buffer = (int**) malloc(sizeof(int*)*n); } void in(int* e) { *(buffer + end) = e; end++; } int* out() { return *(buffer + (begin++)); } void flush() { begin = 0, end = 0; } }; int main() { int n, m; // sizes of the labyrinth int **t; scanf("%d %d", &n, &m); n=n+2; m=m+2; t = (int**) malloc(n*sizeof(int*)); for(int i=0;i<n;i++) { *(t+i) = (int*) malloc(m*sizeof(int)); } // surrounding the labyrinth with -1s for(int i=0;i<m;i++) { *(*(t+0)+i) = -1; *(*(t+n-1)+i) = -1; } for(int i=0;i<n;i++) { *(*(t+i)+0) = -1; *(*(t+i)+m-1) = -1; } // getting the positions of start & exit int s1, s2, e1, e2; scanf("%d %d %d %d", &s1, &s2, &e1, &e2); s1 = s1+1; s2 = s2+1; e1 = e1+1; e2 = e2+1; // checking if start/exit outside the labyrinth if(e1 < 0 or e2 < 0 or e1 > n-2 or e2 > m-2 or s1 < 0 or s2 < 0 or s1 > n-2 or s2 > m-2) { printf("You won't escape"); } else { // getting the labyrinth for(int i=1;i<n-1;i++) { for(int j=1;j<m-1;j++) { scanf("%d", *(t+i)+j); } } // creating the queue FIFOQueue queue(n * m); queue.flush(); // adding the start to the queue int *start = *(t+s1)+s2; *start = -4; queue.in(start); int counter = 0; int *equidistance = start; //pointer which says where the zone of equal distances ends while(1) { if(queue.end - queue.begin > 0) // if queue not empty { int* element = queue.out(); *element = 2; int* left = element - 1; int* right = element + 1; int* up = element - m; int* down = element + m; if(*left == -5 or *right == -5 or *up == -5 or *down == -5) // if exit break { counter++; printf("%d", counter); break; } if(*left != 2 or *left != -1) // checking if a corridor or exit and aading to the queue { queue.in(left); } if(*right != 2 or *right != -1) { queue.in(right); } if(*up != 2 or *up != -1) { queue.in(up); } if(*up != 2 or *down != -1) { queue.in(down); } if(element == equidistance) // increasing the distance { equidistance = *(queue.buffer + queue.end - 1); counter++; } } else { // if the queue is empty printf and break; printf("You won't escape"); break; } } } return 0;
This program would be easier to read if you used array semantics instead of pointers. So replace this: Code: *(*(t+0)+i) = -1; with whichever of the following two statements is correct: Code: t[0][i] = -1; t[i][0] = -1; and repeat throughout the program. Don't do this: Code: if(e1 < 0 or e2 < 0 or e1 > n-2 or e2 > m-2 or... Just learn that || means OR and use || instead of some dumb #define that makes the program look more like Pascal or whatever (because it doesn't, and just ends up making your code harder to read). This code is perfectly readable: Code: if(e1 < 0 || e2 < 0 || e1 > n-2 || e2 > m-2 ||... If your program has crashed then there's a good chance you've stuffed up your memory usage. So for example you might have confused: Code: int t[6][20]; with Code: int t[20][6]; If you really want to manage your arrays at a low level instead of using [][] syntax then just allocate 6*20 ints and address them with x+y*20 syntax instead of all this *(t+this)+that+(*)the**(*)+other*(*) or whatever shit you have to type to do what the compiler does automatically for you when you do t[a].
