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我确信这个错误很明显,但我确实很难找到它。

基本上我正在尝试通过二维数组制作棋盘。我正在通过 8 个皇后测试来测试它的功能……它不起作用。

不知何故,我的整数值之一变得异常,如图gdb所示:

....
(gdb) cont
Continuing.

Program received signal SIGSEGV, Segmentation fault.
0x080487f8 in diagnols (PARAMETER_ONE=0, PARAMETER_TWO=0) at eq1.c:77
77            if (key->board[abc][bcd] == 1) {
(gdb) print abc
4424  // "SHOULD BE" ONE
(gdb) print bcd
4424  // "SHOULD BE" ONE
(gdb) backtrace
#0  0x080487f8 in diagnols (PARAMETER_ONE=0, PARAMETER_TWO=0) at eq1.c:77
#1  0x08048873 in check (param1=0, param2=0) at eq1.c:91
#2  0x08048510 in recur (DEPTH=0, WIDTH=0) at eq1.c:99
#3  0x08048919 in main () at eq1.c:152
(gdb) 

然后是diagnols(...),位于:

int recur(struct chessboard* key, int DEPTH, int WIDTH) {
    /* other functions above diagnols(...) */

诊断(...):

int diagnols(int PARAMETER_ONE, int PARAMETER_TWO) { // returns 0 if good    

    int abc = 0;
    int bcd = 0;
    int counter = 0; // keeps track of conflicting piece occurrences 

    // OTHER CHECKS FIRST... DELETED TO SAVE ROOM

    // checkign diagnol down and to the left
    abc = PARAMETER_ONE+1;
    bcd = PARAMETER_TWO-1;
    while (  (abc>=0)&&(bcd>=0)  ) {
      if (key->board[abc][bcd] == 1) {
        counter++;
      } abc++;
      bcd--;
    }

    // ERROR IN THIS PART
    // checking diagnol down and to the right
    abc = PARAMETER_ONE+1;
    bcd = PARAMETER_TWO+1;
    while (  (abc>=0)&&(bcd>=0)  ) {
      if (key->board[abc][bcd] == 1) { // ERROR
        counter++;
      } abc++;
      bcd++;
    }

    return counter;
  }

并在以下函数中diagnols(...)调用:recur(...)

  int check(int param1, int param2) { // if okay returns 2 
    // other functions
    d = diagnols(param1, param2);
    int total = 0;
    total = (h + v + d); // if okay, equals 2
    return total;
  }

为了更好地衡量,这是我的结构:

struct chessboard {
  int board[7][7];
};

主要:

int main() {
  struct chessboard* master = malloc(sizeof(struct chessboard));
  /* i set the board to zero here. used calloc() before */
  recur(master, 0, 0);
  // stuff
}

是的,我意识到诊断不是拼写对角线;)

4

1 回答 1

2
while (  (abc>=0)&&(bcd>=0)  ) {
      if (key->board[abc][bcd] == 1) { // ERROR
        counter++;
      } abc++;
      bcd++;
    }

由于您正在增加两个索引,因此该条件似乎将始终为真。

您的意思是 < 一些限制吗?

于 2013-04-02T23:05:50.233 回答