在OS X上的合并排序实现中的总线错误10

Question

我想在C中实现合并排序。我编写的代码适用于100,000个数字的列表，但是当我在1,000,000列表上运行它时， “总线错误：10”。

错误发生在我评论过“BUS ERROR HERE”的地方。发生错误时，tmp_list_i == 65920和pws-> merge_cursor == 32776.函数merge会合并任意数量的子数组，因为我也使用它来合并由不同线程排序的子数组。但即使我只使用单个线程时（即，一次只需要合并两个子阵列），总线错误也正在发生。

任何想法？

// Represents a sub-array in the list. 
typedef struct 
{ 
    int begin_i; // inclusive 
    int end_i; // exclusive 
    int already_sorted; // if the partition was sorted before runtime 
    pthread_t tid; // thread associated with this partition, if any 
    int merge_cursor; // index used for merging 
} Partition; 

// O(n log(n)) 
// n = number of comparisons in a merge 
// log(n) = number of merges 
void* merge_sort(void* partition) 
{ 
    Partition* part = (Partition*) partition; 

    // Base case. One item, so partition is sorted 
    int len = part->end_i - part->begin_i; 
    if (len < 2) 
    { 
     part->already_sorted = TRUE; 
     return 0; 
    } 

    // Recursion 
    Partition left_part; 
    left_part.begin_i = part->begin_i; 
    left_part.end_i = part->begin_i + (len/2); 
    left_part.merge_cursor = left_part.begin_i; 

    Partition right_part; 
    right_part.begin_i = part->begin_i + (len/2); 
    right_part.end_i = part->end_i; 
    right_part.merge_cursor = right_part.begin_i; 

    merge_sort(&left_part); 
    merge_sort(&right_part); 

    if (left_part.already_sorted && right_part.already_sorted) 
     part->already_sorted = TRUE; 

    // Create parts array to pass to merge 
    Partition* parts[] = {&left_part, &right_part}; 

    if (merge(parts, 2, len) == FALSE) 
     part->already_sorted = FALSE; 

    return 0; 
} 

// O(n) but more specifically O(n * p + n) where p is num_parts 
int merge(Partition* parts[], int num_parts, int total_num) 
{ 
    int already_sorted = TRUE; // whether the partitions were already sorted 

    int tmp_list[total_num]; 
    int tmp_list_i; 
    for (tmp_list_i = 0; tmp_list_i < total_num; tmp_list_i++) 
    { 
     // find (P)artition (W)ith (S)mallest number under its merge cursor 
     Partition* pws = NULL; 

     int parts_i; 
     for (parts_i = 0; parts_i < num_parts; parts_i++) 
     { 
      Partition* this_part = parts[parts_i]; 

      if (this_part->merge_cursor == MERGE_CURSOR_DONE) 
       continue; 

      if (pws == NULL) 
       pws = this_part; 

      int this_part_num = list[this_part->merge_cursor]; 
      int smallest_part_num = list[pws->merge_cursor]; 

      if (this_part_num < smallest_part_num) 
      { 
       pws = this_part; 
       already_sorted = FALSE; 
      } 
     } 

     // add the smallest of the numbers to current spot in tmp array 
     tmp_list[tmp_list_i] = list[pws->merge_cursor]; // BUS ERROR HERE 

     // increment the merge cursor for pws and set to NULL if done 
     (pws->merge_cursor)++; 
     if (pws->merge_cursor == pws->end_i) 
      pws->merge_cursor = MERGE_CURSOR_DONE; 
    } 

    // Copy back to list from tmp_list. Costs an extra n. 
    int list_i = parts[0]->begin_i; // start where we should in list 
    for (tmp_list_i = 0; tmp_list_i < total_num; tmp_list_i++) 
    { 
     list[list_i] = tmp_list[tmp_list_i]; 
     list_i++; 
    } 

    return already_sorted; 
} 

编辑： 当在堆上分配，而不是堆的一切，我得到一个不同的问题。分配int this_part_num = list[this_part->merge_cursor];似乎并没有被正确评估，最终我获得了SIG故障：

141    int this_part_num = list[this_part->merge_cursor]; 
(gdb) s 
142    int smallest_part_num = list[pws->merge_cursor]; 
(gdb) print this_part_num 
$5 = 1 
(gdb) print list[this_part->merge_cursor] 
$6 = 6 

Answer 1

想通了。列表在单独的文件中声明为int* list，但在merge_sort函数为extern int list[]的文件中声明。