c - 调试简单的c代码，双链表复制

Question

[我的文件在这里(链接已删除)]

我的复制功能不会复制整个列表。

根据@Mat的要求，这里是“复制”功能，但您仍然需要阅读文件的其余部分来理解我的结构。

Line copyLineText(Line l){
  Line newl = malloc (sizeof( struct node ));
  checkMalloc(newl,"copyLineText");
  newl->length = l->length;
  newl->buffer = malloc((newl->length)* sizeof(char));
  checkMalloc(newl->buffer,"copyLineText buffer");
  strncpy(newl->buffer, l->buffer, newl->length);
  return newl;
}
/* Copy the lines between and including 'from' and 'to' of the textbuffer
 * 'tb'.
 *
 * - The result is a new textbuffer (much as one created with newTB()).
 * - The textbuffer 'tb' will remain unmodified.
 * - The program is to abort() with an error message if 'from' or 'to' is out
 *   of range. 
 */
TB copyTB (TB tb, int from, int to){
    if (from > to || from < 0 || to >= tb->numLines){
        printf("Error: the from to is out of range\n");
        abort();
    }
    Line line = tb->begin;
    while(line->linePosition != from){
        line = line->next;
    }

    TB tbCopy = malloc (sizeof( struct textbuffer ));

    Line last = NULL, curr = line, currCopy;

    while( curr != NULL && (curr->linePosition != to + 1) ){
        currCopy = copyLineText(curr);

        if(last == NULL){
          tbCopy->begin  = currCopy;
        }
        currCopy->prev = last;
        if(last != NULL){
            last->next = currCopy;
            //printf("362debug: last->next = currCopy\n");
        }
        last = curr;
        //printf("364debug: %d\n",curr->linePosition);
        curr = curr->next;
    }

        currCopy->next = NULL;
        tbCopy->end  = currCopy;
        tbCopy->numLines  = to - from + 1;
        return tbCopy;
}

如果你运行我的代码，你会看到:

tb2:2 lines
abcde
fg


tb2copy:2 lines
abcde

通过这个简单的测试，我的函数生成的副本比原始结构少了一行。

Answer 1

在 Question.c 的第 161 行，当您复制链表时，您将原始链表中的元素分配给last，而不是分配新创建列表中的上一个。

将该行更改为:

last = curr;

至:

last = currCopy;