Question

我在创建concat();函数时遇到了问题，但是当我继续并释放列表时，我收到malloc错误...我很肯定我正在释放我的列表正确地说，当我不运行我的concat();函数时，列表会被正确释放。我有理由相信我在这个函数中做错了...

typedef struct node {
    ElemType val;
    struct node *next;
} NODE;


struct list_struct {
    NODE *front;
    NODE *back;
};


   void lst_concat(LIST *a, LIST *b) {
if( !( a->front && b->front)) return ;

if (a->front == NULL && b->back == NULL) {

    return;
}

if (a->front == NULL && b->front != NULL) {
    a->front = b->front;
    return;
}
else{
NODE *tmp = a->front;
    if( a->back == NULL )return;
    if( a->front == NULL )return;
    if( b->back == NULL )return;
    if( b->front == NULL )return;
    else{
    NODE *tempPtr = a->back;

    tempPtr->next=b->front;
    tempPtr = b->front;
    a->back = b->back;
    }
}}
void lst_free(LIST *l) {
NODE *p = l->front;
NODE *pnext;

  while(p!= NULL) {
    pnext = p->next;   // keeps us from de-referencing a freed ptr
    free(p);
    p = pnext;
  }
  // now free the LIST 
  free(l);
}

**更新，这个更新列表正确，但是，我仍然可以释放两个列表，或者我可以只加一个，因为它们已加入？我的lst_free()在上面。

Answer 1

这是我能想象的唯一理性方式（到目前为止）......

typedef struct NODE         NODE;
typedef struct list_struct  LIST;


typedef struct node {
    ElemType    val;
    NODE        *next;
} NODE;


struct list_struct {
    NODE        *front;
    NODE        *back;
};


// get the ending node
NODE *LastNode(NODE *root){
    NODE *tmp = NULL;
    while(root){
        tmp  = root;
        root = root->next;
   }
   return tmp;
}


// this will move node from pnode
// and put it at the end of proot nodes
void move_node(NODE **proot,NODE **pnode){

    if(! (pnode && proot) )
        return; // no comments

    if(pnode == proot)
        return; // no comments

    NODE *node = *pnode;
    NODE *root = *proot;
    NODE *tmp;

    if(!node) return;

    if(!root){
        // this node becomes the first one
        *proot      = node;
    }else{
        // this node becomes the last node
        LastNode(root)->next = node;
    }

    // now nodes blongs to other node
    // set this to NULL
    *pnode = NULL;

}


// this will move a nodes to b;
// 
void lst_concat(LIST *a, LIST *b) {
    move_node(&a->front, &b->front);
    move_node(&a->back, &b->back);
    // at this point b has no front, and no back
    // all has been moved to a

}


void free_nodes(NODE **proot){
    if ( !(proot && *proot) )
        return;

    NODE *root = *proot;

    if( root->next)
        free_nodes(&root->next);

    free(root);
    *proot=NULL;
}

void lst_free(LIST *l) {
    if( l ) {
        free_nodes( &l->front);
        free_nodes( &l->back);
        free(l);
    }
}

Answer 2

您在知道a这里是

之前解除引用null

void concat(LIST *a, LIST *b)
{
    if (a->front == NULL) { // !!!! what if a is NULL? Crash...
        a->front = b->front; 
        a->back = b->back;
        return;
    }

首先检查a NULL。

在知道b是否null之前，你也取消引用void concat(LIST *a, LIST *b) { if (a->front == NULL) { a->front = b->front; // !!!! what if b is NULL? Crash... a->back = b->back; return; }：

void concat(LIST *a, LIST *b)
{
    if (a == NULL)  // This needs to go first, i.e. before dereferencing a
    {
         // some code
    }

    // other code ....
}

至少做：

TFramedTransport

C - 链接列表连接运行到malloc错误

2 个答案: