C 指向结构数组并获取内存的指针

链接是指向节点的指针

typedef struct node * link;

在main（）中，我有以下代码（config->m只是一个整数）：

但是我的友好邻域编译器告诉我赋值中的

不兼容类型

     int i;
     // we give the first node a val of zero
     na[0].val = 0;
     // and a null next pointer
     na[0].next = 0 ;

我知道我可以用

struct node heads[config->m];

但我想用指针来做这件事

和往常一样，有人会问我这是否是家庭作业的一部分，答案是肯定的（有点）。但这段特定的代码与实际的赋值无关；这是为了我自己的启示。但是谢谢你的提问：|

link heads[config->m]; 
link buffer = malloc(sizeof(struct node) * config->m);

for(i = 0; i < config->m; i++)
  heads[i] = &buffer[i];

....
free(buffer);

现在我们添加我们的include和config->m：

  #include <stdio.h>
  #include <stdlib.h>
  // your config->m
  const int m = 10 ; 

现在我们声明一个指向节点的指针：

     // na is a node pointer
     struct node* na;

和malloc up m个节点。malloc返回数组的地址，它也是数组中第一个节点的地址。我们将na设置为malloc返回的地址：

      na = malloc(sizeof(struct node) * m);

现在我们将使用na，一个指针，就像它是一个数组一样。这是因为C将

array[offset]

定义为

*（array+offset*sizeof（element））

现在，我们将遍历阵列的其余部分，并将每个节点的位置设置为阵列中前一个节点的旁边：

     for(i = 1; i < m; i++) {
        na[i].val = i ;
        // na[ offset ] is *(na + offset)
        // we don't want to dereference, 
        // we want the address, so we use the 
        // address-of operator ("&")
        na[i].next = &na[ i - 1 ];
     }

我们将打印每个节点的地址。它应该是它的

next

节点指针+

sizeof（struct node）

的地址，因为每个节点都是（在数组中）它在列表中的下一个之后的节点（列表是“反向”数组）

我们将其强制转换为char*以获得以字节为单位的结果。如果不进行强制转换，我们将以

truct node*

为单位获得结果（该单位应始终为1）

在我的系统上，它提供以下输出：

  sizeof( struct node ) = 48
  val 9, address of current 0x804a1b8,  address of current->next 0x804a188,  distance from next: in bytes 48,  in struct nodes 1
  val 8, address of current 0x804a188,  address of current->next 0x804a158,  distance from next: in bytes 48,  in struct nodes 1
  val 7, address of current 0x804a158,  address of current->next 0x804a128,  distance from next: in bytes 48,  in struct nodes 1
  val 6, address of current 0x804a128,  address of current->next 0x804a0f8,  distance from next: in bytes 48,  in struct nodes 1
  val 5, address of current 0x804a0f8,  address of current->next 0x804a0c8,  distance from next: in bytes 48,  in struct nodes 1
  val 4, address of current 0x804a0c8,  address of current->next 0x804a098,  distance from next: in bytes 48,  in struct nodes 1
  val 3, address of current 0x804a098,  address of current->next 0x804a068,  distance from next: in bytes 48,  in struct nodes 1
  val 2, address of current 0x804a068,  address of current->next 0x804a038,  distance from next: in bytes 48,  in struct nodes 1
  val 1, address of current 0x804a038,  address of current->next 0x804a008,  distance from next: in bytes 48,  in struct nodes 1
  val 0, address of current 0x804a008,  address of current->next (nil),

---

不，你需要回路。您的heads数组本质上是一个二维数组。您至少需要两次分配。第一个是指针数组：

link * heads = (link*)malloc (config->m * sizeof (link));

第二个是heads数组的每个成员指向的内存：

link buf = (link)malloc(sizeof(struct node) * config->m);
for(i = 0; i < config->m; i++)
    heads[i] = &buf[i];

---

嗯，我想这回答了我的问题。谢谢。那不应该是heads[i]=&buf[i]；在for循环中？您希望头指针具有节点的地址，对吗？谢谢tpdi。实际上，原始heads数组应该是指针的link*（或node**）数组。因此发生了错误。我喜欢C，直到我在节点结构中添加了一些填充，我的测试程序才崩溃。实际上这里还有一个问题：。我完全同意第一个答案。typedef blah*这件事打乱了我的间接感；）。我最初出于同样的原因犯了同样的错误。我从未见过像这样使用内存缓冲区。我认为，如果您可以在调用中获得所有内存，那么最好避免对malloc进行大量调用。谢谢。不过我把事情搞砸了

     for(i = 1; i < m; i++) {
        na[i].val = i ;
        // na[ offset ] is *(na + offset)
        // we don't want to dereference, 
        // we want the address, so we use the 
        // address-of operator ("&")
        na[i].next = &na[ i - 1 ];
     }

     struct node* current = &na[ m - 1 ];

     while( current ) {
        printf( "val %i, address of current %p, ", current->val, current) ;
        printf( " address of current->next %p, ", current->next ) ;
        if( current->next ) {
           printf( " distance from next: ");
           printf( "in bytes %i, ", 
              ( (char*) current)  - (char*) current->next ) ;
           printf( " in struct nodes %i", current  - current->next ) ;
        }
        printf( "\n" );
        current = current->next;   
     }

     return 0;
  }

  sizeof( struct node ) = 48
  val 9, address of current 0x804a1b8,  address of current->next 0x804a188,  distance from next: in bytes 48,  in struct nodes 1
  val 8, address of current 0x804a188,  address of current->next 0x804a158,  distance from next: in bytes 48,  in struct nodes 1
  val 7, address of current 0x804a158,  address of current->next 0x804a128,  distance from next: in bytes 48,  in struct nodes 1
  val 6, address of current 0x804a128,  address of current->next 0x804a0f8,  distance from next: in bytes 48,  in struct nodes 1
  val 5, address of current 0x804a0f8,  address of current->next 0x804a0c8,  distance from next: in bytes 48,  in struct nodes 1
  val 4, address of current 0x804a0c8,  address of current->next 0x804a098,  distance from next: in bytes 48,  in struct nodes 1
  val 3, address of current 0x804a098,  address of current->next 0x804a068,  distance from next: in bytes 48,  in struct nodes 1
  val 2, address of current 0x804a068,  address of current->next 0x804a038,  distance from next: in bytes 48,  in struct nodes 1
  val 1, address of current 0x804a038,  address of current->next 0x804a008,  distance from next: in bytes 48,  in struct nodes 1
  val 0, address of current 0x804a008,  address of current->next (nil),

link * heads = (link*)malloc (config->m * sizeof (link));

link buf = (link)malloc(sizeof(struct node) * config->m);
for(i = 0; i < config->m; i++)
    heads[i] = &buf[i];

free(heads);
free(buf);