c - C中双指针的使用

Question

void main() {
   struct buf *head = NULL;
   foo(&head);
}

void foo(struct buf **ptr_B) {
   struct buf **ptr_A = ptr_B;
   *ptr_B = NULL;
   *ptr_A = malloc(sizeof(struct buff));
   //here checked *ptr_A != NULL, so malloc is done successfully
   if (*ptr_A == NULL) {
      return;
   }
   *ptr_B->item = 8; 
}

是否有可能*ptr_B仍然等于NULL？

我运行了一个自动检查工具，它总是在这里发出警告，说这是 *ptr_B->item = 8; 行中的前向空风险

Answer 1

我们来分析一下代码:

    struct buf **ptr_B;
    // advance 1 stack slot for ptr_B, ptr_B value wasnot initialized
    struct buf **ptr_A = ptr_B;
    // advance 1 stack slot for ptr_A, set ptr_A by value of ptr_B (uninitialized)

    *ptr_B = NULL;
    // dereference ptr_B (ptr_B uninitialized) and set that memory block into NULL
    *ptr_A = malloc();
    // dereference ptr_A (ptr_A uninitialized) and set that memory block into malloc()
    *ptr_B->item = 8; 
    // dereference ptr_B (ptr_B uninitialized, *ptr_B uninitialized) and use it as "struct buf"

正确代码:

    struct buf **ptr_B;
    struct buf **ptr_A;

    // allocate a pointer of `struct buf*` for both `ptr_A` and `ptr_B` point to
    struct buf** temp_ptr1 = malloc(sizeof(struct buf*));
    ptr_A = temp_ptr1;
    ptr_B = temp_ptr1;

    // set the shared pointer to NULL
    *ptr_B = NULL;

    // malloc
    struct buf* temp_ptr2 = malloc(sizeof(struct buf*));
    *ptr_A = temp_ptr2;

    // Set value from `ptr_B`
    (*ptr_B)->item = 8;

    // clean up
    free(*ptr_B);
    free(ptr_B);