c - 结构内存分配给它的成员

Question

我见过像这样为结构分配内存

   ///enter code here
   for(i = 0; i<10; i++)
     s->a[i].mem = malloc(sizeof(*a[0]));

这是什么意思???它正在为a或mem分配内存。我很困惑。

如果它分配给指针mem？我在代码中看到了，他们通过

mem[0] = bla...bla...
mem[1] = bla...bla...

是否合法？？？

Answer 1

a[]应该是一个结构数组，其中每个都a[i]表示一个结构。

在结构mem中应该是结构类型的指针（自指向）。所以表情。

s->a[i].mem = malloc(sizeof(*a[0])); 

使用 malloc 动态分配内存并将地址分配给mem

就像是：

struct  xyx{
   int data;
   struct  xyx   *mem;
};

 struct  xyx a[10];

所以：a[0]的类型struct xyx。

和表达：

a[i].mem = malloc( sizeof(*a[0])); 

如同：

a[i].mem = malloc( sizeof(struct xyz); 

第二： 现在，什么是 s->a[i]？

s又是一个其他结构类型的指针变量，其中有一个 struct 类型的a[]数组xyz。就像是：

struct abc{ 
  struct  xyx a[10];
}; 

应该s是：

struct abc  x;
struct abc  *s = &x;

或内存s应该动态分配。所以s -> a[i]要访问成员。

所以表达：

s -> a[i].mem =  malloc(sizeof(*a[0]));
^    ^     ^ 
|    |     is pointer to struct of type that same as type of a[0]   
|    array of struct's elements,    
s is pointer to a struct in which `a[]` is datamember  

Answer 2

This will allocate memory of size *a[0] and pointer to the start of this memory block will be saved in mem pointer.

Answer 3

It means allocate a block of memory the same size as the thing that a[0] points to

This memory allocation is pointed to by s->a[i].mem

So, I'd guess that elsewhere in the program a[0] is set up and then this fragment uses the contents of a[0] to find the size for the other members of a[]

Answer 4

a[0]该代码分配了一个与指向的对象具有相同大小的块。然后它将指向该块的指针存储在s->a[i].mem.