How does the struct of a linked list know the type of next pointer of its own type?

Question

struct node
{
    int     data;
    node*   pointerToNextNode;
};

Here pointerToNextNode is the type of struct node , and it is declared inside the struct. How does the struct know the type of next pointer of its own type - when it itself hasn't been formed yet?

There is no keyword extern used. How does this work?

Answer 1

It doesn't need to know the structure, it's enough to know the type name, namely struct node — and that has already been defined.

Same result you can obtain by forward type declaration:

struct node;            // declare the struct not defining it
struct node *pointer;   // declare variable

void foo()
{
    if(pointer != NULL)        // OK, we use the pointer only
        if(pointer->x == 0)    // invalid use - struct contents unknown yet
            return;
}

struct node {           // supply a definition
    int x;
};

void bar()
{
    if(pointer != NULL)
        if(pointer->x == 0)    // OK - struct contents already known
            return;
}

Answer 2

Here pointerToNextNode is the type of struct node

No, it's not. It's of type struct node *

struct node* pointerToNextNode; allocates memory for a pointer variable of type struct node .
It does not allocate memory for struct node , so, till point, it does not need to know about the size and representation of struct node . Only the (data)type name is sufficient.

Also, it's worthy to mention, without a typedef in place, node* pointerToNextNode; should not be valid. It should be written like below

typedef struct node node; 

struct node 
{ 
    int data; 
    node* pointerToNextNode; 
};

BTW, private: is not C thing, if i'm not wrong.

Answer 3

for me this doesn't compile using CC -- and exactly because of what you said. you would have to use struct node * to make the compiler aware you want memory for a pointer