Why does sizeof(*node) give the size of the structure and not size of the pointer

Question

In the below code:

typedef struct{int data1; int data2} node;
node n1;
node* n2;

sizeof(n1) returns 8 // size of the struct node
sizeof(n2) returns 4 // since n2 is a pointer it returns the size of the pointer
sizeof(*n2) returns 8 // HOW DOES THIS WORK ?

How does sizeof actually work ? In the above case *n2 boils down to providing the address of where n2 is pointing. n2 in this case is still a dangling pointer as we have neither allocated memory nor are we pointing it to some valid address. How does it give the size of the structure correctly ?

Answer 1

You need to understand two things:

First, what's the type of *n2 ? The type of n2 is a pointer to node , so the type of *n2 is node .

Second, you are right n2 is a danging pointer, it doesn't point to a valid place, but the magic of sizeof is, it's a compile time operator (except when the operand is a C99 variable length array), sizeof(*n2) is evaluated as the same as sizeof(node) in compilation time.

Answer 2

Basically, you can read *n2 as "the thing that n2 is pointing at".

The thing that n2 is pointing at is a node, and the sizeof a node is 8. Simple as that... it doesn't matter whether it's been allocated or not: the type of the thing that n2 is pointing at is a node, and the size of a node is 8.

Answer 3

When you do *n2 , where n2 is defined as node* n2 you are basically telling it to read the data at the address n2 as if it had the type node .

It does not matter what is written on that address. Consider adding these lines to your example:

void *n3 = n2; // copies the address, but no information about the data there
int *n4 = (int *)n3; // again, copies the address

sizeof(*n4) returns sizeof(int)

So basically, to summarize, if you have:

X* a;
sizeof(a); // will always return 4, the size of a pointer
sizeof(*a); // will always return sizeof(X), no matter if the address is set.