Sizeof operator not giving correct size of pointer in C

Question

I created a structure with 3 integer members and a pointer variable ptr

 ptr=(struct node*)calloc(15,sizeof(struct node*));
  printf("%ld",sizeof(ptr));

however using the sizeof operator to get the size of ptr is printing a size less than 15(i.e. 8). What should be done to just get the size of the pointer variable if this is incorrect.

It *is* returning the size of the pointer: a pointer on your system is 8 bytes, regardless of what it points to. What you want is the allocated size of the buffer that the pointer points to, and C doesn't give you a way to query that; you have to keep track for yourself how much memory you allocated. — Nate Eldredge, Dec 24 '20 at 19:36
Sizeof returns the numbers of bytes, it doesn't work as you think. — Matteo Pinna, Dec 24 '20 at 19:37
By the way, passing `sizeof(struct node *)` to `calloc` is almost certainly a mistake; if `ptr` is to point to an array of `struct node`s, then you want space for 15 `struct`s, not 15 pointers. You should have `sizeof(struct node)` there instead. — Nate Eldredge, Dec 24 '20 at 19:38
Best to use specifier `"%zu"`, not `"%ld"`, when printing the result of `sizeof` — chux - Reinstate Monica, Dec 24 '20 at 21:51

cocool97 · Answer 1 · 2020-12-25T17:01:20.040

2

The result you have obtained is totally as expected.
Indeed, sizeof(ptr) returns the size of a pointer, which depends on your processor architecture (As it returns 8 you are very likely on a 64-bit processor).

edited Dec 25 '20 at 17:01

answered Dec 24 '20 at 19:44

cocool97

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or 2 or 3. There are smaller architectures than 16bits – 0___________ Dec 24 '20 at 20:18
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@P__JsupportswomeninPoland "3" ? - most rare/unusual. PIC24? – chux - Reinstate Monica Dec 24 '20 at 21:49
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@chux-ReinstateMonica AVR if named address spaces are used. – 0___________ Dec 24 '20 at 22:08
***`__memx This is a 24-bit address space that linearizes flash and RAM:` – 0___________ Dec 24 '20 at 22:14
@P__JsupportswomeninPoland That's absolutely right yes ! Updated the answer – cocool97 Dec 25 '20 at 17:00

score 0 · Answer 2 · answered Dec 24 '20 at 19:48

The size of a pointer will depend on your hardware... A pointer stores an address in memory that represents the beggining of that thing you are allocating. It doesn't how matter how big that thing is: It's initial address will have the same number of bytes as any other variable.

The line ptr=(struct node*)calloc(15,sizeof(struct node*)); is allocating space for 15 pointers, not 15 structs, you probably want to remove the * in the sizeof.

After you fix that, you can add the two lines to see the difference:

printf("size of a pointer to node* is: %d\n", sizeof(ptr));
printf("size of a struct pointed by a node* is :%d\n", sizeof(*ptr));

Sizeof operator not giving correct size of pointer in C

2 Answers2