malloc and heap in c

Question

consider the code below:

#include "list.h"
struct List
{
   int size;
   int* data;
};
List *list_create()
{
   List *list;
   printf("%d %d",sizeof(list),sizeof(List));
   list = malloc(sizeof(list));
   assert(list != NULL);
   if (list != NULL) {
      list->size = 0;
   }
   return list;
}

The number printed out is "4 8", i assume this is the 4 bytes taken by "int size" in List object?and the size of "int* data" is 0 cause nothing has assigned to data? the size of int pointer is also 4 bytes so the type List take 8 bytes in total? or there are some thing else going on? Can some one help me understand all this in detail?

then the malloc() get 4 bytes from the heap and assign the address to the pointer list? later in main if i do "list->data[i]=1;" this will give me a run time error why? Is it because I cant change contents in the heap? but if i do "list->size++" this would work, isn't the whole list object is in the heap?

really need some help here

Thanks in advance.

Your runtime error is because you haven't initialized the pointer to anything. You need to allocate space for it, too. — Ry-, Nov 06 '11 at 15:41
This is not valid C because you omitted the `struct` in `sizeof(List)`. Are you using a C++ compiler? — Fred Foo, Nov 06 '11 at 15:42
possible duplicate of [sizeof question](http://stackoverflow.com/questions/5463325/sizeof-question) — Hogan, Nov 06 '11 at 15:43
sorry i have a head file #include "list.h" contains the typedef thing — pythoniku, Nov 06 '11 at 15:46
@SalvatorePreviti: Even when you add the code `list->data[i] = 1` as noted in the post? — Ry-, Nov 06 '11 at 16:29
Even, sizeof is not related to what you are doing with variables. To allocate a variable length array you need to use again malloc. — Salvatore Previti, Nov 06 '11 at 16:31
@JonathanGrynspan Why is the void necessary? I have never used it in all my coding and no warnings (even on pre c90). — chacham15, Nov 07 '11 at 00:47
@chacham15 In C, `(void)` means "this function has no arguments," while `()` means "this function accepts an unspecified argument list." They have different behaviours. — Jonathan Grynspan, Nov 07 '11 at 00:55
@JonathanGrynspan What are the different behaviors? I haven't ever come across that difference before. Do you have a link? — chacham15, Nov 07 '11 at 01:17
@chacham15 I just told you what they are. For more info, see other SO questions such as http://stackoverflow.com/questions/51032/is-there-a-difference-between-foovoid-and-foo-in-c-or-c — Jonathan Grynspan, Nov 07 '11 at 04:54

Salvatore Previti · Accepted Answer · 2011-11-06T16:11:38.910

5

sizeof(List*) is the size of a pointer to a List struct.

sizeof(list) in your case, since variable list is of type List* is the same as sizeof(List*).

sizeof(List) instead is the size of the struct List, it contains two 32 bit variables (I assume you are using a 32 bit compiler obviously), an integer and a pointer and your compiler decided that the right size for your struct is 8 bytes.

Pointers to types are usually 4 byte in 32 bit compilers and 8 bytes in 64 bit compilers.

As a side note, reading your code however i read you never initialize list->data, you should initialize it to something somewhere i guess.

This is C++ however, you should write

typedef struct { ... } List; // This is C.

Sizeof operator is evaluated at compile time, not at runtime, it gives only information of the size of a type. You cannot, for example, know how much elements are in a dynamic array with sizeof, if you were trying to accomplish this, sizeof(pointer) will give you the size in byte of the pointer type.

As something to read about what is a pointer and what is an array i would suggest you to read http://www.lysator.liu.se/c/c-faq/c-2.html or http://pw1.netcom.com/~tjensen/ptr/pointers.htm

edited Nov 06 '11 at 16:11

answered Nov 06 '11 at 15:40

Salvatore Previti

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ERROR: pointer size is determined by the compiler, not the system. It isnt a rare thing to have a 32 bit compilation on a 64 bit system meaning that sizeof(void*) = 4 – chacham15 Nov 06 '11 at 15:45
You also forget that alignment affects the sizeof a struct – chacham15 Nov 06 '11 at 15:56
Packing too indeed, but it would be a little out of topic. – Salvatore Previti Nov 06 '11 at 15:59
so if i wanna use data as an int array i should do "data=(int*)malloc(somenumber)"? and pointer type of int and int array are the same in syntax "int*"? i do got a run time error when i was trying use the variable data as a array – pythoniku Nov 06 '11 at 16:08

chacham15 · Answer 2 · 2011-11-06T15:55:10.953

Technically your code has an error in it.

The code should read: sizeof(struct List) or have typedef struct List List; somewhere.

But yes, sizeof(list) is the size of the variable list. Since list is a pointer it is equivalent to sizeof(void*) which on your system/compiler is 4.

sizeof(struct List) is the size of the struct which is sizeof(int)+sizeof(int*)+any alignment issues. The alignment thing is often forgotten but is very important as it can change the size of the struct in unexpected ways.

malloc and heap in c

2 Answers2