sizeof(struct), sizeof(string), sizeof(array)

Question

I am confused as to what the sizeof function returns.

Suppose I enter the following into a program for sizeof, and this is what is returned:

sizeof(char) ==  1
sizeof(int) == 4
sizeof(float) == 4
sizeof(long) ==  8
sizeof(double) == 8
sizeof(void *) == 8

sizeof Pointer is 8.
So if I malloc a long array:

long *array = (long *)malloc(64)

Would this sizeof() inserted into the malloc of my long array be equivalent?

long *array = (long *)malloc(8*sizeof(long))

Now if I have a struct:

typedef struct treenode {
    struct treenode *left;
    struct treenode *right;
    void *content;
} Treenode;

And I run sizeof(Treenode), do I add all the elements of the struct together?

Pointer + Pointer + void * = 8 + 8 + 8 = 24

I tried running this in a program, however I get sizeof(Treenode) to return 12.

I am also wondering why when I create a new struct like this:

typedef struct country {
    char city[40];
    int population;
    int rank;
}

and try to malloc(sizeof(country *)) only 8 bytes are allocated instead of the bytes of the struct.

Finally when I do sizeof("Korea"), why is my return value 6 and not 5? Should it not be 5 because there are only 5 chars in the string?

What is the definition of `Pointer`? BTW all of your questions have been answered multiple times here on SO. You should be able to find them. — Steve Fallows, Nov 22 '14 at 20:14
You are missing ***A LOT*** of very basic concepts of C. For example, the fact that the size of a pointer is not the same as the size of its pointed object. And that strings contain an extra 0-terminator. You should first familiarize yourself with the fundamentals of the language before trying to do advanced stuff (like dynamic memory management). — The Paramagnetic Croissant, Nov 22 '14 at 21:20
Please read the help text on how to ask question here, again: don't ask several questions in one go, take more care in the formulation your question and search the site before asking. — Jens Gustedt, Nov 22 '14 at 22:20

score 2 · Answer 1 · answered Nov 22 '14 at 20:22

I will address each question separately:

These are equivalent in the context you have described
```
long *array = (long *)malloc(64)
long *array = (long *)malloc(8*sizeof(long))
```
However, this is not necessarily correct on all machines (if say on one machine a long is 16 bytes), therefore this code is not portable. You should only ever use the second method which is portable.
I'm not sure why the size is claiming to be 12 - Sorry. (are you sure all pointer's sizes are 8?)
while a country struct may be very large, a pointer to a country is just a pointer and as such it's size is in your case 8.
In C strings are null terminated so that a string with five letters has a trailing null which is presumably part of the string's size.

score 1 · Answer 2 · answered Nov 22 '14 at 20:19

If sizeof(Treenode) is 12, then you are not using 64-bit pointers/memory. If you got different results elsewhere (sizeof(void *) == 8) then you must have different targets.
sizeof(country *) is the size of a pointer to a country, that is -- the same as any other pointer. Use malloc(sizeof(country)) instead.
sizeof("Korea") includes the size of the string's null terminator.

score 0 · Answer 3 · edited Dec 12 '14 at 13:06

It is all matter of which architecture you are using.
on 32 bit, each pointer size is 4 bytes.
on 64 bit each pointer size is 8 bytes.

So if you run the following code:

long *array = (long *)malloc(64) ;

and the following code:

long *array = (long *)malloc(8*sizeof(long))

it would be the same on 64-bit but not in 32 bit.

malloc(sizeof(country *)) allocates only 8 bytes because (country *) is type of a pointer (pointer to your country struct) and each pointer is 8 bytes long in 64-bit.

sizeof(struct), sizeof(string), sizeof(array)

3 Answers3