🔒 Locked

malloc question

Started by AsOne Aug 31, 2005 at 3:03 PM 9 replies 2k views

Original Post

Aug 31, 2005 at 3:03 PM

Can somone please tell me why this works?


#include <stdio.h>
#include <stdlib.h>

typedef struct _vertex
{
    float x;
    float y;
    float z;
} Vertex;

int main()
{
    Vertex *vertex = malloc(sizeof(Vertex));
    vertex->x = 10;
    vertex->y = 20;
    vertex->z = 30;
    printf("x = %f, y = %f, z = %f\n", vertex->x, vertex->y, vertex->z);
    printf("the memory allocated for the vertex data is %d bytes.\n", sizeof(vertex));
    printf("shouldn't the above be %d bytes?\n", sizeof(Vertex));
    free(vertex);
    return 0;
}

First it prints that the vertex is 4 bytes in size, but the size of the Vertex struct is 12 bytes (which it should be). I've also changed

malloc(sizeof(Vertex))

malloc(0);

and it worked perfectly fine. Is my compiler automatically 'fixing' this? This seems very odd to me.

Enigma

Aug 31, 2005 at 3:12 PM

sizeof(vertex) will give you the size of the pointer, not the size of the object pointed to. Use sizeof(*vertex) to get the size of the pointed-to object.

malloc is allowed to treat zero-byte requests as a request for some non-zero number of bytes, however you are not allowed to access the allocated memory. Basically you're writing to semi-random memory that probably just happens to be valid from the operating-systems point of view, just like if you were overwriting the bounds of an array. Don't do it!

Enigma

Codemonger

Aug 31, 2005 at 3:14 PM

The variable vertex is a pointer .. what it points to , who cares ? you are just telling the compiler and yourself that it points to a structure Vertex (which is 12 bytes). All pointers with 32 bit compiler are 4 bytes, which is simply storage for a memory address.

if malloc(0) worked then I am assuming that you are assigning values with a bad pointer (no pointer). You would eventually crash your program if you continue to access memory that you have not allocated.

*** Edit

Enigma beat me to it lol :)

AsOne

Aug 31, 2005 at 3:22 PM

Quote:

sizeof(vertex) will give you the size of the pointer, not the size of the object pointed to. Use sizeof(*vertex) to get the size of the pointed-to object.

Silly me, I should have realized that :S

So it was luck then that malloc(0) returned a location which was valid?

Enigma

Aug 31, 2005 at 3:47 PM

Quote:
Original post by AsOne
So it was luck then that malloc(0) returned a location which was valid?

It wasn't a valid location. Using that memory had undefined behaviour. It just so happens that sometimes undefined behaviour means it seems to work.

Enigma

Anon Mike

Aug 31, 2005 at 3:48 PM

No, malloc(0) is perfectly legal. You will be give a valid pointer to a 0-byte piece of memory that is guaranteed to be unique relative to any other malloc(0). It's no different than a malloc of any other size. When you're done with it you have to call free on it. Since it's 0 bytes you can't actually use the pointer for anything. Trying to will give you undefined behavior exactly like going off the end of an array will.

There are a couple reasons why it works. The first is that it makes things a little simpler for the caller. The second is that internally malloc has to attach some bookkeeping information anyway so your malloc(0) gets turned into _internal_malloc(0 + sizeof(bookkeeping_info)).

Enigma

Aug 31, 2005 at 3:53 PM

Anon Mike: According to the final draft of the C99 standard the behaviour of malloc on zero-byte allocation requests is implementation defined. Implementations are allowed to behave as you say, but they are also allowed to simply return a null pointer.

Enigma

AsOne

Aug 31, 2005 at 5:19 PM

Thanks for the replies :)

SiCrane

Aug 31, 2005 at 5:21 PM

Quote:
Original post by Anon Mike
No, malloc(0) is perfectly legal. /snip/

You might be thinking about new in C++, which requires that the return value for operator new(0) is a non-null value.

Enigma

Aug 31, 2005 at 5:25 PM

Just to clarify, when I said that the return value of malloc(0) "wasn't a valid location" I didn't mean it wasn't a valid pointer - it may or may not have been. What I meant was that it was not a valid chunk of memory in which you could store a vertex.

Enigma

Anon Mike

Aug 31, 2005 at 7:21 PM

Quote:
Original post by Enigma
Anon Mike: According to the final draft of the C99 standard the behaviour of malloc on zero-byte allocation requests is implementation defined. Implementations are allowed to behave as you say, but they are also allowed to simply return a null pointer.

Ah ok. My C is a bit rusty and my compiler doesn't support C99 anyway, although it does return a valid pointer for malloc(0).

Topic Locked

This topic has been locked by a moderator. New replies are not allowed.