Dynamic Memory Allocation

Hi,

Can anyone explain to me why this code doesn't work well?

Code:

---

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

void teste(int* n)
{
        n = (int*) malloc(sizeof(int));
        *n = 5;
}

int main()
{
        int* n = 0;

        teste(n);

        printf("Teste: %d", *n);

        return 0;
}

---

I would like to alloc memory inside a procedure passing the pointer I declared before in the main. Is there any way to make this thing work like I expect?

Your code doesn't modify the initial pointer in main(): either pass the pointer address (int**) or simply declare an integer in main() and pass its address (but it doesn't require a malloc() this way).

Code:

---

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

void teste(int** n)
{
  *n = (int*) malloc(sizeof(int));
  **n = 5;
}

int main()
{
  int* n = 0;
  teste(&n);
  printf("Teste: &#37;d", *n);
  return 0;
}

---

When you pass 'n' to the function, you are passing the address of an int declared in main. Now, when you use malloc, teste's 'n' will be the address to the newly allocated data, which is not the same pointed by the 'n' in main.

So your printf in main will not print 5, it will print the value that main's 'n' points at.

Well, what you pass is a copy of the original pointer.
So there's no way it can actually change the address of the original pointer because you don't have a pointer to that pointer (sounds familiar?).

Thanks all... It was a little bit confusing, but now I understand why It isn't working like I was expecting.

Thanks root4, samus250 and Elysia!

Quote:

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So your printf in main will not print 5, it will print the value that main's 'n' points at.

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In fact the initial n (in main) is not modified at all and the result is only a SEGV (as *n is deferencing the address 0).

Another way would be to simply return an int* from teste(), which would make the function itself rather pointless...

And besides:
- casting mallocs return value is pointless since a void* is implicitly converted to other pointer types
- avoid using sizeof on types, use sizeof on real objects instead. (sizeof *n) in this case
- free() your memory when done

Quote:

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Originally Posted by TactX

- casting mallocs return value is pointless since a void* is implicitly converted to other pointer types

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And to add to that - it's also dangerous because it can mask a very serious error.
If you don't include the correct header, the compiler won't see the declaration of malloc, and you will get an "implicit" call to the function, which is very, very bad.
If you cast the return from malloc, you will hide this warning (the compiler won't warn you). Therefore it is very bad to do so in C.

Quote:

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Originally Posted by root4

In fact the initial n (in main) is not modified at all and the result is only a SEGV (as *n is deferencing the address 0).

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Yes I know that it doesn't get modified, that's kind of what I tried to say. But I did forget about the initialization to 0 :-) .

What would address 0 hold? I guess its something creepy.

Typically operating systems prevent you from dereferencing address 0. So you get an access violation instead. Which typically results in a crash.

Thanks again! Wow! That was a very nice explanation! Very complete also...

Thanks to all who answered this thread!

Quote:

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Originally Posted by Elysia

And to add to that - it's also dangerous because it can mask a very serious error.
If you don't include the correct header, the compiler won't see the declaration of malloc, and you will get an "implicit" call to the function, which is very, very bad.
If you cast the return from malloc, you will hide this warning (the compiler won't warn you). Therefore it is very bad to do so in C.

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ooops, I've always been casting malloc's return.

Quote:

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Originally Posted by samus250

ooops, I've always been casting malloc's return.

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Note that if you ever use malloc() in a C++ program, it will require casting, since void pointers are not automatically converted to other pointer types in C++.

Of course, you should probably use new....

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Mats