Freeing char pointers

Hi I created a char* pointer and malloc a certain size to it. Then I pass this pointer to a function which fills up the memory allocated.

Now i want to free the memory allocated but I keep getting some error. Heap corruption detected

Here is what i did

Code:

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char* buffer;
buffer = (char*)malloc(100);
fillupbuffer(&buffer);

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the freeing part i did this

Code:

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free(buffer);

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and there comes the error.

Why do you need to pass a char** to fillupbuffer when you have already allocated memory for the buffer? Normally, if you pass a char**, it means that fillupbuffer itself will be doing a malloc. I suggest that you show us the declaration of fillupbuffer. Also, check that you #include <stdlib.h> because the unnecessary cast of the return value of malloc would mask an error associated with failing to include that header.

oh the fillupbuffer function basically looks like this

Code:

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freebuffer(char** buffer)
{
uint16_t a= 1234;
memcpy(*buffer,a,sizeof(uint16_t));
}

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If I do the malloc inside the fill buffer function, how do I free it outside the function?

Yes i included stdlib.h already.

Sorry for the noobness

Quote:

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Originally Posted by Edelweiss

If I do the malloc inside the fill buffer function, how do I free it outside the function?

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Since you don't need malloc there, just pass a char*.

Quote:

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Originally Posted by Edelweiss

oh the fillupbuffer function basically looks like this

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It seems you renamed it ;)
You probably want to pass the address of a, e.g.,

Code:

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void fillupbuffer(char *buffer)
{
    uint16_t a = 1234;
    memcpy(buffer, &a, sizeof(a));
}

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Of course, now fillupbuffer(&buffer) should be fillupbuffer(buffer). Incidentally, are you trying to fill the buffer with "1234" instead?

Oh hahaha sorry for the renaming, morning blues.

But if i declare a char* and allocate memory and pass the address of that pointer to a function, what advantages does it has?

Compared the above to declaring a char* and pass address of that pointer to function and allocating memory in that function,

and compared to just passing the address in.

If I have 2 memcpy actions, in which both copies to different memory addresses in the buffer, which of the above will be better?

Oh i actually have a lot of information to memcpy to the buffer, but its the concept which I am confused with.

Quote:

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Originally Posted by Edelweiss

But if i declare a char* and allocate memory and pass the address of that pointer to a function, what advantages does it has?

Compared the above to declaring a char* and pass address of that pointer to function and allocating memory in that function,

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If you take my suggestion of just having a char* parameter, your function is then more reusable: you can use it even if buffer is an actual array instead of a pointer. It is also typically easier to work with fewer levels of indirection. If you use a char** parameter and malloc from within the function, you are then constrained with working with just such a dynamic array.

Quote:

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Originally Posted by Edelweiss

If I have 2 memcpy actions, in which both copies to different memory addresses in the buffer, which of the above will be better?

Oh i actually have a lot of information to memcpy to the buffer, but its the concept which I am confused with.

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This does not affect the choice.