malloc and printf of its size

[simone.marras]

Hi there,

I am allocating a 1D array of integers by using the malloc function (I report the code down here).
Once allocated the memory, I would like to print to screen the size of such array by means of sizeof(), but it seems that what I obtain does not correspond to the number of elements of the newly allocated array, but the size of one element of type int.

To make it easy: I allocate 13 elements inside array _esup2, but when I printf to screen the sizeof(_esup2), I obtain 4 instead of 13.

What am I doing that is not correct? I need to verify this because the bogger code within this is enclosed crashes for wrong memory allocation.

Here the code, I hope someone can help
Thank you in advance,

All the best
S.M.

       int nnode = 13;
	int *_esup2;

	_esup2 = (int*) malloc((sizeof(int[nnode+1])));

	printf("size(ESUP2) %d\n", sizeof(_esup2));
	;

[tabstop]

sizeof(_esup2) is the sizeof the pointer. It's your job (not the system's) to know how much memory is allocated to a pointer.

[Elysia]

The answer is that you cannot.
Firstly, you are printing the size of the pointer, not the data.
But even if you dereference, you will get only the size of one element, because sizeof is compile-time and the compiler cannot know how big a dynamic array is.

[simone.marras]

I see, thank you very much to both for replying so quickly. I wasn sure if that was the reason.

Given that I am allocating correctly then, I have a question regarding a variable-size array that becomes larger within a certain loop. Should I keep asking this here or should I open a new thread on this (I havent been able to find anything on that)?
Thank you very much in advance
S.M.

[tabstop]

If your new question fits under the heading "malloc and printf of its size" then you can ask here, otherwise there's no harm in starting a new thread.

[simone.marras]

Hi Tabstop,

thank you, I will start a thread then.
All the best
S.M.