What does the pointer of a dynamically allocated memory points to after calling the free() function.
Does the pointer points to NULL, or it still points to the same place it pointed before the deallocation.
does the implementation of free() has some kind of standard for this, or it’s implemented differently in different platforms.
uint8_t * pointer = malloc(12);
printf("%p", pointer); // The current address the pointer points to
free (pointer);
printf("%p", pointer); // must it be NULL or the same value as before ?
Edit:
I know that the printf will produce the same results, I just want to know if I can count on that on different implementations.
According to the standard (6.2.4/2 of C99):
In practice, all implementations I know of will print the same value twice for your example code. However, it is permitted that when you free the memory, the pointer value itself becomes a trap representation, and the implementation does something strange when you try to use the value of the pointer (even though you don’t dereference it).
Supposing that an implementation wants to do something unusual, a hardware exception or program abort would be the most plausible I think. You probably have to imagine an implementation/hardware that does a lot of extra work, though, so that every time a pointer value is loaded into a register, it somehow checks whether the address is valid. That could be by checking it in the memory map (in which case I suppose my hypothetical implementation would only trap if the whole page containing the allocation has been released and unmapped), or some other means.