>Array memory can sit on either the stack or the heap.
No, if we are using the definition of an array that is like int c[] = ..., that is always going to be on the stack. Heap continuous memory =/= array. You can use the [] operator to access it like an array, but fundamentally, as far as structures in C language are concerned, those 2 are different, because they get treated by compiler differently.
>but the size is actually always stored and accessible, and this is virtually mandated by the standard: otherwise, `free(arr)` couldn't realistically work,
That would only be true if each element in the array was a char.
The dynamic data structure stores total amount of memory allocated by address, it has no info about the size of the element, so it can't infer the actual number of items at runtime. You could write your own malloc that does this, but generally, that is left to the user for flexibility. For example, a really good practice in C coding that basically solves any double free is a mempool that allocates all the memory up front. That way, you never really even have to call free, and the memory you allocate can be partitioned any way you chose dynamically.
> No, if we are using the definition of an array that is like int c[] = ..., that is always going to be on the stack. Heap continuous memory =/= array. You can use the [] operator to access it like an array, but fundamentally, as far as structures in C language are concerned, those 2 are different, because they get treated by compiler differently.
Well, not necessarily. For one thing, if we have a function foo(int c[]), it's debatable if c is an array variable or a pointer variable. However, what's not debatable is that you can allocate a struct on the heap, and that struct can have an array member - e.g. `struct foo { int a[10]; }; [...] struct foo *x = malloc(sizeof(struct foo));` would allocate an array on the heap as part of the struct.
> That would only be true if each element in the array was a char.
That's why I said that it depends on what exactly you mean by the size of the array. It's also true that in today's world at least, malloc() will often allocate more memory than you actually ask for, to optimize against fragmentation - and then the internally stored size is the size of the actual allocation, not the logical size that you requested - which may not even fit into a whole number of array elements. So, I was being a little overly simplistic (lying) for dramatic effect.
> For example, a really good practice in C coding that basically solves any double free is a mempool that allocates all the memory up front.
While this is a very valid technique for certain purposes, especially when dynamic allocation is needed in very high performance code, it's very much not a valid solution for memory safety - quite the contrary, it's a terrible practice for that. In particular, this is almost exactly the issue that caused the infamous HeartBleed vulnerability in OpenSSL to stay hidden for so long: the use of a memory pool for the buffers used to store TLS packets was hiding the buffer overflow from UBSan and valgrind and similar tools, since the reads were perfectly valid from a language perspective (they were never reading from free()d/unallocated memory, only from memory that had been released to the memory pool).
> that is always going to be on the stack.
Unless your C implementation doesn't use a stack for data storage. Which existed, you know: IIRC the C compilers for Cray machines used linked lists to hold activation frames. And of course, there are PIC microcontrollers where you can't really use the hardware stack for anything except return addresses.