In C programming, you can pass any kind of pointer you like as an argument to free, how does it know the size of the allocated memory to free? Whenever I pass a pointer to some function, I have to also pass the size (ie an array of 10 elements needs to receive 10 as a parameter to know the size of the array), but I do not have to pass the size to the free function. Why not, and can I use this same technique in my own functions to save me from needing to cart around the extra variable of the array's length?
11 Answers
This answer is relocated from How does free() know how much memory to deallocate? where I was abrubtly prevented from answering by an apparent duplicate question. This answer then should be relevant to this duplicate:
For the case of malloc, the heap allocator stores a mapping of the original returned pointer, to relevant details needed for freeing the memory later. This typically involves storing the size of the memory region in whatever form relevant to the allocator in use, for example raw size, or a node in a binary tree used to track allocations, or a count of memory "units" in use.
free will not fail if you "rename" the pointer, or duplicate it in any way. It is not however reference counted, and only the first free will be correct. Additional frees are "double free" errors.
Attempting to free any pointer with a value different to those returned by previous mallocs, and as yet unfreed is an error. It is not possible to partially free memory regions returned from malloc.
to answer the second question, yes you could (kind of) use the same technique as malloc()
by simply assigning the first cell inside every array to the size of the array.
that lets you send the array without sending an additional size argument.
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