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I have been recently trying to learn how to program in the C programming language.

I am currently having trouble understanding how memory is deallocated by free() in C.

What does it mean to free or release the memory?

For instance, if I have the following pointer:

int *p = malloc(sizeof(int));

When I deallocate it using free(p), what does it do? Does it somehow flag it as "deallocated", so the application may use it for new allocations?

Does it deallocates only the pointer address, or the address being pointed is also deallocated too?

I would do some experiments myself to better understand this, but I am so newbie in the subject that I don't know even how to debug a C program yet (I'm not using any IDE).

Also, what if int *p is actually a pointer to an array of int?

If I call free(p), does it deallocate the whole array or only the element it is pointing to?

I'm so eager to finally understand this, I would very much appreciate any help!

Matias Cicero
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3 Answers3

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What does it mean to free or release the memory?

It means that you're done with the memory and are ready to give it back to the memory allocator.

When I deallocate it using free(p), what does it do?

The specifics are implementation dependent, but for a typical allocator it puts the block back on the free list. The allocator maintains a list of blocks that are available for use. When you ask for a chunk of memory (by calling malloc() or similar) the allocator finds an appropriate block in the list of free blocks, removes it (so it's no longer available), and gives you a pointer to the block. When you call free(), the process is reversed -- the block is put back on the free list and thereby becomes available to be allocated again.

Importantly, once you call free() on a pointer, you must not dereference that pointer again. A common source of memory-related errors is using a pointer after it has been freed. For that reason, some consider it a helpful practice to set a pointer to nil immediately after freeing it. Similarly, you should avoid calling free() on a pointer that you didn't originally get from the allocator (e.g. don't free a pointer to a local variable), and it's never a good idea to call free() twice on the same pointer.

Does it deallocates only the pointer address, or the address being pointed is also deallocated too?

When you request a block of memory from the allocator, you specify the size of the block you want. The allocator keeps track of the size of the block so that when you free the block, it knows both the starting address and the block size. When you call free(p), the block that p points to is deallocated; nothing happens to the pointer p itself.

Also, what if int *p is actually a pointer to an array of int?

An array in C is a contiguous block of memory, so a pointer to the first element of the array is also a pointer to the entire block. Freeing that block will properly deallocate the entire array.

I'm so eager to finally understand this, I would very much appreciate any help!

There are a number of good pages about memory allocation in C that you should read for a much more detailed understanding. One place you could start is with the GNU C Library manual section on memory allocation.

As alluded to above and in the other answers, the actual behavior of the allocator depends on the implementation. Your code shouldn't have any particular expectations about how memory allocation works beyond what's documented in the standard library, i.e. call malloc(), calloc(), etc. to get a block of memory, and call free() to give it back when you're done so that it can be reused.

Caleb
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malloc and free do whatever they want. Their expected behaviour is that malloc allocates a block of desired size in dynamic memory and returns a pointer to it. free must be able to receive one such pointer and correctly deallocate the block. How they keep track of the block size is irrelevant.

Is int *p a pointer to an array of ints ? Maybe. If you allocated sufficient space for several ints, yes.

Quentin
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  • I see. If I had the pointer `int *p = malloc(sizeof(int) * 6)`, would `free(p)` deallocate the whole block of 6 `int` then? – Matias Cicero Nov 18 '14 at 18:32
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There is a fixed and limited amount of memory in your computer, and everybody wants some. The Operating system is charged with the task of assigning ownership to pieces of memory and keeping track of it all to assure that no one messes with anyone else's.

When you ask for memory with malloc(), you're asking the system (the C runtime and the OS) to give you the address of a block of memory that is now yours. You are free to write to it and read from it at will, and the system promises that no one else will mess with it while you own it. When you de-allocate it with free(), nothing happens to the memory itself, it's just no longer yours. What happens to it is none of your business. The system may keep it around for future allocations, it may give it to some other process.

The details of how this happens vary from one system to another, but they really don't concern the programmer (unless you're the one writing the code for malloc/free). Just use the memory while it's yours, and keep your hands off while it's not.

Lee Daniel Crocker
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