malloc implementation?

c, free, malloc, memory-management

Solution

The easiest way to do it is to keep a linked list of free block. In `malloc`, if the list is not empty, you search for a block large enough to satisfy the request and return it. If the list is empty or if no such block can be found, you call `sbrk` to allocate some memory from the operating system. in `free`, you simply add the memory chunk to the list of free block. As bonus, you can try to merge contiguous freed block, and you can change the policy for choosing the block to return (first fit, best fit, ...). You can also choose to split the block if it is larger than the request.

Some sample implementation (it is not tested, and is obviously not thread-safe, use at your own risk):

typedef struct free_block {
    size_t size;
    struct free_block* next;
} free_block;

static free_block free_block_list_head = { 0, 0 };
static const size_t overhead = sizeof(size_t);
static const size_t align_to = 16;

void* malloc(size_t size) {
    size = (size + sizeof(size_t) + (align_to - 1)) & ~ (align_to - 1);
    free_block* block = free_block_list_head.next;
    free_block** head = &(free_block_list_head.next);
    while (block != 0) {
        if (block->size >= size) {
            *head = block->next;
            return ((char*)block) + sizeof(size_t);
        }
        head = &(block->next);
        block = block->next;
    }

    block = (free_block*)sbrk(size);
    block->size = size;

    return ((char*)block) + sizeof(size_t);
}

void free(void* ptr) {
    free_block* block = (free_block*)(((char*)ptr) - sizeof(size_t));
    block->next = free_block_list_head.next;
    free_block_list_head.next = block;
}

Note: `(n + align_to - 1) & ~ (align_to - 1)` is a trick to round `n` to the nearest multiple of `align_to` that is larger than `n`. This only works when `align_to` is a power of two and depends on the binary representation of numbers.

When `align_to` is a power of two, it only has one bit set, and thus `align_to - 1` has all the lowest bit sets (ie. `align_to` is of the form 000...010...0, and `align_to - 1` is of the form `000...001...1`). This means that `~ (align_to - 1)` has all the high bit set, and the low bit unset (ie. it is of the form `111...110...0`). So `x & ~ (align_to - 1)` will set to zero all the low bits of `x` and round it down to the nearest multiple of `align_to`.

Finally, adding `align_to - 1` to `size` ensure that we round-up to the nearest multiple of `align_to` (unless `size` is already a multiple of `align_to` in which case we want to get `size`).

Problem

I'm trying to implement `malloc` and `free` for C, and I am not sure how to reuse memory. I currently have a `struct` that looks like this: ``` typedef struct _mem_dictionary { void *addr; size_t size; int freed; } mem_dictionary; ``` My `malloc` looks like this: ``` void *malloc(size_t size) { void *return_ptr = sbrk(size); if (dictionary == NULL) dictionary = sbrk(1024 * sizeof(mem_dictionary)); dictionary[dictionary_ct].addr = return_ptr; dictionary[dictionary_ct].size = size; dictionary[dictionary_ct].freed = 1; dictionary_ct++; return return_ptr; } ``` When I free memory, I would just mark the address as `0` (that would indicate that it is free). In my `malloc`, I would then use a for loop to look for any value in the array to equal `0` and then allocate memory to that address. I'm kind of confused how to implement this.

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