malloc implementation?

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).