malloc implementation?

Question

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:

Answer 1

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).

Answer 2

I am borrowing code from Sylvain's response. He seems to have missed calculating the size of the free_block* ini calculating the overhead.

In overall the code works by prepending this free_block as a header to the allocated memory. 1. When user calls malloc, malloc returns the address of the payload, right after this header. 2. when free is called, the address of the starting of the header for the block is calculated (by subtracting the header size from the block address) and that is added to the free block pool.

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(free_block) + (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(free_block);
        }
        head = &(block->next);
        block = block->next;
    }

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

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

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

Answer 3

You don't want to set the size field of the dictionary entry to zero -- you will need that information for re-use. Instead, set freed=1 only when the block is freed.

You cannot coalesce adjacent blocks because there may have been intervening calls to sbrk(), so that makes this easier. You just need a for loop which searches for a large enough freed block:

typedef struct _mem_dictionary
{
    void *addr;
    size_t size;
    int freed;
} mem_dictionary;


void *malloc(size_t size)
{
     void *return_ptr = NULL;
     int i;

     if (dictionary == NULL) {
         dictionary = sbrk(1024 * sizeof(mem_dictionary));
         memset(dictionary, 0, 1024 * sizeof(mem_dictionary));
     }

     for (i = 0; i < dictionary_ct; i++)
         if (dictionary[i].size >= size
          && dictionary[i].freed)
     {
         dictionary[i].freed = 0;
         return dictionary[i].addr;
     }

     return_ptr = sbrk(size);

     dictionary[dictionary_ct].addr = return_ptr;
     dictionary[dictionary_ct].size = size;
     dictionary[dictionary_ct].freed = 0;
     dictionary_ct++;

     return return_ptr;
}

void free(void *ptr)
{
    int i;

    if (!dictionary)
        return;

    for (i = 0; i < dictionary_ct; i++ )
    {
        if (dictionary[i].addr == ptr)
        {
            dictionary[i].freed = 1;
            return;
        }
    }
}

This is not a great malloc() implementation. In fact, most malloc/free implementations will allocate a small header for each block returned by malloc. The header might start at the address eight (8) bytes less than the returned pointer, for example. In those bytes you can store a pointer to the mem_dictionary entry owning the block. This avoids the O(N) operation in free. You can avoid the O(N) in malloc() by implementing a priority queue of freed blocks. Consider using a binomial heap, with block size as the index.