The memory allocation problem Define the memory allocation problem Memory organization and memory allocation schemes.

Up to this point: Process management provides the illusion of an infinite number of CPUs –On a single processor machine Memory management provides a different set of illusions –Protected memory –Infinite amount of memory

Memory allocation problem A sub-problem in memory management. –A large block of memory (e.g. physical memory), say N bytes. –Requests for sub-blocks come in an unpredictable fashion (e.g. processes start and need memory). Each request may ask for 1 to N bytes. –Allocate continuous memory to a request when available. The memory block is in use for a while and is then returned to the system (free). –Goal: to satisfy as many requests as possible –constraints: CPU and memory overheads should be low.

Memory allocation problem Example –10KB memory to be managed –r1 = req(1K); –r2 = req (2K); –r3 = req(4k); –free(r2); –free(r1); –r4 = req(4k); How to do it makes a difference!! Internal fragment: unused memory within a block –Asking for 100 bytes and get a 512 bytes block External fragment: unused memory between blocks –Even when the total available memory is more than a request, the request cannot be satisfied as in the example.

Variations of the memory allocation problem occur in different situations. Disk space management heap management –new and delete in C++ –malloc and free in C.

Two issues in the memory allocation problem: –Memory organization: how to divide the memory into blocks for allocation? Static method: divide the memory once before the memory are allocated. Dynamic method: divide it up as the memory is allocated. –Memory allocation: select which piece of memory for a request. –Memory organization and memory allocation are close related.

Static memory organization: –Statically divide memory into fixed size subblocks, each for a request. –Advantages: easy to implement. Good when the sizes for memory requests are fixed. –Can be extended for handle memory requests for different sizes known a prior. Example: 500, 000 bytes, each request for either 50,000 or 200,000. Two 50,000 bytes blocks, Two 200,000 byte blocks. –Data structure: a linked list for each type of blocks.

Static memory organization: –Worst case complexity: new and free O(1) for both –Disadvantage: cannot handle variable-size requests effectively. Might need to use a large block to satisfy a request for small size. –Internal fragmentation

Buddy system: –Allow to use larger blocks to satisfy smaller requests without wasting more than half of the block. –Maintain a free block list for each power of two size. E.g. memory has 512K bytes. The system will maintain free block list for blocks of 512K, 256K, 128K, 64K, 32K, 16K, 8K, 4K, 2K, 1K, 512bytes, 256 bytes, 128bytes, 64bytes, 32bytes, 16bytes, 8bytes, 4 bytes, 2bytes, 1bytes. –All free lists start off empty except for the free block list for the largest block which contains one free block.

Buddy system: –When receiving a request, round it up to the next power of two and look for that list. If that block list is empty, look for the next larger power of two and divide a free block there into two blocks (buddy). If still fail, keep going up until you find one. –When freeing a block, look whether its buddy is free, if yes, merge them, otherwise, insert the free block into the appropriate free block list.

Example: –memory = 32 bytes, free block list for 32, 16, 8, 4, 2, 1 bytes. –Initial state of the free block list? –What is the memory state after the sequence: request 1( 1 byte), free (request 1), request 2 (4 bytes), request 3 (8 bytes), free request 2, request 4 (1 byte), request 5 (2 bytes), free request 3.

–Buddy system is a semi-dynamic method keeps the simplicity of the statically allocated blocks and handles the difficult cases of different request sizes. –Worst case complexity? –Drawback: –Can still have internal fragments

Dynamic memory allocation: –Grant only the size requested (different from static memory allocation and buddy system). –Example: total 512 bytes: allocate(r1, 100), allocate(r2, 200), allocate(r3, 200), free(r2), allocate(r4, 10), free(r1), allocate(r5, 200) –Fragmentation: memory is divided up into small blocks that none of them can be used to satisfy any requests. Static allocation -- internal fragment. dynamic allocation -- external fragment.

Issues in dynamic memory allocation. –Where are the free memory blocks? How to keep track of the free/used memory blocks –Which memory blocks to allocate? There may exist multiple free memory blocks that can satisfy a request. Which block to use? –Fragments must be reduced.

–Keeping track of the blocks: the list method. Keep a link list of all blocks of memory(block list). Example: total 512 bytes: allocate(r1, 100), allocate(r2, 200), allocate(r3, 200), free(r2), allocate(r4, 10), free(r1), allocate(r5, 200) How to do the allocation operation? How to do the free operation? What information is to be kept in the list node?

–What is the information to be kept in the list node? Address: Block start address size: size of the block. Allocated: whether the block is free or allocated. Struct list_node { int address; int size; int allocated; struct list_node *next; struct list_node *prev; };

–Do we have to keep track of allocated block in the list? Not necessary. What is good and bad about keeping the allocated blocks information? –Can check invalid free operations. –Where to keep the block list? Reserving space for the block list at the beginning of the memory. –Limited number of block list nodes, constant space overhead (memory space taken away for memory management). Use space within each block. Each block has a header for block_list node.

–Keeping track of the memory blocks: the bitmap method. Memory is divided into blocks, use one bit to represent whether one block is allocated or not. A bitmap is a string of bits. –E.g 4Mbytes memory, each block has 1 byte. How many bits are needed to keep track of the status of the memory? How about each block has 1KB? How to allocate? How to free? Commonly used in disk.

Comparing the list method and bitmap method: list method(block header) bitmap method space: # of blocks fixed static time finding free blocks? O(B) string matching free a block? O(B) O(1)

Which free block to allocate? –First fit: search from the beginning, use the first free block. –next fit: search from the current position, use the first free block. –best fit: use the smallest block that can fit. –worst fit: use the largest block that can fit. Which algorithm is faster? Which is slower? Which one is the best (satisfies most requests)? –Depend the programs

One question What would be a simple and effective heap management scheme for most applications? –Hints: most applications do not use heap that much!!!