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Transcript of Dynamic Memory Allocation I Topics Basic representation and alignment (mainly for static memory...

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Dynamic Memory Allocation I Topics Basic representation and alignment (mainly for static memory allocation, main concepts carry over to dynamic memory allocation) Simple explicit allocators Data structures Policies MechanismsReadings Bryant & OHallaron textbook section 10.9: Dynamic Memory Allocation Slide 2 2 Data Representations Sizes of C Objects (in Bytes) C Data Type Intel IA32 int 4 (1 word) long int4 char1 short2 float4 double 8 (2 words) char *4 Or any other pointer A word is 4 bytes Slide 3 3 Alignment (recap) Aligned Data Primitive data type requires K bytes Address must be multiple of K Required on some machines; advised on Intel Motivation for Aligning Data Memory accessed by double words Inefficient to load or store datum that spans double word boundariesCompiler Inserts gaps in structure to ensure correct alignment of fields Slide 4 4 struct rec { int i; int a[3]; int *p; }; Structures Concept Contiguously-allocated region of memory Refer to members within structure by names Members may be of different types Memory Layout iap 0416 20 Slide 5 5 Specific Cases of Alignment Size of Primitive Data Type: 1 byte (e.g., char) no restrictions on address 2 bytes (e.g., short ) lowest 1 bit of address is 0 (2) 4 bytes (e.g., int, float, char *, etc.) lowest 2 bits of address - 00 (2) 8 bytes (e.g., double ) lowest 3 bits of address 000 (2) Slide 6 6 Satisfying Alignment with Structures Offsets Within Structure Must satisfy elements alignment requirement Overall Structure Placement Each structure has alignment requirement K Largest alignment of any element Initial address & structure length must be multiples of K Slide 7 7 Example (under Linux) K = 4, due to int element K = 4, due to int element struct S1 { char c; int i[2]; } *p; ci[0]i[1] p+0p+4p+8 Multiple of 4 p+12 Slide 8 8 Overall Alignment Requirement struct S2{ float x[2]; int i[2]; char c; } *p; ci[0]i[1] p+0p+12p+8p+16p+20 x[0]x[1] p+4 p must be multiple of 4 Slide 9 9 Arrays of Structures Principle Allocated by repeating allocation for array type a[0] a+0 a[1]a[2] a+12a+24a+36 a+12a+20a+16a+24 struct S3 { short i; float v; short j; } a[10]; a[1].ia[1].ja[1].v Slide 10 10 Misc: Boolean (Bitwise) Operators Operate on numbers as Bit Vectors Operations applied bitwise 01101001 & 01010101 01000001 01101001 | 01010101 01111101 01101001 ^ 01010101 00111100 ~ 01010101 10101010 Slide 11 11 Slide 12 12 Dynamic Memory Allocation Explicit vs. Implicit Memory Allocator Explicit: application allocates and frees space E.g., malloc and free in C Implicit: application allocates, but does not free space E.g. garbage collection in Java Application Dynamic Memory Allocator Heap Memory Slide 13 13 Dynamic Memory Allocation Explicit vs. Implicit Memory Allocator Explicit: application allocates and frees space E.g., malloc and free in C Implicit: application allocates, but does not free space E.g. garbage collection in JavaAllocation In both cases the memory allocator provides an abstraction of memory as a set of blocks Doles out free memory blocks to application Will discuss simple explicit memory allocation today Application Dynamic Memory Allocator Heap Memory Slide 14 14 Process Memory Image kernel virtual memory run-time heap (via malloc ) program text (. text ) initialized data (. data ) uninitialized data (. bss ) stack 0 %esp memory invisible to user code the brk ptr Allocators request additional heap memory from the operating system using the sbrk function. Slide 15 15 Malloc Package #include #include void *malloc(size_t size) If successful: Returns a pointer to a memory block of at least size bytes, (typically) aligned to 8-byte boundary. If size == 0, returns NULL If unsuccessful: returns NULL (0) and sets errno. void free(void *p) Returns the block pointed at by p to pool of available memory p must come from a previous call to malloc or realloc. void *realloc(void *p, size_t size) Changes size of block p and returns pointer to new block. Contents of new block unchanged up to min of old and new size. Slide 16 16 Malloc Example Slide 17 17 Assumptions Assumptions made in this lecture Memory is word addressed Allocated block (4 words) Free block (3 words) Free word Allocated word Slide 18 18 Allocation Examples p1 = malloc(4) p2 = malloc(5) p3 = malloc(6) free(p2) p4 = malloc(2) Slide 19 19 Constraints Applications: Can issue arbitrary sequence of allocation and free requests Free requests must correspond to an allocated block Slide 20 20 Constraints Applications: Can issue arbitrary sequence of allocation and free requests Free requests must correspond to an allocated blockAllocators Must respond immediately to all allocation requests i.e., cant reorder or buffer requests Must align blocks so they satisfy all alignment requirements 8 byte alignment for GNU malloc ( libc malloc) on Linux boxes Cant move the allocated blocks once they are allocated i.e., compaction is not allowed Can only manipulate and modify free memory Slide 21 21 Goals of Good malloc/free Primary goals Good time performance for malloc and free Ideally should take constant time (not always possible) Should certainly not take linear time in the number of blocks Good space utilization Want to minimize fragmentation. Some other goals Good locality properties Structures allocated close in time should be close in space Slide 22 22 Performance Goals: Throughput Given some sequence of malloc and free requests: R 0, R 1,..., R k,..., R n-1 Want to maximize throughput and peak memory utilization. These goals are often conflictingThroughput: Number of completed requests per unit time Example: 5,000 malloc calls and 5,000 free calls in 10 seconds Throughput is 1,000 operations/second. Slide 23 23 Performance Goals: Peak Memory Utilization Def: Aggregate payload P k : malloc(p) results in a block with a payload of p bytes.. Aggregate payload P k is the sum of currently allocated payloads. Def: Current heap size is denoted by H k Def: Peak memory utilization: After k requests, peak memory utilization is: U k = ( max i Slide 24 24 Internal Fragmentation Poor memory utilization caused by fragmentation. Comes in two forms: internal and external fragmentation Internal fragmentation Depends only on the pattern of previous requests, and thus is easy to measure. payload Internal fragmentation block Internal fragmentation Slide 25 25 External Fragmentation p1 = malloc(4) p2 = malloc(5) p3 = malloc(6) free(p2) p4 = malloc(7) oops! Occurs when there is enough aggregate heap memory, but no single free block is large enough External fragmentation depends on the pattern of future requests, and thus is difficult to measure. Slide 26 26 Implementation Issues How do we know how much memory to free just given a pointer? How do we know how much memory to free just given a pointer? How do we keep track of the free blocks? How do we keep track of the free blocks? What do we do with the extra space ? What do we do with the extra space ? How do we pick a block to use ? How do we pick a block to use ? How do we reinsert freed block? How do we reinsert freed block? p0 Slide 27 27 Knowing How Much to Free Standard method Keep the length of a block in the word preceding the block. This word is often called the header Requires an extra word for every allocated block free(p0) p0 = malloc(4)p0 Block sizedata 5 Slide 28 28 Boolean (Bitwise) Operators Operate on numbers as Bit Vectors Operations applied bitwise 01101001 & 01010101 01000001 01101001 | 01010101 01111101 01101001 ^ 01010101 00111100 ~ 01010101 10101010 Slide 29 29 Keeping Track of Free Blocks Method 1: Implicit list using lengths -- links all blocks Method 2: Explicit list among the free blocks using pointers within the free blocks Method 3: Segregated free list Different free lists for different size classes Method 4: Blocks sorted by size Can use a balanced tree (e.g. Red-Black tree) with pointers within each free block, and the length used as a key 5 426 5 426 Slide 30 30 Method 1: Implicit List Need to identify whether each block is free or allocated Can use extra bit Bit can be put in the same word as the size if block sizes are always multiples of two (mask out low order bit when reading size). size 1 word Format of allocated and free blocks payload a = 1: allocated block a = 0: free block size: block size payload: application data (allocated blocks only) a optional padding Slide 31 31 Implicit List: Finding a Free Block First fit: Search list from beginning, choose first free block that fits Can take linear time in total number of blocks (allocated and free) In practice it can cause splinters at beginning of list Slide 32 32 Slide 33 33 Implicit List: Finding a Free Block Next fit: Like first-fit, but search list from location of end of previous search Research suggests that fragmentation is worse Best fit: Search the list, choose the free block with the closest size that fits Keeps fragments small --- usually helps fragmentation Will typically run slower than first-fit Slide 34 34 Implicit List: Allocating in Free Block Allocating in a free block - splitting Since allocated space might be smaller than free space, we might want to split the block 4426 p Slide 35 35 Implicit List: Allocating in Free Block 4426 42 p 4 addblock(p, 2) Slide 36 36 Slide 37 37 Implicit List: Freeing a Block Simplest implementation: Only need to clear allocated flag void free_block(ptr p) { *p = *p & -2} But can lead to false fragmentation There is enough free space, but the allocator wont be able to find it 424 2 free(p) p 442 4 4 2 malloc(5) Oops! Slide 38 38 Implicit List: Coalescing Join (coalesce) with next and/or previous block if they are free Coalescing with next block 424 2 p 442 4 Slide 39 39 Implicit List: Coalescing Join (coalesce) with next and/or previous block if they are free Coalescing with next block But h