inst eecs berkeley edu cs61c CS61C Machine Structures Lecture 6 C Memory Management 2005 01 31 Lecturer PSOE Dan Garcia www cs berkeley edu ddgarcia Geek Rhythms Geeksta rap album by Rajeeve Bajaj is getting lots of play for highlighting that it s cool to be a geek www rlpkrecords com CS61C L06 C Memory Management 1 Garcia Spring 2005 UCB Lecture clarifications Thanks to Rehan Waliany unix program arg1 arg2 argc is 3 not 2 as I said in lecture P 53 contains a precedence table useful for answering these questions x p p p 1 x p CS61C L06 C Memory Management 2 Garcia Spring 2005 UCB Where allocated Structure declaration does not allocate memory Variable declaration does allocate memory If declare outside a procedure allocated in static storage If declare inside procedure allocated on the stack and freed when int myGlobal procedure returns main NB main is a int myTemp procedure CS61C L06 C Memory Management 3 Garcia Spring 2005 UCB The Stack Stack frame includes Return address Parameters Space for other local variables SP Stack frames contiguous blocks of memory stack pointer tells where top stack frame is When procedure ends stack frame is tossed off the stack frees memory for future stack frames CS61C L06 C Memory Management 4 frame frame frame frame Garcia Spring 2005 UCB Stack Last In First Out LIFO memory stack usage main a 0 void a int m b 1 void b int n c 2 void c int o d 3 void d int p CS61C L06 C Memory Management 5 Stack Pointer Stack Pointer Stack Pointer Stack Pointer Stack Pointer Garcia Spring 2005 UCB Who cares about stack management Pointers in C allow access to deallocated memory leading to hard to find bugs int ptr main main main int y SP y 3 ptr printf return y y 3 y SP SP main int stackAddr content stackAddr ptr content stackAddr printf d content 3 content stackAddr printf d content 13451514 CS61C L06 C Memory Management 6 Garcia Spring 2005 UCB C Memory Management C has 3 pools of memory Static storage global variable storage basically permanent entire program run The Stack local variable storage parameters return address location of activation records in Java or stack frame in C The Heap dynamic storage data lives until deallocated by programmer C requires knowing where objects are in memory otherwise things don t work as expected Java hides location of objects CS61C L06 C Memory Management 7 Garcia Spring 2005 UCB The Heap Dynamic memory Large pool of memory not allocated in contiguous order back to back requests for heap memory could result blocks very far apart where Java new command allocates memory In C specify number of bytes of memory explicitly to allocate item int ptr ptr int malloc sizeof int malloc returns type void so need to cast to right type malloc Allocates raw uninitialized memory from heap CS61C L06 C Memory Management 8 Garcia Spring 2005 UCB Review Normal C Memory Management FFFF FFFFhex A program s address space contains 4 regions stack stack local variables grows downward heap space requested for pointers via malloc resizes dynamically grows upward heap static data static data variables code declared outside main does not grow or shrink 0 For now OS somehow code loaded when prevents accesses between program starts does not stack and heap gray hash change lines Wait for virtual memory hex CS61C L06 C Memory Management 9 Garcia Spring 2005 UCB Intel 80x86 C Memory Management A C program s 80x86 address space heap space requested for pointers via malloc resizes dynamically grows upward static data variables declared outside main 08000000 does not grow or shrink heap static data hex code stack code loaded when program starts does not change stack local variables grows downward CS61C L06 C Memory Management 10 Garcia Spring 2005 UCB Memory Management How do we manage memory Code Static storage are easy they never grow or shrink Stack space is also easy stack frames are created and destroyed in last in first out LIFO order Managing the heap is tricky memory can be allocated deallocated at any time CS61C L06 C Memory Management 11 Garcia Spring 2005 UCB Heap Management Requirements Want malloc and free to run quickly Want minimal memory overhead Want to avoid fragmentation when most of our free memory is in many small chunks In this case we might have many free bytes but not be able to satisfy a large request since the free bytes are not contiguous in memory CS61C L06 C Memory Management 12 Garcia Spring 2005 UCB Heap Management An example Request R1 for 100 bytes R1 100 bytes Request R2 for 1 byte R2 1 byte Memory from R1 is freed Request R3 for 50 bytes CS61C L06 C Memory Management 13 Garcia Spring 2005 UCB Heap Management An example Request R1 for 100 bytes Request R2 for 1 byte R2 1 byte Memory from R1 is freed Request R3 for 50 bytes CS61C L06 C Memory Management 14 R3 R3 Garcia Spring 2005 UCB K R Malloc Free Implementation From Section 8 7 of K R Code in the book uses some C language features we haven t discussed and is written in a very terse style don t worry if you can t decipher the code Each block of memory is preceded by a header that has two fields size of the block and a pointer to the next block All free blocks are kept in a linked list the pointer field is unused in an allocated block CS61C L06 C Memory Management 15 Garcia Spring 2005 UCB K R Implementation malloc searches the free list for a block that is big enough If none is found more memory is requested from the operating system If what it gets can t satisfy the request it fails free checks if the blocks adjacent to the freed block are also free If so adjacent free blocks are merged coalesced into a single larger free block Otherwise the freed block is just added to the free list CS61C L06 C Memory Management 16 Garcia Spring 2005 UCB Choosing a block in malloc If there are multiple free blocks of memory that are big enough for some request how do we choose which one to use best fit choose the smallest block that is big enough for the request first fit choose the first block we see that is big enough next fit like first fit but remember where we finished searching and resume searching from there CS61C L06 C Memory Management 17 Garcia Spring 2005 UCB Peer Instruction Pros and Cons of fits A The con of first fit is that it results in many small blocks at the beginning of the free list B The con of next fit is it is slower than first fit since it takes longer in steady state to find a match C The con of best fit is that it leaves lots of tiny blocks CS61C L06 C Memory
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