1. CS 240 – Lecture 21
Alloc Implementation

2. Discussions on hw9 Memory Allocation Exercise 3 functions:
void initalloc(void);
char * alloc(int n);
void freef(char *p);
You are asked to write the freef() function
The alloctest.c program is provided to test the functions

3. initalloc function Initializes a large free buffer F allocbuf
struct blockl{ unsigned int tag:8; unsigned int size :24; struct blockl *nextp; struct blockl *prevp;};
Blockrp = allocbuf + ALLOCSIZE -TAGSIZE
Cursorp = freep = allocbuf F
allocbuf
TAGSIZE
ALLOCSIZE
Struct blockr{
unsigned int tag:8;
unsigned int size:24;};
TAGSIZE = sizeof(blockr)
MINFREESIZE= sizeof(blockl) + sizeof(blockr)
USEDTAG = 0xaa
FREETAG = 0x55

4. Function alloc(n) U U Allocate a buffer of
allocsize = n (in multiple of 4 ) + 2*TAGSIZE
bytes from the free buffer pool
The tags for both ends of the allocated buffer are of type struct blockr. They are not kept using a linked list (no need for the char * elements)
Function returns a pointer = holdp + TAGSIZE
holdp
pointer returned by alloc U U
n bytes in multiple of 4
TAGSIZE
TAGSIZE

5. Allocation Algorithm
I) cursorp->size < allocsize=4*((n-1)/4 +1) +2*TAGSIZE
return null if all free blocks are not big enough
II) cursorp->size <allocsize + MINFREESIZE
return pointer =holdp + TAGSIZE
III) cursor->size > allocsize + MINFREESIZE
p=cursorp=holdp
blockrp = p +holdp->size -TAGSIZE U
allocsize U
blockrp =p + allocsize-TAGSIZE
cursorp
allocsize F U
new free size
holdp = p
blockrp = p-TAGSIZE

6. Enchaining Free Buffers
Allocated buffers are not chained
An example of chaining 4 free buffers
Most recent
freep
cursorp
Least recent
BUF 4
BUF 3
BUF 2
BUF 1
nextp =3
prevp =1
nextp =2
nextp =1
nextp =4
prevp =4
prevp =3
prevp =2

7. Unchaining free buffers
Most recent
Least recent
BUF 4
BUF 3
BUF 2
BUF 1
nextp =3
prevp =1
nextp =2
nextp =1
nextp =4
prevp =4
prevp =3
prevp =2
freep
cursorp
BUF 4
BUF 2
BUF 1
nextp =2
prevp =1
nextp =1
nextp =4
prevp =4
prevp =2

8. Multiple cases of freef
Left block is not free
Case 1:
Left block is free F F U U F F
Case 3:
Both left and right blocks are free

9. Pseudocode for freef(char *p)
p is the same pointer obtained from alloc()
Change tag to FREETAG on blockl
Change tag to FREETAG on blockr
No need to update size on blockl and blockr because alloc() sets the sizes correctly
Check if block on left is also free?
case 1: yes, coalesce free block on left and p
case 2: no, enchain buffer
Check if block on right is also free?
case 3: yes, unchain the block on right and coalesce p with block on right.
return