The following problem concerns dynamic storage allocation.
Consider an allocator that uses an implicit free list. The layout of each allocated and free memory block is as follows:
31 2 1 0
__________________________________
Header | Block Size (bytes) | |
|____________________________|_____|
| |
| |
| |
| |
| |
|__________________________________|
Footer | Block Size (bytes) | |
|____________________________|_____|
Each memory block, either allocated or free, has a size that is a multiple of eight bytes. Thus, only the 29 higher order bits in the header and footer are needed to record block size, which includes the header and footer. The usage of the remaining 3 lower order bits is as follows:
- bit 0 indicates the use of the current block: 1 for allocated, 0 for free.
- bit 1 indicates the use of the previous adjacent block: 1 for allocated, 0 for free.
- bit 2 indicates the use of the next adjacent block: 1 for allocated, 0 for free.
Given the contents of the heap shown on the left, show the new contents of the heap (in the right table) after a call to free(0x600a010) is executed. Your answers should be given as hex values. Note that the address grows from bottom up. Assume that the allocator uses immediate coalescing, that is, adjacent free blocks are merged immediately each time a block is freed.
| Address | |
|---|---|
| 0x600a028 | 0x00000016 |
| 0x600a024 | 0x600b611c |
| 0x600a020 | 0x600b512c |
| 0x600a01c | 0x00000016 |
| 0x600a018 | 0x00000011 |
| 0x600a014 | 0x600b511c |
| 0x600a010 | 0x600b601c |
| 0x600a00c | 0x00000011 |
| 0x600a008 | 0x00000016 |
| 0x600a004 | 0x600b601c |
| 0x600a000 | 0x600b511c |
| 0x6009ffc | 0x00000016 |
| Address | |
|---|---|
| 0x600a028 | |
| 0x600a024 | 0x600b611c |
| 0x600a020 | 0x600b512c |
| 0x600a01c | |
| 0x600a018 | |
| 0x600a014 | 0x600b511c |
| 0x600a010 | 0x600b601c |
| 0x600a00c | |
| 0x600a008 | |
| 0x600a004 | 0x600b601c |
| 0x600a000 | 0x600b511c |
| 0x6009ffc |