Suppose a malloc implementation returns 8-byte aligned addresses and uses an explicit free list where the next and previous pointers are each 32-bits. Blocks have a 32-bit header and 32-bit footer, where the low-order bit of the header and footer are used to indicate whether the block is allocated (1) or free (0). Furthermore, the block size (which includes the header, payload, footer, and any necessary padding) is rounded up to the nearest multiple of 8, and this size (in bytes) is stored in the header and footer. Assume any padding must be between the payload and the footer. If we call malloc(1), what block size will be allocated, in bytes? With the same conditions as #1, and assuming we've already called malloc(1), if the heap used by malloc starts at address 0x4000 (16384 in decimal), what address would be returned if we then called malloc(32)?
Suppose a malloc implementation returns 8-byte aligned addresses and uses an explicit free list where the next and previous pointers are each 32-bits. Blocks have a 32-bit header and 32-bit footer, where the low-order bit of the header and footer are used to indicate whether the block is allocated (1) or free (0). Furthermore, the block size (which includes the header, payload, footer, and any necessary padding) is rounded up to the nearest multiple of 8, and this size (in bytes) is stored in the header and footer. Assume any padding must be between the payload and the footer.
If we call malloc(1), what block size will be allocated, in bytes?
With the same conditions as #1, and assuming we've already called malloc(1), if the heap used by malloc starts at address 0x4000 (16384 in decimal), what address would be returned if we then called malloc(32)?
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