问题
http://www.alex-ionescu.com/?p=50.
I read the above post. The author explains why Windows x64 supports only 44-bit virtual memory address with singly linked list example.
struct { // 8-byte header ULONGLONG Depth:16; ULONGLONG Sequence:9; ULONGLONG NextEntry:39; } Header8;The first sacrifice to make was to reduce the space for the sequence number to 9 bits instead of 16 bits, reducing the maximum sequence number the list could achieve. This still only left 39 bits for the pointer — a mediocre improvement over 32 bits. By forcing the structure to be 16-byte aligned when allocated, 4 more bits could be won, since the bottom bits could now always be assumed to be 0.
Oh, I can't understand.
What "By forcing the structure to be 16-byte aligned when allocated, 4 more bits could be won, since the bottom bits could now always be assumed to be 0." means?
回答1:
For a 2^N-byte aligned pointer, its address is always divisible by 2^N - which means that lower N bits are always zero. You can store additional information in them:
encode ptr payload = ptr | payload
decode_ptr data = data & ~mask
decode_payload data = data & mask
where mask is (1 << N) - 1 - i.e. a number with low N bits set.
This trick is often used to save space in low-level code (payload can be GC flags, type tag, etc.)
In effect you're not storing a pointer, but a number from which a pointer can be extracted. Of course, care should be taken to not dereference the number as a pointer without decoding.
回答2:
16 is 0010000 in binary
32 is 0100000 in binary
64 is 1000000 in binary
etc
You can see that for all numbers that are multiples of 16, the last four bits are always zero. So, instead of storing these bits you can leave them out and add them back in when it's time to use the pointer.
来源：https://stackoverflow.com/questions/4563297/behind-windows-x64s-44-bit-virtual-memory-address-limit