On 07/27/2018 03:33 PM, John Reiser wrote:
On 07/27/2018 08:21 UTC, Florian Weimer wrote:
On 07/27/2018 05:10 AM, John Reiser wrote:
Always requiring 16-byte alignment on x86_64 can waste too much space
due to internal fragmentation. The rule should be:
The required alignment is at least min(16, max_p2_divisor_of_size)
where the second argument max_p2_divisor_of_size is the maximum
power of 2
that divides the requested size [when 0!= size].
Thus a request for 6 bytes may be satisfied by a block whose address is
divisible by 2.
But this rule is wrong. Consider this:
struct string
{
size_t length;
char data[];
};
To allocate a one-byte string, malloc would be called with an argument
of sizeof (size_t) + 1. Your rule gives alignment 1, but in reality,
alignment 4 or 8 is required.
(Flexible struct members are not strictly required to implement this,
of course, so it's not a new problem at all.)
As written, sizeof(struct string) is 4 on a 32-bit machine, or 8 on a
64-bit machine.
Actually storing something in data[] requires more bytes; the question is,
"How many more?". The answer is round_up(size_for_data,
sizeof(string.length))
in order to guarantee alignment for string.length;
That's simply not true for top-level allocations which are not part of
an array.
The key principle is that sizeof(foo) must be the stride of an array of
foo,
and the array must guarantee alignment of each element in the array.
Why do you think that? If some documentation claims this is the case
for individual objects, we need to fix it.
Thanks,
Florian
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