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Subject: Re: [boost] [optional] memory use for optional refs and ptrs
From: Stewart, Robert (Robert.Stewart_at_[hidden])
Date: 2010-10-07 09:22:41

Sebastian Redl wrote:
> On 07.10.2010 13:22, Stewart, Robert wrote:
> > That's not what I was suggesting. short can have alignment
> requirements, but bool and char do not. I was suggesting that this:
> >
> > struct P
> > {
> > short s;
> > char c;
> > short t;
> > char d;
> > };
> >
> > could occupy less space than if the chars and shorts were
> > reversed. That is, that P::c could occupy padding between
> > P::s and P::t and P::d could occupy padding between a P
> > instance and something following it in another composite.
> > Doing so would not violate an ABI, unless the ABI
> > specifically disallowed it, because it can be established as
> > the expected layout in those cases.
> >
> True, but that requires a type where alignof(T) > sizeof(T),
> which only happens if you override alignment requirements with
> attributes or something similar. Which aligned_storage actually

OK, I see that.

> But you're missing my point. The whole exercise is useless. For
> layout purposes, boost::optional is a struct containing two
> members. It can be written either as { T t; bool b; } or { bool
> b; T t; }. Assuming alignof(bool) == sizeof(bool) == 1 (which
> is actually not the case in some old ABIs), it holds true in
> both cases that alignof(optional) == alignof(T) and
> sizeof(optional) == 2 * sizeof(T). Because layout algorithms
> treat member types as opaque, it doesn't matter where the
> padding in optional is - no compiler will embed members of a
> containing struct in the padding of the optional.

Got it.

> Let me give you an example. Here's a struct that actually
> shrinks if you reorder the members:
> struct A {
> char c1;
> int i;
> char c2;
> }; // sizeof == 12
> struct A_Opt {
> int i;
> char c1;
> char c2;
> }; // sizeof == 8
> But this only works because all members are top-level.

Right. That's what I was thinking of, but went too far in applying it to the case of there being just one small field.

> Contained structs are opaque! This means that if I change A_Opt
> to pack the first two members together, the size optimization
> is lost!
> struct A_Bad {
> struct {
> int i;
> char c1;
> } bad;
> char c2;
> }; // sizeof == 12
> This is exactly the problem you face when trying to optimize
> optional. It doesn't matter that you move the padding to the
> end of the struct. The compiler still won't use it.

That much I didn't expect, but I was thinking that the size of optional could be sizeof(T) + sizeof(bool) rather than 2 * sizeof(T), which I thought could have been useful in some cases. I understand the reasons why that would not be so now.

Rob Stewart robert.stewart_at_[hidden]
Software Engineer, Core Software using std::disclaimer;
Susquehanna International Group, LLP

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