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Re: std::vector move assign patch

On Tue, 25 Apr 2017, Jonathan Wakely wrote:

On 25/04/17 17:23 +0200, Marc Glisse wrote:
On Tue, 25 Apr 2017, Jonathan Wakely wrote:

On 24/04/17 22:10 +0200, Marc Glisse wrote:
It seems that this patch had 2 consequences that may or may not have been planned. Consider this example (from PR64601)

#include <vector>
typedef std::vector<int> V;
void f(V&v,V&w){ V(std::move(w)).swap(v); }
void g(V&v,V&w){ v=std::move(w); }

1) We generate shorter code for f than for g, probably since the fix for PR59738. g ends up zeroing v, copying w to v, and finally zeroing w, and for weird reasons (and because we swap the members one by one) the standard prevents us from assuming that v and w do not overlap in weird ways so we cannot optimize as much as one might expect.

f has an additional precondition (that the allocators of the vectors
being swapped must propagate on swap or be equal) and so the swap code
doesn't have to worry about non-equal allocators.

g has to be able to cope with the case where the allocator doesn't
propagate and isn't equal, and so is more complicated.

However, the propagation trait is known at compile-time, and for the
common case so is the equality condition, so it's unfortunate if that
can't be simplified (I'm sure you've analysed it carefully already

The code isn't horrible. With f, we get:

       movq    (%rsi), %r8
       movq    8(%rsi), %rcx
       movq    $0, (%rsi)
       movq    $0, 8(%rsi)
       movq    16(%rsi), %rdx
       movq    $0, 16(%rsi)
       movq    (%rdi), %rax
       movq    %rcx, 8(%rdi)
       movq    %r8, (%rdi)
       movq    %rdx, 16(%rdi)
       testq   %rax, %rax

which seems quite optimal: read each pointer from w, write them to v, write 0s in w, that's 9 memory operations, +1 to read the pointer from w and possibly call delete on it.

Of course I should have mentioned that writing 0s is done between the read and the other write, which differentiates it from g.

With g:

       movq    $0, 8(%rdi)
       movq    (%rdi), %rax
       movq    $0, 16(%rdi)
       movq    $0, (%rdi)
       movq    (%rsi), %rdx
       movq    %rdx, (%rdi)
       movq    8(%rsi), %rcx
       movq    $0, (%rsi)
       movq    8(%rdi), %rdx
       movq    %rcx, 8(%rdi)
       movq    16(%rsi), %rcx
       movq    %rdx, 8(%rsi)
       movq    16(%rdi), %rdx
       movq    %rcx, 16(%rdi)
       movq    %rdx, 16(%rsi)
       testq   %rax, %rax

That's only 5 more memory operations. If I tweak vector swapping to avoid calling swap on each member (which drops type-based aliasing information, that was the topic of PR64601)

I didn't really understand the discussion in the PR. I find that's
true of most TBAA discussions.

std::swap(T& x, T& y) is hard to optimise because we don't know that
the dynamic type of the thing at &x is the same type as T?

std::swap<int*> only knows that it is dealing with pointers to integers, so _M_start from one vector might be in the same location as _M_finish from some other vector...

When I access __x._M_start directly, I am accessing some part of a _Vector_impl, and 2 _Vector_impl can only be the same or disjoint, they cannot partially overlap.

       void _M_swap_data(_Vector_impl& __x) _GLIBCXX_NOEXCEPT
         pointer tmp;
#define MARC(x,y) tmp=x; x=y; y=tmp
         MARC(_M_start, __x._M_start);
         MARC(_M_finish, __x._M_finish);
         MARC(_M_end_of_storage, __x._M_end_of_storage);

this gets down to 13, which is kind of sensible
* 0 the elements of v -> 3 ops
* read the elements of w -> 3 ops
* write them to v -> 3 ops
* 0 the elements of w -> 3 ops
(+1 to get the pointer that we might call delete on)

The first step of zeroing the elements of v is redundant
* if v and w don't alias, we are going to overwrite those 0s in step 3 without ever reading them
* if v and w are the same, we are going to write those 0s in step 4 anyway

but that's hard for the optimizers to notice.

I didn't try hard to find a nice C++ way to get an equivalent of g that generates the optimal number of operations, but it would be a little ugly to write in operator=
this->_M_impl._M_finish = x._M_impl._M_finish; x._M_impl._M_finish = 0;
same for _M_end_of_storage and _M_start, and remembering to use the original this->_M_impl._M_start for delete.

I'm not opposed to writing it out by hand. Operations on std::vector
should be as fast as possible, and move-assignment should be cheap.

I am not sure it would make a noticable difference. I might do it at some point in the future...

By the way, do we have a policy on writing if(p)delete p; vs directly delete p; which does nothing for p==0? There are cases where the extra shortcut is a nice optimization, others where it is redundant work. At -Os, the middle-end could optimize if(p)free(p) to free(p) in the future.

Marc Glisse

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