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Re: std::vector move assign patch


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
though!)

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.

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)

        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.

2) g(v,v) seems to turn v into a nice empty vector,

Yes.

while f(v,v) turns it into an invalid vector pointing at released memory.

Does it?! I don't see that happening, and it's a bug if it does.

Er, my fault, you are right. It shuffles things around and amounts to a NOP, v remains as it was before the call to f. Which could be considered less desirable than clearing the vector by some, but not as much as getting something invalid of course :-)

--
Marc Glisse


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