Re: Implicit move constructor rules in c++0x still badly broken?

"Martin B." <>
Wed, 23 Feb 2011 09:46:26 CST
On 22.02.2011 20:16, Howard Hinnant wrote:

On Feb 22, 6:39 am, Stephen Howe<> wrote:

The ironic thing is that the link you site has not yet offered a code
example that gets broken.


Follow the links Howard.

You arrive at Scott Meyers


which does break in the destructor as v[0] is undefined.

This link was written prior to the 2010 Fall standards meeting in
Batavia which fixed the problem in this link.

No, it did *not* fix it.

This thread is no longer about the implicit generation of move
constructors. It is about implicitly considering the expression of a
return statement when that expression is eligible for RVO.

This thread here - maybe. But not the post by David Abrahams.

There is *way* too much confusion and mis-information in this thread.

I agree about the confusion.

I *strongly* encourage all participants to post code, to post the
results from having actually compiled/tested that code, and to post
what compiler was used to obtain those results.

Which compiler (version) that I could easily use on Windows already
implements the lates std draft?

I stand by my statement:

has not posted code that actually breaks.

I do not claim that such code doesn't exist. I only wish to have this
discussion based in fact, not in fud.

lists the example tweak#2 - class has no ctor but still a class
invariant. (I'll re-post the code at the end.)

Per the new rules

     If the definition of a class X does not explicitly
     declare a move constructor, one will be implicitly
     declared as defaulted if and only if
       * X does not have a user-declared copy constructor,
       * X does not have a user-declared copy assignment operator,
       * X does not have a user-declared move assignment operator,
       * X does not have a user-declared destructor, and
       * the move constructor would not be implicitly defined as deleted.

this example *still* breaks.

Note again: I'm not saying implicit-move-ctors should necessarily be
removed completely, I'm just saying that Dave *does* have a point.


- - -
Code Example Tweak#2 from David's post @
- - -
#include <iostream>
#include <vector>

// An always-initialized wrapper for unsigned int
struct Number
     Number(unsigned x = 0) : value(x) {}
     operator unsigned() const { return value; }
     unsigned value;

struct Y
     // Invariant: length == values.size(). Default ctor is fine.

     // Maintains the invariant
     void resize(unsigned n)
         std::vector<int> s(n);
         length = Number(n);

     bool operator==(Y const& rhs) const
         return this->values == rhs.values;

     friend std::ostream& operator<<(std::ostream& s, Y const& a)
         for (unsigned i = 0; i < a.length; ++i)
             std::cout << a.values[i] << " ";
         return s;

     std::vector<int> values;
     Number length;

int main()
     std::vector<Y> z(1, Y());

     Y a;

     std::remove(z.begin(), z.end(), Y());
     std::cout << z[1] << std::endl;
- - -

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