Re: Improve insert/emplace robustness to self insertion

[Jonathan Wakely <jwakely at redhat dot com>]

On 04/07/16 15:55 +0100, Jonathan Wakely wrote:
> I'm getting nervous about the smart insertion trick to avoid making a
> copy, I have a devious testcase in mind which will break with that
> change. I'll share the testcase later today.

Here's a testcase which passes with libstdc++ but fails with libc++
because libc++ doesn't make a copy when inserting a T lvalue into
std::vector<T>:

#include <vector>
#include <memory>
#include <cassert>

struct T
{
 T(int v = 0) : value(v) { }
 T(const T& t);
 T& operator=(const T& t);
 void make_child() { child = std::make_unique<T>(value + 10); }
 std::unique_ptr<T> child;
 int value;
};

T::T(const T& t) : value(t.value)
{
 if (t.child)
   child.reset(new T(*t.child));
}

T& T::operator=(const T& t)
{
 value = t.value;
 if (t.child)
 {
   if (child)
     *child = *t.child;
   else
     child.reset(new T(*t.child));
 }
 else
   child.reset();
 return *this;
}

int main()
{
 std::vector<T> v;
 v.reserve(3);
 v.push_back(T(1));
 v.back().make_child();
 v.push_back(T(2));
 v.back().make_child();

 assert(v[1].child->value == 12);
 assert(v[1].child->child == nullptr);

 v.insert(v.begin(), *v[1].child);

 assert(v[0].value == 12);
 assert(v[0].child == nullptr);
}

The problem is that the object being inserted (*v[1].child) is not an
element of the vector, so the optimization assumes it is unchanged by
shuffling the existing elements. That assumption is wrong.

As far as I can see, this program is perfectly valid. It's slightly
contrived to prove a point, but it's not entirely unrealistic code.