Re: user-defined op= for type with reference member

Paul Bibbings <>
Sun, 13 Jun 2010 08:04:42 -0700 (PDT)
On Jun 13, 3:58 pm, Paul Bibbings <> wrote:

On Jun 13, 1:30 pm, "Leigh Johnston" <> wrote:

It is UB to reseat a reference. You are attempting to reseat a refer=

ence if

you call destructor and placement new in a class's assignment operator =


"after the lifetime of an object has ended and before the storage which=


object occupied is reused or released". Read the above section from =


standard again. Violating a requirement in the standard is UB and ar=


this point is pointless.


I don't disagree, on the whole, with the conclusions reached in this
thread, but I am having some problem with it being stated in terms of
being "UB to reseat a reference." To my mind, "if you call destructor
and placement new in a class's assignment operator" you are *not*
thereby "reseat[ing] a reference." You *are* destructing an object
creating a new one in its place, with the attendant UB that has been
identified in the examples given; however, there is, to my mind, no
reference-reseating simply because, with the destruction of the
object, it's reference-member is destroyed with it and a *new* one
created in its place.

Now, if you consider the following:

   #include <new>

   class HasRefMem {
      HasRefMem(int& i)
         : i_(i)
      { }
      HasRefMem& operator=(const HasRefMem& other)
         if (this != &other)
            new (this) HasRefMem(other);
         return *this;
      operator int&() { return i_; }
      int& i_;

   int main()
      int i, j;
      HasRefMem hrmi(i);
      HasRefMem hrmj(j);
      int& i_ref = hrmi; // #1
      hrmi = hrmj; // #2

then I might be able to consider i_ref (in line #1) as having been
`reseated' in line #2.

Well, actually ... no (replying to myself). i_ref will continue to
refer to i.


Paul Bibbings

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