bool x = (pdl == pdr); //false

My understanding is that pdl and pdr were both uninitialized and hence
the result should be undefined, based on what I know about pointers.
But in paragraph's context I don't see any reason. Can someone explain?

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Shark wrote:
> bool x = (pdl == pdr); //false
>
> My understanding is that pdl and pdr were both uninitialized and hence
> the result should be undefined, based on what I know about pointers.
> But in paragraph's context I don't see any reason. Can someone explain?

First of all the example is on page 88 in the 2003 version of the C++
Standard. It was on page 87 in the 1998 version. You should always refer
to the latest version otherwise you might ask for something that has
already been fixed.

Back to the question, let me quote the entire example (with added comments):

  struct B {
  int f();
  };
  struct L : B { };
  struct R : B { };
  struct D : L, R { };
  int (B::*pb)() = &B::f;
  int (L::*pl)() = pb;
  int (R::*pr)() = pb;
  int (D::*pdl)() = pl;  // <<<<<<<<<<<<<<<<<<<<<<<<
  int (D::*pdr)() = pr;  // <<<<<<<<<<<<<<<<<<<<<<<<
  bool x = (pdl == pdr); // false

the two lines marked with "// <<<<<<<<<<<<<<<<<<<<<<<<" clearly
initialize both pdl and pdr, so what makes you think that they are both
uninitialized?

Ganesh

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Shark wrote:
> bool x = (pdl == pdr); //false
>
> My understanding is that pdl and pdr were both uninitialized and hence

What???  'pdl' is initialised with the value of 'pl', which is in turn
initialised from the value of 'pb', which is initialised from '&B::f'.
The whole point of the example is to show that the conversions are applied
and while both of the pointers are "growing" from the same expression, the
values change because of the conversions.

> the result should be undefined, based on what I know about pointers.
> But in paragraph's context I don't see any reason. Can someone explain?

See above.

V

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===================================== MODERATOR'S COMMENT:

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Victor Bazarov wrote:
> Shark wrote:
> > bool x = (pdl == pdr); //false
> >
> > My understanding is that pdl and pdr were both uninitialized and hence
>
> What???  'pdl' is initialised with the value of 'pl', which is in turn
> initialised from the value of 'pb', which is initialised from '&B::f'.
> The whole point of the example is to show that the conversions are applied
> and while both of the pointers are "growing" from the same expression, the
> values change because of the conversions.
>

Oh, well my bad. I'll be more careful next time. Thanks Ganesh and
Victor

> > the result should be undefined, based on what I know about pointers.
> > But in paragraph's context I don't see any reason. Can someone explain?
>
> See above.
>
> V
>
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