Re: regarding Auto and decltype

From:
Stuart Golodetz <sgolodetz@NdOiSaPlA.pMiPpLeExA.ScEom>
Newsgroups:
comp.lang.c++.moderated
Date:
Tue, 25 May 2010 12:56:25 CST
Message-ID:
<LoCdnSgHZ9aTZ2bWnZ2dnUVZ7vSdnZ2d@pipex.net>
Paul Bibbings wrote:

ManicQin <manicqin@gmail.com> writes:

Hello everybody.
In Scott Meyers lecture notes he states that one of the differences
between Auto and decltype is that the decltype does not evaluate the
expression.

I have a question regarding the evaluation of the expression, in the
next scenario what should I expect:

class B
{
public:
    B()
    {

    }
    virtual B* Clone()
    {
        cout << "B" << endl;
        return new B();
    }
};

class D : public B
{
public:
    D(){}
    virtual D* Clone()
    {
        cout << "D" << endl;
        return new D();
    }
};

int main()
{
        B* tmp = new D();
    auto test1 = tmp->Clone(); //returns D*!!!
    decltype(tmp->Clone()) test2 = tmp->Clone();
return 0;
}

Please note That D::Clone overloads with a different return type.

In my understanding if the "auto" is evaluating so it means that the
type of test1 should be D*, but VS10 understands different :) what am
I missing?


To my understanding, the type of test2 will be B* owing to:

      [dcl.type.simple]/4
   "The type denoted by decltype(e) is defined as follows:
    // ...
    - otherwise, if e is a function call (5.2.2) or an invocation of an
   overloaded operator (parentheses around e are ignored), decltype(e)
   is the return type of the statically chosen function."

Now, when it comes to your application of the auto specifier,
effectively we need to apply a process corresponding template argument
deduction (see [dcl.spec.auto]/6) to follow it through, one which, IIUC,
requires that, for:

   auto test1 = tmp->Clone();

   "The type of [test1] is the deduced type of the parameter u in the
   call f(expr) [for expr = tmp->Clone()] of the following invented
   function template:

      template <class U> void f(const U& u)"

Unfortunately, I am not yet able to find my way through the details
fully in the C++0x FCD, but applying this by way of example:

      22:12:33 Paul Bibbings@JIJOU
   /cygdrive/d/CPPProjects/CLCPPM $cat auto_decl.cpp
   class B
   {
   public:
      virtual B * Clone() { return new B(); }
   };

   class D : public B
   {
   public:
      virtual D * Clone() { return new D(); }
   };

   template <class U> void f(const U& u) { }

   int main()
   {
      B* b = new D();
      f(b->Clone());
   }

   22:12:38 Paul Bibbings@JIJOU
   /cygdrive/d/CPPProjects/CLCPPM $gcc -std=c++0x -c auto_decl.cpp

   22:12:54 Paul Bibbings@JIJOU
   /cygdrive/d/CPPProjects/CLCPPM $nm --demangle auto_decl.o | grep f\<
   00000000 T void f<B*>(B* const&)

you'll see that this process of template argument deduction again
deduces according to the return type considered statically.

If this is the case, as it apparently it is, it would seem that the
notion of decltype(expr) providing an unevaluated context whereas auto
*does* `evaluate' in some sense (although I'm not quite sure of the exact
sense of Meyers' purported idea here) is not actually the factor that
decides it.

Having said this, again, a quote from Meyers would help here.

Regards

Paul Bibbings


As I understand it, all that's being said is that in:

auto var1 = expr1;
decltype(expr2) var2 = expr3;

expr2 is *not* evaluated at run-time. However, both expr1 and expr3
*are* evaluated. Evaluation of the values of the expressions at run-time
is not the same as deduction of their types. The *types* of expr1 and
expr2 are worked out at compile-time and given to var1 and var2,
respectively.

These two give var the same type:

1) auto var = e;
2) decltype(e) var = e;

In both 1) and 2), e is evaluated at run-time (i.e. the e on the RHS of
the assignment gets evaluated in each case). However, the e within
decltype(e) in 2) doesn't get evaluated - it is only used at
*compile-time* to deduce the type of var. (And the deduction process
does not involve evaluating it, natch.)

Phew! Think that makes sense...

Stu

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