Re: regarding Auto and decltype

From:

CornedBee <wasti.redl@gmx.net>

Newsgroups:

comp.lang.c++.moderated

Date:

Tue, 25 May 2010 12:55:21 CST

Message-ID:

<77ad6a6b-e66e-45ed-8c90-deeaf719f16a@o1g2000vbe.googlegroups.com>

On May 24, 5:45 pm, ManicQin <manic...@gmail.com> wrote:

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:

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?

You misunderstand what "evaluating" means. Both auto and decltype
determine types at compile time, so they can only use static type
information. Static type information says that the return type of tmp-

Clone() is a B*, so both test1 and test2 are B*.

Evaluating is about whether the expression is actually computed at run
time. In your example, Clone() will be called twice during the
execution of the program. Once for the initializer of test1, once more
for the initializer of test2, but *not* for the expression within
decltype().

Sebastian

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