Re: auto and decltype

From:
"Johannes Schaub (litb)" <schaub-johannes@web.de>
Newsgroups:
comp.lang.c++.moderated
Date:
Thu, 11 Mar 2010 18:53:56 CST
Message-ID:
<hnbosk$tf5$00$1@news.t-online.com>
restor wrote:

Hi,
In Boost, BOOST_AUTO is simply defined in terms of BOOST_TYPEOF:

  #define BOOST_AUTO(Var, Expr) BOOST_TYPEOF(Expr) Var = Expr

I already know that the relation between c++0x auto and decltype is
not so simple. But I am wondering if auto can be expressed in terms of
decltype in any (even more complex) way. E.g. does replacing keyword
auto (in the meaning of type deduction, excluding function syntax)
with:

  std::remove_cv<std::remove_reference<typeof(expr)>>

always produce the same result? Given all the CV and reference
combinations. Are the 4 variables' types deduced exactly the same in:

 auto x1 = expr1;
 const auto x2 = exp2;
 auto const& x3 = expr3;
 auto && x4 = expr4;

as in:

 std::remove_cv<std::remove_reference<typeof(expr)>>x1 = expr1;
 const std::remove_cv<std::remove_reference<typeof(expr)>>x2 = exp2;
 std::remove_cv<std::remove_reference<typeof(expr)>>const& x3 = expr3;
 std::remove_cv<std::remove_reference<typeof(expr)>>&& x4 = expr4;

? If not, is there any other (even more sophisticated way) to express
the "relation" between auto and decltype?


I think there is no way to express auto with decltype, because auto may be
used like

auto *p = expression; // declarator is more complicated.
auto p = { 1, 2 }; // initializer clause is not an expression

In the last case, "p" has type initializer_list<int>.

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