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Subject: Re: [boost] Implicit cast rules
From: Vicente Botet (vicente.botet_at_[hidden])
Date: 2011-05-24 10:41:29

Gregory Crosswhite wrote:
> Hey everyone,
> Would someone please explain to me the implicit cast rule that prevents
> the following from compiling, and/or if there is a way around it?
> #include <boost/optional.hpp>
> #include <iostream>
> using namespace boost;
> using namespace std;
> struct C {
> operator optional<int>() { return none; }
> };
> int main() {
> C c;
> cout << static_cast&lt;optional&lt;int&gt; >(c) << endl;
> return 0;
> }
> Error message:
> In file included from /usr/include/boost/optional.hpp:15,
> from cast-to-optional.cpp:1:
> /usr/include/boost/optional/optional.hpp: In member function ‘void
> boost::optional_detail::optional_base<T>::construct(const Expr&, const
> void*) [with Expr = C, T = int]’:
> /usr/include/boost/optional/optional.hpp:262: instantiated from
> ‘boost::optional_detail::optional_base<T>::optional_base(const Expr&,
> const Expr*) [with Expr = C, T = int]’
> /usr/include/boost/optional/optional.hpp:559: instantiated from
> ‘boost::optional<T>::optional(const Expr&) [with Expr = C, T = int]’
> cast-to-optional.cpp:13: instantiated from here
> /usr/include/boost/optional/optional.hpp:392: error: cannot convert
> ‘const C’ to ‘int’ in initialization
> I can tell what is happening; it would seem that the implicit cast rule
> selects the optional() constructor before it even looks at my more
> specific cast operator. I'm just confused about why this should be,
> since I see no reason why constructor casts should be prioritized over
> operator casts. Is this just the way it is, a GCC quirk, something I'm
> doing wrong, etc?


from N3242 5.2.9

"4 Otherwise, an expression e can be explicitly converted to a type T using
a static_cast of the form static_cast<T>(e) if the declaration T t(e); is
well-formed, for some invented temporary variable t (8.5). The effect of
such an explicit conversion is the same as performing the declaration and
initialization and then using the temporary variable as the result of the
conversion. The expression e is used as a glvalue if and only if the
initialization uses it as a glvalue."

I don't know why the standard uses a temporary variable T t(e) and return it
instead of returning directly T(e).


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