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/**
* @file ll_template_cast.h
* @author Nat Goodspeed
* @date 2009-11-21
* @brief Define ll_template_cast function
*
* $LicenseInfo:firstyear=2009&license=viewergpl$
* Copyright (c) 2009, Linden Research, Inc.
* $/LicenseInfo$
*/
#if ! defined(LL_LL_TEMPLATE_CAST_H)
#define LL_LL_TEMPLATE_CAST_H
/**
* Implementation for ll_template_cast() (q.v.).
*
* Default implementation: trying to cast two completely unrelated types
* returns 0. Typically you'd specify T and U as pointer types, but in fact T
* can be any type that can be initialized with 0.
*/
template <typename T, typename U>
struct ll_template_cast_impl
{
T operator()(U)
{
return 0;
}
};
/**
* ll_template_cast<T>(some_value) is for use in a template function when
* some_value might be of arbitrary type, but you want to recognize type T
* specially.
*
* It's designed for use with pointer types. Example:
* @code
* struct SpecialClass
* {
* void someMethod(const std::string&) const;
* };
*
* template <class REALCLASS>
* void somefunc(const REALCLASS& instance)
* {
* const SpecialClass* ptr = ll_template_cast<const SpecialClass*>(&instance);
* if (ptr)
* {
* ptr->someMethod("Call method only available on SpecialClass");
* }
* }
* @endcode
*
* Why is this better than dynamic_cast<>? Because unless OtherClass is
* polymorphic, the following won't even compile (gcc 4.0.1):
* @code
* OtherClass other;
* SpecialClass* ptr = dynamic_cast<SpecialClass*>(&other);
* @endcode
* to say nothing of this:
* @code
* void function(int);
* SpecialClass* ptr = dynamic_cast<SpecialClass*>(&function);
* @endcode
* ll_template_cast handles these kinds of cases by returning 0.
*/
template <typename T, typename U>
T ll_template_cast(U value)
{
return ll_template_cast_impl<T, U>()(value);
}
/**
* Implementation for ll_template_cast() (q.v.).
*
* Implementation for identical types: return same value.
*/
template <typename T>
struct ll_template_cast_impl<T, T>
{
T operator()(T value)
{
return value;
}
};
/**
* LL_TEMPLATE_CONVERTIBLE(dest, source) asserts that, for a value @c s of
* type @c source, <tt>ll_template_cast<dest>(s)</tt> will return @c s --
* presuming that @c source can be converted to @c dest by the normal rules of
* C++.
*
* By default, <tt>ll_template_cast<dest>(s)</tt> will return 0 unless @c s's
* type is literally identical to @c dest. (This is because of the
* straightforward application of template specialization rules.) That can
* lead to surprising results, e.g.:
*
* @code
* Foo myFoo;
* const Foo* fooptr = ll_template_cast<const Foo*>(&myFoo);
* @endcode
*
* Here @c fooptr will be 0 because <tt>&myFoo</tt> is of type <tt>Foo*</tt>
* -- @em not <tt>const Foo*</tt>. (Declaring <tt>const Foo myFoo;</tt> would
* force the compiler to do the right thing.)
*
* More disappointingly:
* @code
* struct Base {};
* struct Subclass: public Base {};
* Subclass object;
* Base* ptr = ll_template_cast<Base*>(&object);
* @endcode
*
* Here @c ptr will be 0 because <tt>&object</tt> is of type
* <tt>Subclass*</tt> rather than <tt>Base*</tt>. We @em want this cast to
* succeed, but without our help ll_template_cast can't recognize it.
*
* The following would suffice:
* @code
* LL_TEMPLATE_CONVERTIBLE(Base*, Subclass*);
* ...
* Base* ptr = ll_template_cast<Base*>(&object);
* @endcode
*
* However, as noted earlier, this is easily fooled:
* @code
* const Base* ptr = ll_template_cast<const Base*>(&object);
* @endcode
* would still produce 0 because we haven't yet seen:
* @code
* LL_TEMPLATE_CONVERTIBLE(const Base*, Subclass*);
* @endcode
*
* @TODO
* This macro should use Boost type_traits facilities for stripping and
* re-adding @c const and @c volatile qualifiers so that invoking
* LL_TEMPLATE_CONVERTIBLE(dest, source) will automatically generate all
* permitted permutations. It's really not fair to the coder to require
* separate:
* @code
* LL_TEMPLATE_CONVERTIBLE(Base*, Subclass*);
* LL_TEMPLATE_CONVERTIBLE(const Base*, Subclass*);
* LL_TEMPLATE_CONVERTIBLE(const Base*, const Subclass*);
* @endcode
*
* (Naturally we omit <tt>LL_TEMPLATE_CONVERTIBLE(Base*, const Subclass*)</tt>
* because that's not permitted by normal C++ assignment anyway.)
*/
#define LL_TEMPLATE_CONVERTIBLE(DEST, SOURCE) \
template <> \
struct ll_template_cast_impl<DEST, SOURCE> \
{ \
DEST operator()(SOURCE wrapper) \
{ \
return wrapper; \
} \
}
#endif /* ! defined(LL_LL_TEMPLATE_CAST_H) */
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