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diff --git a/indra/llcommon/ll_template_cast.h b/indra/llcommon/ll_template_cast.h
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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) */