Make variant constructor from std::shared_ptr a template

For some reason, when the variant constructor only accepts
std::shared_ptr<const variant_callable>, MSVC2013 gets confused about which
constructor it should call when it has, say,
std::shared_ptr<ai::attack_analysis>. Making the constructor a template
fixes it.

src/formula/variant.cpp

@@ -162,12 +162,6 @@ variant::variant(double n, variant::DECIMAL_VARIANT_TYPE)
 	assert(value_.get());
 }
 
-variant::variant(const_formula_callable_ptr callable)
-	: value_(std::make_shared<variant_callable>(callable))
-{
-	assert(value_.get());
-}
-
 variant::variant(const std::vector<variant>& vec)
     : value_((std::make_shared<variant_list>(vec)))
 {

src/formula/variant.hpp

@@ -34,11 +34,17 @@ class variant
 	explicit variant(int n);
 	variant(int n, DECIMAL_VARIANT_TYPE /*type*/);
 	variant(double n, DECIMAL_VARIANT_TYPE /*type*/);
-	variant(const_formula_callable_ptr callable);
 	explicit variant(const std::vector<variant>& array);
 	explicit variant(const std::string& str);
 	explicit variant(const std::map<variant, variant>& map);
 
+	template<typename T>
+	variant(std::shared_ptr<T> callable)
+		: value_(std::make_shared<variant_callable>(callable))
+	{
+		assert(value_.get());
+	}
+
 	variant& operator=(const variant& v);
 
 	variant operator[](size_t n) const;