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Make variant constructor from std::shared_ptr a template

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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.
This commit is contained in:
Jyrki Vesterinen 2017-04-06 20:16:25 +03:00 • committed by Celtic Minstrel
parent d536d6ab76
commit 0b02ba0876
2 changed files with 7 additions and 7 deletions
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6
src/formula/variant.cpp
View file

@ -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)))
{

8
src/formula/variant.hpp
View file

@ -34,11 +34,17 @@ public:
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;