1

Is this the only way to initialize default parameter initial_value as a compile time constant of type TNumber? If not, what is the preferred method of doing so?

template <typename TNumber>
class Widget {
 public:
  Widget(TNumber initial_value = static_cast<TNumber>(1))
    : value_(initial_value) {};

 private:
  TNumber value_;
};
Roman
  • 13
  • 3

2 Answers2

1

The preferred way is to use a type's expected constructor. In C++ 98:

Widget(TNumber initial_value = TNumber(1));

In C++ 11+ you could use the aggregate constructor as well:

Widget(TNumber initial_value = TNumber{1});

This also works when your parameter is a const ref, as in 

Widget(const TNumber& initial_value = TNumber{1});

By convention, the explicit default constructor returns 0 for numerical values, so the two following lines are equivalent for the standard numerical types (int, char, float, etc...)

Widget(const TNumber& initial_value = TNumber{0});

Widget(const TNumber& initial_value = TNumber{});

[EDIT] The same is true for complex numbers (std::complex<>). I suggest you follow the same convention if you ever define your own 'numerical-like' type. It could really save your day at some point.

Michaël Roy
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0

Can you use default member initialization with template types? So, just ...

private: TNumber value_ = 1;

Jay Burcham
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