CODEWITHSUNDEEP
c++templates
NarekMeta
NarekMeta

Reputation: 103

how to provide default template parameter and template function argument default value in the same time?

I Have the following code:

template <typename Iter, typename STREAM>
void print(Iter b, Iter e, STREAM& strm)
{
    while (b != e)
    {
        strm << *b;
        strm << "   ";
        ++b;
    }
    strm << "\n";
}

How to have default value for strm = std::cout ?

Upvotes: 3

Views: 129

Answers (3)

Jarod42
Jarod42

Reputation: 217275

You have to provide default for type and for argument:

template <typename Iter, typename STREAM = std::ostream>
void print(Iter b, Iter e, STREAM& strm = std::cout);

Demo

As std::cout is not a good default value for other STREAM, you probably need to change the default:

template <typename STREAM>
STREAM& DefaultStream;

template <>
std::ostream& DefaultStream<std::ostream> = std::cout;

and then

template <typename Iter, typename STREAM = std::ostream>
void print(Iter b, Iter e, STREAM& strm = DefaultStream<STREAM>);

Upvotes: 2

Picaud Vincent
Picaud Vincent

Reputation: 10982

Note that you can also use

template <typename Iter, typename STREAM = decltype(std::cout)>
void print(Iter b, Iter e, STREAM& strm = std::cout)
{
  while (b != e)
  {
    strm << *b;
    strm << "   ";
    ++b;
  }
  strm << "\n";
}

This works with c++11, but I think that @AsteroidsWithWings's solution is the right way to go (as it is simpler and avoid the necessary redundancy required to define both the template type and argument value)

Upvotes: 1

Asteroids With Wings
Asteroids With Wings

Reputation: 17454

If you were to just add = std::cout, you'd need to provide the template argument std::ostream explicitly when calling the function with two arguments, which is a bit rubbish.

The easiest way to do what you want is to provide a new overload:

template <typename Iter>
void print(Iter b, Iter e)
{
    print(b, e, std::cout);
}

You could also consider getting rid of STREAM entirely; you usually only ever want to stream to a std::ostream. There's a reason a bunch of useful types inherit from it.

Upvotes: 5

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