CODEWITHSUNDEEP
c++castingstdstringimplicit-conversionuser-defined-types
Michele De Pascalis
Michele De Pascalis

Reputation: 952

Is possible to get automatic cast from user-defined type to std::string using cout?

As in the question, if I define a string operator in my class:

class Literal {
  operator string const () {
    return toStr ();
  };

  string toStr () const;
};

and then I use it:

Literal l1 ("fa-2bd2bc3e0");
cout << (string)l1 << " Declared" << endl;

with an explicit cast everything goes right, but if I remove the (string) the compiler says that it needs a cast operator declared in std::string. Shouldn't it cast my type automatically? SOLVED: I'm overloading operator<< (ostream& os, const Literal& l).

Upvotes: 3

Views: 2473

Answers (2)

aschepler
aschepler

Reputation: 72356

Short answer: Keep using a cast or toStr(), or write your own operator<< function. (I would prefer l1.toStr() to (string)l1.)

Long answer: This might work if the Standard Library had a function

std::ostream& operator<<( std::ostream&, std::string const& );

Which it almost does, but not technically. Both ostream and string are really typedefs of template instantiations. And there's a template function for inserting one into the other.

// This is somewhat simplified.  For the real definitions, see the Standard
// and/or your complying implementation's headers.
namespace std {
  typedef basic_string<char> string;
  typedef basic_ostream<char> ostream;

  template <typename CharT>
  basic_ostream<CharT>& operator<<(
    basic_ostream<CharT>&, 
    basic_string<CharT> const&);
}

So when you use cout << str where the type of str is std::string, it can figure out to use that template function, with CharT = char.

But C++ doesn't allow you to have the compiler figure out both an implicit type conversion (Literal to string) and deduce template function template parameters (CharT = char) on the same call.

Upvotes: 5

RC.
RC.

Reputation: 28217

No.. std::string would have to have a constructor that took Literal as an argument.

What you could do is overload operator << for your Literal class and have it cast and insert into the stream in there.

ostream &operator <<(std::ostream &stream, const Literal &rhs)
{
    stream << (string) rhs;
    return stream;
}

Upvotes: 10

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