Re: Implicit conversion to std::ostream

From:
SG <s.gesemann@gmail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Tue, 21 Jul 2009 11:13:49 CST
Message-ID:
<f58479ed-ae8a-4be3-b0f7-479ada343172@32g2000yqj.googlegroups.com>
On 20 Jul., 23:31, Lars Storjord wrote:

Suppose we have the following class:

class ImplicitOstream {
public:
     ImplicitOstream(std::ostream& strm) : strm(strm) {}
     operator std::ostream&() { return strm; }
     std::ostream& strm() { return strm; }
private:
     std::ostream& strm;
};

Then the following won't work:

ImplicitStream implstrm;
implstrm << "Hello, world!\n";

Neither will operator<<(implstrm, cout);

However, the following does work:

void f(std::ostream& os)
{
     os << "Hello\n";
}

f(implstrm);

Why won't the call to operator<<() succeed? Is it supposed to be like
this? I would assume that implicit type conversion wasn't invoked due
to the reference-to-non-const, but then it shouldn't have worked for f
() either, should it? Is there something I have overlooked?


Yes. The operator<< you want to call is a *member* of ostream. So, the
correct explicit syntax would be

   ostream& myostream = implstrm;
   myostream.operator<<("Hello World!");

If this operator<< had been a free non-template function it would have
worked like you wanted. But it even won't work for function templates:

   template<typename T>
   void bar(ostream& os, T const& x) {os << x;}

   bar(implstrm,42);

because types have to "match exactly" (which includes lvalue
transformations and qualification adjustments)

Cheers!
SG

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