Re: stream operator << overload resolution: temporaries vs non-tem

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=?Utf-8?B?UGF1bA==?= <vhr@newsgroups.nospam>

Newsgroups:

microsoft.public.vc.language

Date:

Sat, 19 Aug 2006 07:49:02 -0700

Message-ID:

<F8312E9E-01F7-4278-BE98-E0A1A2B86EDF@microsoft.com>

"Igor Tandetnik" wrote:

This is a well known problem. The C++ standard actually requires this
behavior. To avoid surprises, don't use stream objects as temporaries.

The issue is that some overloads of operator<< are member functions of
ostream, and others are standalone functions. The standalone versions
take ostream parameter by non-const reference. It looks roughly like
this (drastically simplified):

class ostream {
public:
ostream& operator<<(const void*); // (1): print a pointer value
};

ostream& operator<<(ostream&, const char*); // (2): print a string

Now consider

std::ofstream("test.txt") << "Hello!";

According to C++ rules, a temporary cannot bind to a non-const
reference, so (2) is not a viable match and is not considered by
overloading resolution. But a non-const member function can be called on
a temporary, so (1) is chosen. Instead of printing the contents of the
string, it prints the address.

With best wishes,
Igor Tandetnik

Thank you, Igor, I have indeed missed on that. But this still does not
explain a couple of things, I think.

1) Why would it work for strings (std::string): string operator functions
too take a reference to a stream:

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

std::ofstream("test.txt") << std::string("Hello!") << std::endl;

(I debugged and checked: it does call the above operator << for strings.)

2) Why would it work on subsequent insertions:

std::ofstream("test.txt") << "Hello!" /*returns address*/ << ' ' << "Hello!"
<< std::endl;

If the stream is treated as a constant object:

const std::ofstream& tmp = std::ostream("test.txt");
tmp << "Hello!";

then the reference to the stream returned from this operation will also be a
constant?

Thank you.

Paul