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From: kipp@warp.esd.sgi.com (Kipp Hickman)
Newsgroups: comp.lang.c++
Subject: Hairy C++ Question
Keywords: constructors, operators
Message-ID: <3734@odin.SGI.COM>
Date: 7 Feb 90 21:34:01 GMT
Sender: news@odin.SGI.COM
Reply-To: kipp@warp.esd.sgi.com (Kipp Hickman)
Organization: Silicon Graphics, Entry Systems Division
Lines: 47

This source code caused C++ to generate a warning:

	Str s4 = s1 | s2 | s3;

"timeStr.c++", line 57: warning: temporary used for non-const TempStr &
argument

("|" means concatenation).  What this turns into is (approx):

	Str s4( or( TempStr(s1,s2), s3 ) )

Now, there is only one temporary here because the "or" function is defined
like thus:

	friend TempStr& operator|(TempStr&, const Str&);

The expansion that uses that operator is what generates the warning.

So here come the questions:

1. Why is this a problem?  Is there a way that the temporary can be reused
   in a single expression that would cause a problem?  If not, then the
   warning is bogus.  If so, could I please see an example?

2. Why is this is a problem?  If the compiler can notice that the temporary
   is being modified, it seems it can notice that another temporary could
   be generated?

On a related subject, I had written my TempStr class without a
TempStr(const TempStr&) constructor.  What the compiler did in terms of
code generation was entirely different, with and without this constructor.
For example:

	class TempStr {
	public:
		friend TempStr operator|(const Str&, const Str&);
	};

The above example generated horrible structure copy code for the implementation
of the operator| code.  When I added the TempStr(const TempStr&) constructor,
it stopped doing the copies, and instead wrote the function with an extra
argument (which is what I wanted).  It turns out that I don't even have to
implement the TempStr(const TempStr&) constructor, because its *never* used
(am I wrong).  What gives?


					kipp