Path: utzoo!attcan!utgpu!jarvis.csri.toronto.edu!rutgers!tut.cis.ohio-state.edu!ucbvax!ulysses!hector!jss
From: jss@hector.UUCP (Jerry Schwarz)
Newsgroups: comp.lang.c++
Subject: Re: istream conversion operator
Keywords: istream conversion operator
Message-ID: <11597@ulysses.homer.nj.att.com>
Date: 25 May 89 20:03:56 GMT
References: <522@atcmpe.atcmp.nl>
Sender: netnews@ulysses.homer.nj.att.com
Reply-To: jss@hector.UUCP (Jerry Schwarz)
Organization: AT&T Bell Laboratories
Lines: 32

In article <522@atcmpe.atcmp.nl> leo@atcmp.nl (Leo  Willems)
considers code that looks like

	while ( cin >> i)  ...

The istream& returned by the extraction operator(>>) gets converted
to a void* by operator void*::istream.

>
>My question is: Why does this conversion take  place? There is no
>assignment involved (i.e. to a void*) from which the type system can decide
>which conversion is appropriate.

In C++ conversions can arise implicitly in many contexts other than
the right hand side of assignment.

In particular in "boolean context", if the type of the expression is
not one that can be tested against zero (arithmetic or pointer type)
then C++ tries to convert it with a user defined conversion operator.
If there is more than one such conversion operator that converts to a
type that can be tested against 0, then there is an ambiguity.

Boolean contexts are (unless I've overlooked something):
	expression in an if statement
	expression in a while statement
	expression in a repeat until statement
	2nd expression in a for statement
	either operand of logical operators (&&, ||)


Jerry Schwarz
AT&T Bell Labs, Murray Hill