Path: utzoo!utgpu!watserv1!watmath!att!dptg!ulysses!andante!alice!ark
From: ark@alice.UUCP (Andrew Koenig)
Newsgroups: comp.lang.c++
Subject: Re: The type yielded by "new T[expression]"
Message-ID: <10605@alice.UUCP>
Date: 21 Mar 90 04:23:52 GMT
References: <26068361.27247@paris.ics.uci.edu>
Organization: AT&T Bell Laboratories, Liberty Corner NJ
Lines: 30

In article <26068361.27247@paris.ics.uci.edu>, rfg@paris.ics.uci.edu (Ronald Guilmette) writes:
> I believe that there exists a ghastly violation of strong typing rules
> currently built into the C++ language.  I'd like somebody to explain
> this to me and defend it (if possible).

Not particularly ghastly, but rather a minor convenience that
is well described in the manual.

> Now the question I'm concerned about is: "What is the type of the value
> yielded by:

> 	new T[size];

The type is `pointer to T' for any `new' statement with this syntax.

> Currently (it seems) the value yielded by an "array new" operation like
> this is treated as being of type "pointer-to-T".  This seems highly
> nonsensical to me.  It would make far more sense if the type of value
> yielded were "pointer-to-array-of-T" when the "new" invocation includes
> a (square-bracketed) size expression.

Impossible, because `size' doesn't need to be a constant.
If you want a pointer to array of, say, 17 T, you can write

	new T[17][1]

which will get you exactly that.
-- 
				--Andrew Koenig
				  ark@europa.att.com