Path: utzoo!attcan!uunet!pilchuck!ssc!happym!polari!rlb
From: rlb@polari.UUCP (rlb)
Newsgroups: comp.lang.c++
Subject: Re: unary-expression syntax
Summary: How does 'new' bind?
Keywords: new,initializer,C++
Message-ID: <459@polari.UUCP>
Date: 23 May 88 13:46:22 GMT
References: <458@polari.UUCP>
Organization: Polarserv, Seattle WA
Lines: 17

In article <458@polari.UUCP>, rlb@polari.UUCP (rlb) writes:
> In Stroustrup's book, section 7.2 of the reference manual contains:
>     unary-expression:
>         ...
>         new type-name initializer.opt
and furthermore, how would something like this bind:
    int *x = new int = 5 + 6;
For 'new' to be treated as an operator with unary binding strength, it looks
like the above would have to bind as:
    int *x = (new int = 5) + 6;
which would be a little bizarre; equal sign ceases to be an operator in an
expression if a certain unary operator appears to the left of it.  On the
other hand, if it binds like this:
    int *x = new int = (5 + 6);
then it looks a whole lot more like a primary-expression than a unary op;
under what circumstances does its precedence and associativity ever matter?
-Ron Burk