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Path: utzoo!mnetor!seismo!rutgers!princeton!allegra!alice!bs
From: bs@alice.UUCP
Newsgroups: comp.lang.c++
Subject: Re: Passing pointers by reference
Message-ID: <6412@alice.uUCp>
Date: Tue, 2-Dec-86 08:58:05 EST
Article-I.D.: alice.6412
Posted: Tue Dec  2 08:58:05 1986
Date-Received: Tue, 2-Dec-86 22:06:05 EST
Organization: AT&T Bell Laboratories, Murray Hill NJ
Lines: 46
Summary: it works, but beware of the declarator syntax

In article <572@sdcc18.ucsd.EDU>, ee161aba@sdcc18.UUCP writes:
> 
> Is it possible to call a function with a pointer by reference?  What I am
> trying to do is pass a pointer and I want to be able to modify the pointer
> value (not what it points to).  Logically, something like:
> 
> 	foo(bar & * junk) 
> 
> should do it. This would be a pointer to bar passed by reference to the
> function foo.  However, this does not work.
...
> I realize I could use a double indirected pointer (and I probably will end
> up doing so) but I'm lazy and the call by reference is an elegant mechanism.
> Now if only it would work!
> 
> 			David L. Smith
> 			UC San Diego
> 			sdcsvax!sdcc18!ee161aba

I think you simply got the declaration wrong
	foo (bar&*);
declares a pointer to a reference, not a reference to a pointer.

How about:	

f(int*& p)
{
	p++;	// change the pointer value	
}

main() {
	int i;
	int* a = &i;
	printf("%d\n",a);	// some value
	f(a);
	printf("%d\n",a);	// some value + sizeof(int*)
}

Yes, the declarator syntax is perverse. Try reading them right to left (that is
easier than the proper ``inside out'' and often sufficient to get them right):

	int*&:	reference to pointer to int
	int&*:	pointer to refernce to int
	const int *:	pointer to integer constant
	int *const:	constant pointer to integer
	etc.