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Path: utzoo!mnetor!seismo!munnari!otc!mikem
From: mikem@otc.OZ (Michael Mowbray)
Newsgroups: comp.lang.c++
Subject: Re: Passing pointers by reference
Message-ID: <49@otc.OZ>
Date: Mon, 1-Dec-86 18:07:51 EST
Article-I.D.: otc.49
Posted: Mon Dec  1 18:07:51 1986
Date-Received: Tue, 2-Dec-86 08:20:00 EST
References: <572@sdcc18.ucsd.EDU>
Organization: O.T.C. Systems Development, Australia
Lines: 48

In article <572@sdcc18.ucsd.EDU>, ee161aba@sdcc18.ucsd.EDU (David L. Smith) writes:

> 	foo(bar & * junk) 
> 
> This would be a pointer to bar passed by reference to the
> function foo.

No it's not. You've made junk a pointer to a reference to a bar, whereas
you wanted a reference to a pointer to a bar. If you use

	foo(bar* &junk)

it works. (This sort of error always catches people. To understand
declarations you have to read them backwards, i.e: right to left.)

Example follows:

    //------------------------------------------------------------
    #include <stream.h>

    struct Bar {
	    int i;
    };

    Bar b1,b2;

    void foo(Bar* &junk)
    {
	junk = &b2;
    }

    main()
    {
	cout << "&b1,&b2: " << (long)&b1 << " " << (long)&b2 << "\n";

	Bar* bp = &b1;

	cout << "bp before foo: " << (long)bp << "\n";
	foo(bp);
	cout << "bp after foo: " << (long)bp << "\n";
    }
    //------------------------------------------------------------


    $ a.out
    &b1,&b2: 16960 16964
    bp before foo: 16960
    bp after foo: 16964