Re: better understanding references

From: David White (no_at_email.provided)
Date: 03/04/04 

Date: Thu, 4 Mar 2004 15:50:39 +1100

"thides" <swatts@globalserve.net> wrote in message
news:67y1c.68104$9j1.25598@nntp-post.primus.ca...
>
> "Gary" <glabowitz@comcast.net> wrote in message
> news:f6-dnecc3btOGdvdRVn-tw@comcast.com...
> > Stop trying to think about pointers, addresses and the rest. Think
> instead:
> >
> > When I write a function with a parameter list containing &x, I am
renaming
> > the value being passed to me as "x." The identifier x is a name for a
> > location in memory containing the value being passed.
> >
> > Caller... z = foo(y);
> >
> > Called... int foo(int &x) { ...
> >
> > The variable addressed by "y" is the same variable addressed by "x" even
> > though the identifiers are in different code blocks.
> >
> > If you understand this, you understand references in general.
> >
> > int a;
> > int &b = a;
> >
> > The two identifiers "a" and "b" each refer to the value in a single
> > variable. The variable addressed by "a" is the same variable addressed
by
> > "b."
>
> I guess if a changes then so will b. And if b changes so will a.
>
> Like this:
>
> int a;
> int &b=a;
>
> b=4; // therefore a=4

Correct.

> But that wont work because b is a reference to a value not a value
itself...

What do you mean "that won't work"?

b is another name for a, or b _refers_ to a (hence it is a reference). If
you assign b to 4, a and b will equal 4, because a and b refer to the same
value. If you assign a to 4, a and b will again equal 4.

> OK I think if I am correct on this point i might be able to move on.

Well, I don't know if you are.

DW

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