Re: better understanding references

From: Greg Comeau (comeau_at_panix.com)
Date: 03/04/04 

Date: 4 Mar 2004 08:34:19 -0500

In article <IBy1c.2736$KS1.102886@nasal.pacific.net.au>,
David White <no@email.provided> wrote:
>"thides" <swatts@globalserve.net> wrote in message
>news:67y1c.68104$9j1.25598@nntp-post.primus.ca...
>> 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"?

I suspect that thides is thinking that:

int a = 99;
int &b = a;

since often a use of a means 99, that the reference is referring
to that and not a, after all, we don't use &a. However, b is
indeed referring to a and not the 99 per se.

>b is another name for a, or b _refers_ to a (hence it is a reference). If
>you assign b to 4,

assign 4 to b

>a and b will equal 4,

Something like that.

>because a and b refer to the same value.

Refer to the same object.

>If you assign a to 4, a and b will again equal 4.

4 to a. 

-- 
Greg Comeau / Comeau C++ 4.3.3, for C++03 core language support
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