if $a =& $b is assignment by reference, why don't you need to dereference it?

Summercool wrote:
> First of all, I think in the very traditional and basic form of
> "reference", it means "pointers".
>
> So that's why in C,
>
> when you say
>
> int *ip;
>
> i is a pointer or "reference" to an integer. and that's why when you
> use it to get back the integer, you need to do *ip and that's called
> "dereference".
>

No. C does not have references. ip is a pointer. Nothing more,
nothing less.

> So in C++, it seems that there is a different kind of reference, and
> that's like an alias type of reference? So in C++, Java, and PHP, you
> can have
>
> int i = 10
> int &j = i
> printf "%d", j and you get 10?
> j = 20
> printf "%d %d", i, j and both are 20 now?
>

j is a reference to i.

> that's different from the traditional pointer reference
>
> a = 10; b = 20
> int *ip = &a // ip pointers to an integer
> int *jp = ip // jp pointers to the same integer
> printf "%d", *jp
> *jp = 20
> printf "%d %d", *ip, *jp
>
>

Pointers ARE NOT REFERENCES!

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Summercool wrote:
> sorry, the message was accidentally posted before it was complete:
>
> First of all, I think in the very traditional and basic form of
> "reference", it means "pointers".
>
> So that's why in C,
>
> when you say
>
> int *ip;
>
> ip is a pointer or "reference" to an integer. and that's why when you
> use it to get back the integer, you need to do *ip and that's called
> "dereference".
>
> So in C++, it seems that there is a different kind of reference, and
> that's like an alias type of reference? So in C++, Java, and PHP, you
> can have
>
> a = 10; b = 20;
> int i = a;
> int &j = i;
> printf "%d", j; and you get 10?
> j = b;
> printf "%d %d", i, j; and both are 20 now?
>
> that's different from the traditional pointer reference
>
> a = 10; b = 20;
> int *ip = &a; // ip pointers to an integer
> int *jp = ip; // jp pointers to the same integer
> printf "%d", *jp; // print 10
> jp = &b; // now jp pointer to a different integer
> printf "%d %d", *ip, *jp; // now it prints 10 and 20
>
>
> the first behavior is the same as PHP's
> $a =& $b
>

That is correct. $a is a reference to $b.

> the second behavior is the same as PHP5's object assignment:
> $obj1 = $obj2
>

In PHP5, objects are automatically references. If you want a new copy,
you need a clone() method.

> so it seems like both are called a reference?
>

Yes, the only difference being the first is explicit, the second implicit.

> Isn't there standard names for their difference? The first one is
> more like "an alias reference". The second one is like "a pointer
> reference". I think calling them the same as "reference" is very
> dangerous as they mean different things and behave differently.
>

They are both just references.

>
> some discussions:
> http://en.wikipedia.org/wiki/C%2B%2B_reference
> http://en.wikipedia.org/wiki/Referen...ter_science%29
>
>
>

I haven't looked at them, but don't believe everything you see in wikipedia.

--
==================
Remove the "x" from my email address
Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
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