pointer and reference confusion

[Scott]

Hello,

I am having some difficulty with the following C++:

I have a declared a pointer to Foo like so:

Foo *pFoo;

Now, I have an API object that I try to use like so:

fooAPI.processF oo( pFoo );

Then, I get a compiler error that the prototype for processFoo is
expecting a Foo&, but is instead getting a Foo*&. So, my question is: how
do I transform my *pFoo to a Foo&, and perhaps someone could also help me
understand exactly what the difference between the two is? I hope this is
a least somewhat clear.

Thank-you.
Scott

[Axter]

Scott wrote:

Hello,

I am having some difficulty with the following C++:

I have a declared a pointer to Foo like so:

Foo *pFoo;

Now, I have an API object that I try to use like so:

fooAPI.processF oo( pFoo );

Then, I get a compiler error that the prototype for processFoo is
expecting a Foo&, but is instead getting a Foo*&. So, my question is: how
do I transform my *pFoo to a Foo&, and perhaps someone could also help me
understand exactly what the difference between the two is? I hope this is
a least somewhat clear.

I believe the function is expecting a reference to the object, and not
a pointer to the object.
If you have a pointer, you can pass it by dereferencing it using *.
Example:
fooAPI.processF oo( *pFoo );

However, you should make sure that pFoo is pointing to a valid object.
Functions that take a reference, will usually expect to be getting a
valid initialized object, and not a NULL value.
Where as functions that take a pointer may expect the pointer to be
NULL.

In C++, experts recommend using reference over pointers when possible,
because it's less ambiguous in general.

[Dennis Jones]

"Scott" <sc***@nospam.c om> wrote in message
news:pa******** *************** *****@nospam.co m...

Hello,

I am having some difficulty with the following C++:

I have a declared a pointer to Foo like so:

Foo *pFoo;

Now, I have an API object that I try to use like so:

fooAPI.processF oo( pFoo );

Then, I get a compiler error that the prototype for processFoo is
expecting a Foo&, but is instead getting a Foo*&. So, my question is: how
do I transform my *pFoo to a Foo&, and perhaps someone could also help me
understand exactly what the difference between the two is? I hope this is
a least somewhat clear.

A reference works *sort of* like a pointer, but without the pointer syntax.
However, unlike a pointer, a reference must always refer to an object
(pointers can be NULL, references cannot), so for example, it is illegal to
write:

int & ref; // illegal -- 'ref' does not refer to anything.

Instead, you would have to write:

int i;
int & ref = i;

In this case, 'ref' refers to 'i' and for all intents and purposes 'i' and
'ref' are the same variable, representing the same space in memory, and both
may be used in the same ways. Thus, when writing:

ref = 6;
i = 6;

....both assignments above accomplish the exact same thing, that is,
assigning the value 6 to the variable 'i'.

Now, to your main question...sinc e 'processFoo' expects a reference to a
'Foo' (Foo&), you cannot pass it a pointer, so you must dereference the
pointer so that the function gets a Foo instead of a Foo*, like this:

fooAPI.processF oo( *pFoo );

Of course, 'pFoo' must be pointing to something reasonable, or dereferencing
it will result in a nasty crash. But then you probably already knew that.

HTH,

- Dennis

[Scott]

On Wed, 24 May 2006 22:47:21 -0700, Axter wrote:

I believe the function is expecting a reference to the object, and not
a pointer to the object.
If you have a pointer, you can pass it by dereferencing it using *.
Example:
fooAPI.processF oo( *pFoo );

Thanks to both responders. That was the ticket!

Best,
Scott